Mixing
Antiderivative of $sin{x}$ when $x$ is given in degrees
$begingroup$
I know that calculus is generally done in radians and that $int{sin{x}}$ ${dx}=-cos{x}+C$ when working with them.
I am trying to find the following definite integral with respect to $x$ from $0$ to $30$ degrees:
$$int_{0}^{30}{sin{x}} ; {dx}$$
What is the antiderivative of $sin{x}$ when $x$ is given in degrees?
When I evaluated $int_{0}^{30}{sin{x}}$ ${dx}$ when my calculator is in degree mode, I get $approx7.676178925$ but this is not equivalent to the corresponding definite integral given when my calculator is radian mode. With a $30$ degree angle having a radian measure of $frac{pi}{6}$, I evaluated $int_{0}^{frac{pi}{6}}{sin{x}}$ ${dx}$ which gave $approx.1339745962$, which makes sense since $1-cos{frac{pi}{6}}approx.1339745962$.
My second question is, what is the meaning of $int_{0}^{30}{sin{x}}$ ${dx} =7.676178925$ when $x$ is given in degrees? I noticed that $frac{7.676178925pi}{180}approx.1339745962$ but am not sure what implication that has for this problem other than the fact that it is $7.676178925$ degrees in radians.
calculus integration trigonometry
edited Jan 5 at 23:56
Eevee Trainer
5,7871936
asked Jan 5 at 23:24
limitsandlogs224limitsandlogs224
476
$endgroup$
add a comment |
$begingroup$
I know that calculus is generally done in radians and that $int{sin{x}}$ ${dx}=-cos{x}+C$ when working with them.
I am trying to find the following definite integral with respect to $x$ from $0$ to $30$ degrees:
$$int_{0}^{30}{sin{x}} ; {dx}$$
What is the antiderivative of $sin{x}$ when $x$ is given in degrees?
When I evaluated $int_{0}^{30}{sin{x}}$ ${dx}$ when my calculator is in degree mode, I get $approx7.676178925$ but this is not equivalent to the corresponding definite integral given when my calculator is radian mode. With a $30$ degree angle having a radian measure of $frac{pi}{6}$, I evaluated $int_{0}^{frac{pi}{6}}{sin{x}}$ ${dx}$ which gave $approx.1339745962$, which makes sense since $1-cos{frac{pi}{6}}approx.1339745962$.
My second question is, what is the meaning of $int_{0}^{30}{sin{x}}$ ${dx} =7.676178925$ when $x$ is given in degrees? I noticed that $frac{7.676178925pi}{180}approx.1339745962$ but am not sure what implication that has for this problem other than the fact that it is $7.676178925$ degrees in radians.
calculus integration trigonometry
edited Jan 5 at 23:56
Eevee Trainer
5,7871936
asked Jan 5 at 23:24
limitsandlogs224limitsandlogs224
476
$endgroup$
$begingroup$
$sin x^circ = sin frac{pi}{180} x$.
$endgroup$
– anomaly
Jan 5 at 23:28
3
$begingroup$
The result of the integral is not an angle.
$endgroup$
– John Douma
Jan 5 at 23:29
$begingroup$
Of course, the result of the integral is an angle (see my answer).
$endgroup$
– Fabian
Jan 6 at 8:19
add a comment |
$begingroup$
I know that calculus is generally done in radians and that $int{sin{x}}$ ${dx}=-cos{x}+C$ when working with them.
I am trying to find the following definite integral with respect to $x$ from $0$ to $30$ degrees:
$$int_{0}^{30}{sin{x}} ; {dx}$$
What is the antiderivative of $sin{x}$ when $x$ is given in degrees?
When I evaluated $int_{0}^{30}{sin{x}}$ ${dx}$ when my calculator is in degree mode, I get $approx7.676178925$ but this is not equivalent to the corresponding definite integral given when my calculator is radian mode. With a $30$ degree angle having a radian measure of $frac{pi}{6}$, I evaluated $int_{0}^{frac{pi}{6}}{sin{x}}$ ${dx}$ which gave $approx.1339745962$, which makes sense since $1-cos{frac{pi}{6}}approx.1339745962$.
My second question is, what is the meaning of $int_{0}^{30}{sin{x}}$ ${dx} =7.676178925$ when $x$ is given in degrees? I noticed that $frac{7.676178925pi}{180}approx.1339745962$ but am not sure what implication that has for this problem other than the fact that it is $7.676178925$ degrees in radians.
calculus integration trigonometry
edited Jan 5 at 23:56
Eevee Trainer
5,7871936
asked Jan 5 at 23:24
limitsandlogs224limitsandlogs224
476
$endgroup$
I know that calculus is generally done in radians and that $int{sin{x}}$ ${dx}=-cos{x}+C$ when working with them.
I am trying to find the following definite integral with respect to $x$ from $0$ to $30$ degrees:
$$int_{0}^{30}{sin{x}} ; {dx}$$
What is the antiderivative of $sin{x}$ when $x$ is given in degrees?
When I evaluated $int_{0}^{30}{sin{x}}$ ${dx}$ when my calculator is in degree mode, I get $approx7.676178925$ but this is not equivalent to the corresponding definite integral given when my calculator is radian mode. With a $30$ degree angle having a radian measure of $frac{pi}{6}$, I evaluated $int_{0}^{frac{pi}{6}}{sin{x}}$ ${dx}$ which gave $approx.1339745962$, which makes sense since $1-cos{frac{pi}{6}}approx.1339745962$.
My second question is, what is the meaning of $int_{0}^{30}{sin{x}}$ ${dx} =7.676178925$ when $x$ is given in degrees? I noticed that $frac{7.676178925pi}{180}approx.1339745962$ but am not sure what implication that has for this problem other than the fact that it is $7.676178925$ degrees in radians.
calculus integration trigonometry
calculus integration trigonometry
edited Jan 5 at 23:56
Eevee Trainer
5,7871936
asked Jan 5 at 23:24
limitsandlogs224limitsandlogs224
476
edited Jan 5 at 23:56
Eevee Trainer
5,7871936
asked Jan 5 at 23:24
limitsandlogs224limitsandlogs224
476
edited Jan 5 at 23:56
Eevee Trainer
5,7871936
edited Jan 5 at 23:56
Eevee Trainer
5,7871936
edited Jan 5 at 23:56
Eevee Trainer
5,7871936
5,7871936
asked Jan 5 at 23:24
limitsandlogs224limitsandlogs224
476
asked Jan 5 at 23:24
limitsandlogs224limitsandlogs224
476
asked Jan 5 at 23:24
limitsandlogs224limitsandlogs224
476
476
$begingroup$
$sin x^circ = sin frac{pi}{180} x$.
$endgroup$
– anomaly
Jan 5 at 23:28
3
$begingroup$
The result of the integral is not an angle.
$endgroup$
– John Douma
Jan 5 at 23:29
$begingroup$
Of course, the result of the integral is an angle (see my answer).
$endgroup$
– Fabian
Jan 6 at 8:19
add a comment |
$begingroup$
$sin x^circ = sin frac{pi}{180} x$.
$endgroup$
– anomaly
Jan 5 at 23:28
3
$begingroup$
The result of the integral is not an angle.
$endgroup$
– John Douma
Jan 5 at 23:29
$begingroup$
Of course, the result of the integral is an angle (see my answer).
$endgroup$
– Fabian
Jan 6 at 8:19
$begingroup$
$sin x^circ = sin frac{pi}{180} x$.
$endgroup$
– anomaly
Jan 5 at 23:28
$begingroup$
$sin x^circ = sin frac{pi}{180} x$.
$endgroup$
– anomaly
Jan 5 at 23:28
3
3
$begingroup$
The result of the integral is not an angle.
$endgroup$
– John Douma
Jan 5 at 23:29
$begingroup$
The result of the integral is not an angle.
$endgroup$
– John Douma
Jan 5 at 23:29
$begingroup$
Of course, the result of the integral is an angle (see my answer).
$endgroup$
– Fabian
Jan 6 at 8:19
$begingroup$
Of course, the result of the integral is an angle (see my answer).
$endgroup$
– Fabian
Jan 6 at 8:19
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
To find the integral in degrees, it's better - for clarity's sake, not necessarily any formal reason - to start in radians with the conversion to degrees in the sine function. That is,
$$sin(x^circ) = sin left( frac{pi ; text{radians}}{180^circ} x^circright )$$
We just typically omit the "radians" because "radians" is less a unit of measure like the degree, than just a measure or a ratio like $pi$, but writing it explicitly like above helps. Though going forward I'll omit that, but it should be clear that the argument of the function on the right becomes one in degrees.
Then, making use of typical antidifferentiation techniques and rules, and utilizing $pi = 180^circ$,
$$begin{align}
int sin(x^circ)dx &= int sin left( frac{pi}{180^circ} x^circright )dx \
&= frac{-180^circ}{pi} cos left( frac{pi}{180^circ} x^circ right) +C\
&= frac{-180^circ}{pi} cos left( x^circ right) +C
end{align}$$
Of course, the fundamental theorem of calculus ($int_a^b f(x)dx = F(b)-F(a)$) also applies here. Just remember to be sure $a,b$ are also in degrees. Beyond that, the antiderivative is basically the same as in radians.
As for your second question, the meaning of any integral is essentially the same: the signed area under the curve of the integrand between the two points chosen as the bounds. Note that it is not going to share the same units as the integrand, i.e. it's not going to be in degrees.
answered Jan 5 at 23:54
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
3
$begingroup$
Where did the ratio $180/pi$ vanish? I assume a typo.
$endgroup$
– Paramanand Singh
Jan 6 at 6:59
add a comment |
$begingroup$
Although your calculator is evaluating $sin x$ assuming $x$ is in degrees, it is integrating $x$ from $0$ to $30$. What it is doing would usually be written
$$int_0^{30} sin frac {pi x}{180};dxapprox 7.676$$ where the argument of $sin$ is now in radians. Note that the result of the integral is a pure number, not degrees or radians.
answered Jan 5 at 23:38
Ross MillikanRoss Millikan
294k23198371
$endgroup$
$begingroup$
The value of the integral should be degrees (see my answer). Of course, you can convert it back to radians but then the result (.1339) is the same as when you calculate in radians from the beginning.
$endgroup$
– Fabian
Jan 6 at 8:23
add a comment |
$begingroup$
The antiderivative does not change with the units, as long as the units are the same in the antiderivative and the original. Which is to say, if sin(x) is evaluated with degrees, then the antiderivative is still -cos(x)+C, x still being in degrees. If you want one or the other in radians, you only need to compose in $frac{pi}{180}$ for x to change to radians. The value of the integral $int_0^{30}sin(x)dx$ you got is definitely incorrect, as area should be without units. I'm not sure how your calculator got that answer.
answered Jan 5 at 23:41
H HuangH Huang
8710
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add a comment |
$begingroup$
First of all observe that angle is dimensionless and its measurement (and the units involved) is not like that of length or area.
Further the use of degree is more like a "dozen". Thus if dozen means $12$ a degree means the real number $pi/180$. However the degree is used mostly (I think exclusively) in context of angles.
Based on above we have $sin x^{circ} =sin (pi x/180)$ and thus $$int_{0}^{30}sin x^{circ} , dx=int_{0}^{30}sin(pi x/180),dx=left.-frac{180}{pi}cos(pi x/180)right|_{x=0}^{x=30}$$ The integral evaluates to $$frac{180}{pi}left(1-cosfrac{pi}{6}right)=7.6761789dots$$
answered Jan 6 at 7:18
Paramanand SinghParamanand Singh
49.6k556163
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add a comment |
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I am a bit confused about some of the answers to this question. I believe a lot of the confusion goes back to the fact that the function 'sine' in radians and degrees is not the same function (in the following, I will denote the former with $sin$ and the latter with $operatorname{Sin}$).
So, here is how I understand the issue.
purely dimensional analysis
Having dimensional variables the value of $$int f(x),dx$$ has the units of $f(x)$ times the units of $x$. The reason is very simple to see. The integral measures the area under a curve and indeed the area (of let us say a rectangle) has the product of the units of the two sides.
So the integral $$I= int_{0^circ}^{30^circ} operatorname{Sin} x,dx tag{1}$$
has units degrees (note that the sine function is dimensionless for any measure of the angle).
Whereas the corresponding integral $$J= int_{0}^{pi/6} sin x,dx tag{2}$$ has no units (or radian if you want). The conversion from (1) to (2) can of course be done by converting degrees to radians that is by
$$ frac{pi}{180^circ} I =J,. tag{3}$$
calculation
The two sine functions in (1) and (2) (note that sine in degrees is not the same function as sine in radians) are related by
$$ operatorname{Sin}(x) = sinleft(frac{pi}{180^circ}xright)qquad operatorname{Cos}(x) = cos left(frac{pi}{180^circ}xright),.$$
It is now very simple to show (3) by mathematical arguments (such as the chain rule). In particular, we can find from the antiderivative
$$ frac{d}{dx} cos(x) = sin(x)$$
in radians, the antiderivative
$$ frac{d}{dx} left(frac{180^circ}pi operatorname{Cos}(x) right)= operatorname{Sin}(x)$$
of the sine and cosine functions in degrees. Note that the antiderivative of sine in degrees indeed has the units of degree.
answered Jan 6 at 8:03
FabianFabian
19.6k3674
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add a comment |
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5 Answers
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active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To find the integral in degrees, it's better - for clarity's sake, not necessarily any formal reason - to start in radians with the conversion to degrees in the sine function. That is,
$$sin(x^circ) = sin left( frac{pi ; text{radians}}{180^circ} x^circright )$$
We just typically omit the "radians" because "radians" is less a unit of measure like the degree, than just a measure or a ratio like $pi$, but writing it explicitly like above helps. Though going forward I'll omit that, but it should be clear that the argument of the function on the right becomes one in degrees.
Then, making use of typical antidifferentiation techniques and rules, and utilizing $pi = 180^circ$,
$$begin{align}
int sin(x^circ)dx &= int sin left( frac{pi}{180^circ} x^circright )dx \
&= frac{-180^circ}{pi} cos left( frac{pi}{180^circ} x^circ right) +C\
&= frac{-180^circ}{pi} cos left( x^circ right) +C
end{align}$$
Of course, the fundamental theorem of calculus ($int_a^b f(x)dx = F(b)-F(a)$) also applies here. Just remember to be sure $a,b$ are also in degrees. Beyond that, the antiderivative is basically the same as in radians.
As for your second question, the meaning of any integral is essentially the same: the signed area under the curve of the integrand between the two points chosen as the bounds. Note that it is not going to share the same units as the integrand, i.e. it's not going to be in degrees.
answered Jan 5 at 23:54
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
3
$begingroup$
Where did the ratio $180/pi$ vanish? I assume a typo.
$endgroup$
– Paramanand Singh
Jan 6 at 6:59
add a comment |
$begingroup$
To find the integral in degrees, it's better - for clarity's sake, not necessarily any formal reason - to start in radians with the conversion to degrees in the sine function. That is,
$$sin(x^circ) = sin left( frac{pi ; text{radians}}{180^circ} x^circright )$$
We just typically omit the "radians" because "radians" is less a unit of measure like the degree, than just a measure or a ratio like $pi$, but writing it explicitly like above helps. Though going forward I'll omit that, but it should be clear that the argument of the function on the right becomes one in degrees.
Then, making use of typical antidifferentiation techniques and rules, and utilizing $pi = 180^circ$,
$$begin{align}
int sin(x^circ)dx &= int sin left( frac{pi}{180^circ} x^circright )dx \
&= frac{-180^circ}{pi} cos left( frac{pi}{180^circ} x^circ right) +C\
&= frac{-180^circ}{pi} cos left( x^circ right) +C
end{align}$$
Of course, the fundamental theorem of calculus ($int_a^b f(x)dx = F(b)-F(a)$) also applies here. Just remember to be sure $a,b$ are also in degrees. Beyond that, the antiderivative is basically the same as in radians.
As for your second question, the meaning of any integral is essentially the same: the signed area under the curve of the integrand between the two points chosen as the bounds. Note that it is not going to share the same units as the integrand, i.e. it's not going to be in degrees.
answered Jan 5 at 23:54
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
3
$begingroup$
Where did the ratio $180/pi$ vanish? I assume a typo.
$endgroup$
– Paramanand Singh
Jan 6 at 6:59
add a comment |
$begingroup$
To find the integral in degrees, it's better - for clarity's sake, not necessarily any formal reason - to start in radians with the conversion to degrees in the sine function. That is,
$$sin(x^circ) = sin left( frac{pi ; text{radians}}{180^circ} x^circright )$$
We just typically omit the "radians" because "radians" is less a unit of measure like the degree, than just a measure or a ratio like $pi$, but writing it explicitly like above helps. Though going forward I'll omit that, but it should be clear that the argument of the function on the right becomes one in degrees.
Then, making use of typical antidifferentiation techniques and rules, and utilizing $pi = 180^circ$,
$$begin{align}
int sin(x^circ)dx &= int sin left( frac{pi}{180^circ} x^circright )dx \
&= frac{-180^circ}{pi} cos left( frac{pi}{180^circ} x^circ right) +C\
&= frac{-180^circ}{pi} cos left( x^circ right) +C
end{align}$$
Of course, the fundamental theorem of calculus ($int_a^b f(x)dx = F(b)-F(a)$) also applies here. Just remember to be sure $a,b$ are also in degrees. Beyond that, the antiderivative is basically the same as in radians.
As for your second question, the meaning of any integral is essentially the same: the signed area under the curve of the integrand between the two points chosen as the bounds. Note that it is not going to share the same units as the integrand, i.e. it's not going to be in degrees.
answered Jan 5 at 23:54
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
To find the integral in degrees, it's better - for clarity's sake, not necessarily any formal reason - to start in radians with the conversion to degrees in the sine function. That is,
$$sin(x^circ) = sin left( frac{pi ; text{radians}}{180^circ} x^circright )$$
We just typically omit the "radians" because "radians" is less a unit of measure like the degree, than just a measure or a ratio like $pi$, but writing it explicitly like above helps. Though going forward I'll omit that, but it should be clear that the argument of the function on the right becomes one in degrees.
Then, making use of typical antidifferentiation techniques and rules, and utilizing $pi = 180^circ$,
$$begin{align}
int sin(x^circ)dx &= int sin left( frac{pi}{180^circ} x^circright )dx \
&= frac{-180^circ}{pi} cos left( frac{pi}{180^circ} x^circ right) +C\
&= frac{-180^circ}{pi} cos left( x^circ right) +C
end{align}$$
Of course, the fundamental theorem of calculus ($int_a^b f(x)dx = F(b)-F(a)$) also applies here. Just remember to be sure $a,b$ are also in degrees. Beyond that, the antiderivative is basically the same as in radians.
As for your second question, the meaning of any integral is essentially the same: the signed area under the curve of the integrand between the two points chosen as the bounds. Note that it is not going to share the same units as the integrand, i.e. it's not going to be in degrees.
answered Jan 5 at 23:54
Eevee TrainerEevee Trainer
5,7871936
edited Jan 6 at 18:20
answered Jan 5 at 23:54
Eevee TrainerEevee Trainer
5,7871936
answered Jan 5 at 23:54
Eevee TrainerEevee Trainer
5,7871936
answered Jan 5 at 23:54
Eevee TrainerEevee Trainer
5,7871936
5,7871936
3
$begingroup$
Where did the ratio $180/pi$ vanish? I assume a typo.
$endgroup$
– Paramanand Singh
Jan 6 at 6:59
add a comment |
3
$begingroup$
Where did the ratio $180/pi$ vanish? I assume a typo.
$endgroup$
– Paramanand Singh
Jan 6 at 6:59
3
3
$begingroup$
Where did the ratio $180/pi$ vanish? I assume a typo.
$endgroup$
– Paramanand Singh
Jan 6 at 6:59
$begingroup$
Where did the ratio $180/pi$ vanish? I assume a typo.
$endgroup$
– Paramanand Singh
Jan 6 at 6:59
add a comment |
$begingroup$
Although your calculator is evaluating $sin x$ assuming $x$ is in degrees, it is integrating $x$ from $0$ to $30$. What it is doing would usually be written
$$int_0^{30} sin frac {pi x}{180};dxapprox 7.676$$ where the argument of $sin$ is now in radians. Note that the result of the integral is a pure number, not degrees or radians.
answered Jan 5 at 23:38
Ross MillikanRoss Millikan
294k23198371
$endgroup$
$begingroup$
The value of the integral should be degrees (see my answer). Of course, you can convert it back to radians but then the result (.1339) is the same as when you calculate in radians from the beginning.
$endgroup$
– Fabian
Jan 6 at 8:23
add a comment |
$begingroup$
Although your calculator is evaluating $sin x$ assuming $x$ is in degrees, it is integrating $x$ from $0$ to $30$. What it is doing would usually be written
$$int_0^{30} sin frac {pi x}{180};dxapprox 7.676$$ where the argument of $sin$ is now in radians. Note that the result of the integral is a pure number, not degrees or radians.
answered Jan 5 at 23:38
Ross MillikanRoss Millikan
294k23198371
$endgroup$
$begingroup$
The value of the integral should be degrees (see my answer). Of course, you can convert it back to radians but then the result (.1339) is the same as when you calculate in radians from the beginning.
$endgroup$
– Fabian
Jan 6 at 8:23
add a comment |
$begingroup$
Although your calculator is evaluating $sin x$ assuming $x$ is in degrees, it is integrating $x$ from $0$ to $30$. What it is doing would usually be written
$$int_0^{30} sin frac {pi x}{180};dxapprox 7.676$$ where the argument of $sin$ is now in radians. Note that the result of the integral is a pure number, not degrees or radians.
answered Jan 5 at 23:38
Ross MillikanRoss Millikan
294k23198371
$endgroup$
Although your calculator is evaluating $sin x$ assuming $x$ is in degrees, it is integrating $x$ from $0$ to $30$. What it is doing would usually be written
$$int_0^{30} sin frac {pi x}{180};dxapprox 7.676$$ where the argument of $sin$ is now in radians. Note that the result of the integral is a pure number, not degrees or radians.
answered Jan 5 at 23:38
Ross MillikanRoss Millikan
294k23198371
edited Jan 5 at 23:45
answered Jan 5 at 23:38
Ross MillikanRoss Millikan
294k23198371
answered Jan 5 at 23:38
Ross MillikanRoss Millikan
294k23198371
answered Jan 5 at 23:38
Ross MillikanRoss Millikan
294k23198371
294k23198371
$begingroup$
The value of the integral should be degrees (see my answer). Of course, you can convert it back to radians but then the result (.1339) is the same as when you calculate in radians from the beginning.
$endgroup$
– Fabian
Jan 6 at 8:23
add a comment |
$begingroup$
The value of the integral should be degrees (see my answer). Of course, you can convert it back to radians but then the result (.1339) is the same as when you calculate in radians from the beginning.
$endgroup$
– Fabian
Jan 6 at 8:23
$begingroup$
The value of the integral should be degrees (see my answer). Of course, you can convert it back to radians but then the result (.1339) is the same as when you calculate in radians from the beginning.
$endgroup$
– Fabian
Jan 6 at 8:23
$begingroup$
The value of the integral should be degrees (see my answer). Of course, you can convert it back to radians but then the result (.1339) is the same as when you calculate in radians from the beginning.
$endgroup$
– Fabian
Jan 6 at 8:23
add a comment |
$begingroup$
The antiderivative does not change with the units, as long as the units are the same in the antiderivative and the original. Which is to say, if sin(x) is evaluated with degrees, then the antiderivative is still -cos(x)+C, x still being in degrees. If you want one or the other in radians, you only need to compose in $frac{pi}{180}$ for x to change to radians. The value of the integral $int_0^{30}sin(x)dx$ you got is definitely incorrect, as area should be without units. I'm not sure how your calculator got that answer.
answered Jan 5 at 23:41
H HuangH Huang
8710
$endgroup$
add a comment |
$begingroup$
The antiderivative does not change with the units, as long as the units are the same in the antiderivative and the original. Which is to say, if sin(x) is evaluated with degrees, then the antiderivative is still -cos(x)+C, x still being in degrees. If you want one or the other in radians, you only need to compose in $frac{pi}{180}$ for x to change to radians. The value of the integral $int_0^{30}sin(x)dx$ you got is definitely incorrect, as area should be without units. I'm not sure how your calculator got that answer.
answered Jan 5 at 23:41
H HuangH Huang
8710
$endgroup$
add a comment |
$begingroup$
The antiderivative does not change with the units, as long as the units are the same in the antiderivative and the original. Which is to say, if sin(x) is evaluated with degrees, then the antiderivative is still -cos(x)+C, x still being in degrees. If you want one or the other in radians, you only need to compose in $frac{pi}{180}$ for x to change to radians. The value of the integral $int_0^{30}sin(x)dx$ you got is definitely incorrect, as area should be without units. I'm not sure how your calculator got that answer.
answered Jan 5 at 23:41
H HuangH Huang
8710
$endgroup$
The antiderivative does not change with the units, as long as the units are the same in the antiderivative and the original. Which is to say, if sin(x) is evaluated with degrees, then the antiderivative is still -cos(x)+C, x still being in degrees. If you want one or the other in radians, you only need to compose in $frac{pi}{180}$ for x to change to radians. The value of the integral $int_0^{30}sin(x)dx$ you got is definitely incorrect, as area should be without units. I'm not sure how your calculator got that answer.
answered Jan 5 at 23:41
H HuangH Huang
8710
edited Jan 6 at 0:12
answered Jan 5 at 23:41
H HuangH Huang
8710
answered Jan 5 at 23:41
H HuangH Huang
8710
answered Jan 5 at 23:41
H HuangH Huang
8710
8710
add a comment |
add a comment |
$begingroup$
First of all observe that angle is dimensionless and its measurement (and the units involved) is not like that of length or area.
Further the use of degree is more like a "dozen". Thus if dozen means $12$ a degree means the real number $pi/180$. However the degree is used mostly (I think exclusively) in context of angles.
Based on above we have $sin x^{circ} =sin (pi x/180)$ and thus $$int_{0}^{30}sin x^{circ} , dx=int_{0}^{30}sin(pi x/180),dx=left.-frac{180}{pi}cos(pi x/180)right|_{x=0}^{x=30}$$ The integral evaluates to $$frac{180}{pi}left(1-cosfrac{pi}{6}right)=7.6761789dots$$
answered Jan 6 at 7:18
Paramanand SinghParamanand Singh
49.6k556163
$endgroup$
add a comment |
$begingroup$
First of all observe that angle is dimensionless and its measurement (and the units involved) is not like that of length or area.
Further the use of degree is more like a "dozen". Thus if dozen means $12$ a degree means the real number $pi/180$. However the degree is used mostly (I think exclusively) in context of angles.
Based on above we have $sin x^{circ} =sin (pi x/180)$ and thus $$int_{0}^{30}sin x^{circ} , dx=int_{0}^{30}sin(pi x/180),dx=left.-frac{180}{pi}cos(pi x/180)right|_{x=0}^{x=30}$$ The integral evaluates to $$frac{180}{pi}left(1-cosfrac{pi}{6}right)=7.6761789dots$$
answered Jan 6 at 7:18
Paramanand SinghParamanand Singh
49.6k556163
$endgroup$
add a comment |
$begingroup$
First of all observe that angle is dimensionless and its measurement (and the units involved) is not like that of length or area.
Further the use of degree is more like a "dozen". Thus if dozen means $12$ a degree means the real number $pi/180$. However the degree is used mostly (I think exclusively) in context of angles.
Based on above we have $sin x^{circ} =sin (pi x/180)$ and thus $$int_{0}^{30}sin x^{circ} , dx=int_{0}^{30}sin(pi x/180),dx=left.-frac{180}{pi}cos(pi x/180)right|_{x=0}^{x=30}$$ The integral evaluates to $$frac{180}{pi}left(1-cosfrac{pi}{6}right)=7.6761789dots$$
answered Jan 6 at 7:18
Paramanand SinghParamanand Singh
49.6k556163
$endgroup$
First of all observe that angle is dimensionless and its measurement (and the units involved) is not like that of length or area.
Further the use of degree is more like a "dozen". Thus if dozen means $12$ a degree means the real number $pi/180$. However the degree is used mostly (I think exclusively) in context of angles.
Based on above we have $sin x^{circ} =sin (pi x/180)$ and thus $$int_{0}^{30}sin x^{circ} , dx=int_{0}^{30}sin(pi x/180),dx=left.-frac{180}{pi}cos(pi x/180)right|_{x=0}^{x=30}$$ The integral evaluates to $$frac{180}{pi}left(1-cosfrac{pi}{6}right)=7.6761789dots$$
answered Jan 6 at 7:18
Paramanand SinghParamanand Singh
49.6k556163
answered Jan 6 at 7:18
Paramanand SinghParamanand Singh
49.6k556163
answered Jan 6 at 7:18
Paramanand SinghParamanand Singh
49.6k556163
answered Jan 6 at 7:18
Paramanand SinghParamanand Singh
49.6k556163
49.6k556163
add a comment |
add a comment |
$begingroup$
I am a bit confused about some of the answers to this question. I believe a lot of the confusion goes back to the fact that the function 'sine' in radians and degrees is not the same function (in the following, I will denote the former with $sin$ and the latter with $operatorname{Sin}$).
So, here is how I understand the issue.
purely dimensional analysis
Having dimensional variables the value of $$int f(x),dx$$ has the units of $f(x)$ times the units of $x$. The reason is very simple to see. The integral measures the area under a curve and indeed the area (of let us say a rectangle) has the product of the units of the two sides.
So the integral $$I= int_{0^circ}^{30^circ} operatorname{Sin} x,dx tag{1}$$
has units degrees (note that the sine function is dimensionless for any measure of the angle).
Whereas the corresponding integral $$J= int_{0}^{pi/6} sin x,dx tag{2}$$ has no units (or radian if you want). The conversion from (1) to (2) can of course be done by converting degrees to radians that is by
$$ frac{pi}{180^circ} I =J,. tag{3}$$
calculation
The two sine functions in (1) and (2) (note that sine in degrees is not the same function as sine in radians) are related by
$$ operatorname{Sin}(x) = sinleft(frac{pi}{180^circ}xright)qquad operatorname{Cos}(x) = cos left(frac{pi}{180^circ}xright),.$$
It is now very simple to show (3) by mathematical arguments (such as the chain rule). In particular, we can find from the antiderivative
$$ frac{d}{dx} cos(x) = sin(x)$$
in radians, the antiderivative
$$ frac{d}{dx} left(frac{180^circ}pi operatorname{Cos}(x) right)= operatorname{Sin}(x)$$
of the sine and cosine functions in degrees. Note that the antiderivative of sine in degrees indeed has the units of degree.
answered Jan 6 at 8:03
FabianFabian
19.6k3674
$endgroup$
add a comment |
$begingroup$
I am a bit confused about some of the answers to this question. I believe a lot of the confusion goes back to the fact that the function 'sine' in radians and degrees is not the same function (in the following, I will denote the former with $sin$ and the latter with $operatorname{Sin}$).
So, here is how I understand the issue.
purely dimensional analysis
Having dimensional variables the value of $$int f(x),dx$$ has the units of $f(x)$ times the units of $x$. The reason is very simple to see. The integral measures the area under a curve and indeed the area (of let us say a rectangle) has the product of the units of the two sides.
So the integral $$I= int_{0^circ}^{30^circ} operatorname{Sin} x,dx tag{1}$$
has units degrees (note that the sine function is dimensionless for any measure of the angle).
Whereas the corresponding integral $$J= int_{0}^{pi/6} sin x,dx tag{2}$$ has no units (or radian if you want). The conversion from (1) to (2) can of course be done by converting degrees to radians that is by
$$ frac{pi}{180^circ} I =J,. tag{3}$$
calculation
The two sine functions in (1) and (2) (note that sine in degrees is not the same function as sine in radians) are related by
$$ operatorname{Sin}(x) = sinleft(frac{pi}{180^circ}xright)qquad operatorname{Cos}(x) = cos left(frac{pi}{180^circ}xright),.$$
It is now very simple to show (3) by mathematical arguments (such as the chain rule). In particular, we can find from the antiderivative
$$ frac{d}{dx} cos(x) = sin(x)$$
in radians, the antiderivative
$$ frac{d}{dx} left(frac{180^circ}pi operatorname{Cos}(x) right)= operatorname{Sin}(x)$$
of the sine and cosine functions in degrees. Note that the antiderivative of sine in degrees indeed has the units of degree.
answered Jan 6 at 8:03
FabianFabian
19.6k3674
$endgroup$
add a comment |
$begingroup$
I am a bit confused about some of the answers to this question. I believe a lot of the confusion goes back to the fact that the function 'sine' in radians and degrees is not the same function (in the following, I will denote the former with $sin$ and the latter with $operatorname{Sin}$).
So, here is how I understand the issue.
purely dimensional analysis
Having dimensional variables the value of $$int f(x),dx$$ has the units of $f(x)$ times the units of $x$. The reason is very simple to see. The integral measures the area under a curve and indeed the area (of let us say a rectangle) has the product of the units of the two sides.
So the integral $$I= int_{0^circ}^{30^circ} operatorname{Sin} x,dx tag{1}$$
has units degrees (note that the sine function is dimensionless for any measure of the angle).
Whereas the corresponding integral $$J= int_{0}^{pi/6} sin x,dx tag{2}$$ has no units (or radian if you want). The conversion from (1) to (2) can of course be done by converting degrees to radians that is by
$$ frac{pi}{180^circ} I =J,. tag{3}$$
calculation
The two sine functions in (1) and (2) (note that sine in degrees is not the same function as sine in radians) are related by
$$ operatorname{Sin}(x) = sinleft(frac{pi}{180^circ}xright)qquad operatorname{Cos}(x) = cos left(frac{pi}{180^circ}xright),.$$
It is now very simple to show (3) by mathematical arguments (such as the chain rule). In particular, we can find from the antiderivative
$$ frac{d}{dx} cos(x) = sin(x)$$
in radians, the antiderivative
$$ frac{d}{dx} left(frac{180^circ}pi operatorname{Cos}(x) right)= operatorname{Sin}(x)$$
of the sine and cosine functions in degrees. Note that the antiderivative of sine in degrees indeed has the units of degree.
answered Jan 6 at 8:03
FabianFabian
19.6k3674
$endgroup$
I am a bit confused about some of the answers to this question. I believe a lot of the confusion goes back to the fact that the function 'sine' in radians and degrees is not the same function (in the following, I will denote the former with $sin$ and the latter with $operatorname{Sin}$).
So, here is how I understand the issue.
purely dimensional analysis
Having dimensional variables the value of $$int f(x),dx$$ has the units of $f(x)$ times the units of $x$. The reason is very simple to see. The integral measures the area under a curve and indeed the area (of let us say a rectangle) has the product of the units of the two sides.
So the integral $$I= int_{0^circ}^{30^circ} operatorname{Sin} x,dx tag{1}$$
has units degrees (note that the sine function is dimensionless for any measure of the angle).
Whereas the corresponding integral $$J= int_{0}^{pi/6} sin x,dx tag{2}$$ has no units (or radian if you want). The conversion from (1) to (2) can of course be done by converting degrees to radians that is by
$$ frac{pi}{180^circ} I =J,. tag{3}$$
calculation
The two sine functions in (1) and (2) (note that sine in degrees is not the same function as sine in radians) are related by
$$ operatorname{Sin}(x) = sinleft(frac{pi}{180^circ}xright)qquad operatorname{Cos}(x) = cos left(frac{pi}{180^circ}xright),.$$
It is now very simple to show (3) by mathematical arguments (such as the chain rule). In particular, we can find from the antiderivative
$$ frac{d}{dx} cos(x) = sin(x)$$
in radians, the antiderivative
$$ frac{d}{dx} left(frac{180^circ}pi operatorname{Cos}(x) right)= operatorname{Sin}(x)$$
of the sine and cosine functions in degrees. Note that the antiderivative of sine in degrees indeed has the units of degree.
answered Jan 6 at 8:03
FabianFabian
19.6k3674
edited Jan 6 at 8:33
answered Jan 6 at 8:03
FabianFabian
19.6k3674
answered Jan 6 at 8:03
FabianFabian
19.6k3674
answered Jan 6 at 8:03
FabianFabian
19.6k3674
19.6k3674
add a comment |
add a comment |
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$begingroup$
$sin x^circ = sin frac{pi}{180} x$.
$endgroup$
– anomaly
Jan 5 at 23:28
3
$begingroup$
The result of the integral is not an angle.
$endgroup$
– John Douma
Jan 5 at 23:29
$begingroup$
Of course, the result of the integral is an angle (see my answer).
$endgroup$
– Fabian
Jan 6 at 8:19