Mixing
Calculus with indefinite integral
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Any other solutions(advice) are welcome
calculus
asked Jan 23 at 9:43
mina_worldmina_world
1799
$endgroup$
add a comment |
$begingroup$
Any other solutions(advice) are welcome
calculus
asked Jan 23 at 9:43
mina_worldmina_world
1799
$endgroup$
1
$begingroup$
It is not necessary to demand $cos xneq0$ apartly. The presence of term $tan x$ in the expression already indicates that.
$endgroup$
– drhab
Jan 23 at 9:52
add a comment |
$begingroup$
Any other solutions(advice) are welcome
calculus
asked Jan 23 at 9:43
mina_worldmina_world
1799
$endgroup$
Any other solutions(advice) are welcome
calculus
calculus
asked Jan 23 at 9:43
mina_worldmina_world
1799
asked Jan 23 at 9:43
mina_worldmina_world
1799
edited Jan 23 at 14:23
mina_world
asked Jan 23 at 9:43
mina_worldmina_world
1799
asked Jan 23 at 9:43
mina_worldmina_world
1799
asked Jan 23 at 9:43
mina_worldmina_world
1799
1799
1
$begingroup$
It is not necessary to demand $cos xneq0$ apartly. The presence of term $tan x$ in the expression already indicates that.
$endgroup$
– drhab
Jan 23 at 9:52
add a comment |
1
$begingroup$
It is not necessary to demand $cos xneq0$ apartly. The presence of term $tan x$ in the expression already indicates that.
$endgroup$
– drhab
Jan 23 at 9:52
1
1
$begingroup$
It is not necessary to demand $cos xneq0$ apartly. The presence of term $tan x$ in the expression already indicates that.
$endgroup$
– drhab
Jan 23 at 9:52
$begingroup$
It is not necessary to demand $cos xneq0$ apartly. The presence of term $tan x$ in the expression already indicates that.
$endgroup$
– drhab
Jan 23 at 9:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First of all, there is a mistake in the computation from the second to the third line. The correct answer therefore should be $sec(x) + tan(x) - x$.
Usually in these exercises, it is generally understood that the antiderivative (or the derivative for that matter) is only to be taken where it makes sense. However, you have a point here in the sense that the solution $sec(x) + tan(x) - x$ is incomplete. In fact, the antiderivative of $frac{sin(x)}{1-sin(x)}$ is perfectly well defined in $sin(x) = -1$. Therefore a complete solution would be $$begin{cases} sec(x) + tan(x) - x & text{for } sin(x) notin{pm 1} \ frac{pi}{2} & text{for } sin(x) = -1end{cases}$$
and of course undefined for $sin(x) = 1$. The case $sin(x) = -1$ is just an artefact of the computation method.
answered Jan 23 at 10:42
KlausKlaus
2,30712
$endgroup$
$begingroup$
Yes, I corrected the wrong part you said. Thank you.
$endgroup$
– mina_world
Jan 23 at 11:04
add a comment |
$begingroup$
When you dual with indefinite integral, you always assume implicitly that the function is on a valid interval. For example, in this question, you may assume the function is on $(frac{-3pi}{2},frac{pi}{2})$.
Your arguments about the condition $Sin(x)neq-1$ is correct, the final result obviously does not hold for $Sin(x)=-1$. You can get the complete solution as in Klaus answer. Here I give another solution which can avoid this situation:
begin{align}
int frac{sin(x)}{1-sin(x)}dx
&=int (-1+frac{1}{1-sin(x)})dx \
&=-x+int frac{1}{1-2 sin(frac{x}{2})cos(frac{x}{2})}dx \
&=-x+int frac{1}{(sin(frac{x}{2})-cos(frac{x}{2}))^2}dx \
&=-x+int frac{1}{2 sin^2(frac{x}{2}-frac{pi}{4})}dx \
&=-x- cot(frac{x}{2}-frac{pi}{4})+C
end{align}
Auxiliary angle formula $asin(x)+bcos(x)=sin(x+arctan(frac{b}{a}))$ has been used in second last line
answered Jan 23 at 13:10
DragunityMAXDragunityMAX
1618
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
First of all, there is a mistake in the computation from the second to the third line. The correct answer therefore should be $sec(x) + tan(x) - x$.
Usually in these exercises, it is generally understood that the antiderivative (or the derivative for that matter) is only to be taken where it makes sense. However, you have a point here in the sense that the solution $sec(x) + tan(x) - x$ is incomplete. In fact, the antiderivative of $frac{sin(x)}{1-sin(x)}$ is perfectly well defined in $sin(x) = -1$. Therefore a complete solution would be $$begin{cases} sec(x) + tan(x) - x & text{for } sin(x) notin{pm 1} \ frac{pi}{2} & text{for } sin(x) = -1end{cases}$$
and of course undefined for $sin(x) = 1$. The case $sin(x) = -1$ is just an artefact of the computation method.
answered Jan 23 at 10:42
KlausKlaus
2,30712
$endgroup$
$begingroup$
Yes, I corrected the wrong part you said. Thank you.
$endgroup$
– mina_world
Jan 23 at 11:04
add a comment |
$begingroup$
First of all, there is a mistake in the computation from the second to the third line. The correct answer therefore should be $sec(x) + tan(x) - x$.
Usually in these exercises, it is generally understood that the antiderivative (or the derivative for that matter) is only to be taken where it makes sense. However, you have a point here in the sense that the solution $sec(x) + tan(x) - x$ is incomplete. In fact, the antiderivative of $frac{sin(x)}{1-sin(x)}$ is perfectly well defined in $sin(x) = -1$. Therefore a complete solution would be $$begin{cases} sec(x) + tan(x) - x & text{for } sin(x) notin{pm 1} \ frac{pi}{2} & text{for } sin(x) = -1end{cases}$$
and of course undefined for $sin(x) = 1$. The case $sin(x) = -1$ is just an artefact of the computation method.
answered Jan 23 at 10:42
KlausKlaus
2,30712
$endgroup$
$begingroup$
Yes, I corrected the wrong part you said. Thank you.
$endgroup$
– mina_world
Jan 23 at 11:04
add a comment |
$begingroup$
First of all, there is a mistake in the computation from the second to the third line. The correct answer therefore should be $sec(x) + tan(x) - x$.
Usually in these exercises, it is generally understood that the antiderivative (or the derivative for that matter) is only to be taken where it makes sense. However, you have a point here in the sense that the solution $sec(x) + tan(x) - x$ is incomplete. In fact, the antiderivative of $frac{sin(x)}{1-sin(x)}$ is perfectly well defined in $sin(x) = -1$. Therefore a complete solution would be $$begin{cases} sec(x) + tan(x) - x & text{for } sin(x) notin{pm 1} \ frac{pi}{2} & text{for } sin(x) = -1end{cases}$$
and of course undefined for $sin(x) = 1$. The case $sin(x) = -1$ is just an artefact of the computation method.
answered Jan 23 at 10:42
KlausKlaus
2,30712
$endgroup$
First of all, there is a mistake in the computation from the second to the third line. The correct answer therefore should be $sec(x) + tan(x) - x$.
Usually in these exercises, it is generally understood that the antiderivative (or the derivative for that matter) is only to be taken where it makes sense. However, you have a point here in the sense that the solution $sec(x) + tan(x) - x$ is incomplete. In fact, the antiderivative of $frac{sin(x)}{1-sin(x)}$ is perfectly well defined in $sin(x) = -1$. Therefore a complete solution would be $$begin{cases} sec(x) + tan(x) - x & text{for } sin(x) notin{pm 1} \ frac{pi}{2} & text{for } sin(x) = -1end{cases}$$
and of course undefined for $sin(x) = 1$. The case $sin(x) = -1$ is just an artefact of the computation method.
answered Jan 23 at 10:42
KlausKlaus
2,30712
answered Jan 23 at 10:42
KlausKlaus
2,30712
answered Jan 23 at 10:42
KlausKlaus
2,30712
answered Jan 23 at 10:42
KlausKlaus
2,30712
2,30712
$begingroup$
Yes, I corrected the wrong part you said. Thank you.
$endgroup$
– mina_world
Jan 23 at 11:04
add a comment |
$begingroup$
Yes, I corrected the wrong part you said. Thank you.
$endgroup$
– mina_world
Jan 23 at 11:04
$begingroup$
Yes, I corrected the wrong part you said. Thank you.
$endgroup$
– mina_world
Jan 23 at 11:04
$begingroup$
Yes, I corrected the wrong part you said. Thank you.
$endgroup$
– mina_world
Jan 23 at 11:04
add a comment |
$begingroup$
When you dual with indefinite integral, you always assume implicitly that the function is on a valid interval. For example, in this question, you may assume the function is on $(frac{-3pi}{2},frac{pi}{2})$.
Your arguments about the condition $Sin(x)neq-1$ is correct, the final result obviously does not hold for $Sin(x)=-1$. You can get the complete solution as in Klaus answer. Here I give another solution which can avoid this situation:
begin{align}
int frac{sin(x)}{1-sin(x)}dx
&=int (-1+frac{1}{1-sin(x)})dx \
&=-x+int frac{1}{1-2 sin(frac{x}{2})cos(frac{x}{2})}dx \
&=-x+int frac{1}{(sin(frac{x}{2})-cos(frac{x}{2}))^2}dx \
&=-x+int frac{1}{2 sin^2(frac{x}{2}-frac{pi}{4})}dx \
&=-x- cot(frac{x}{2}-frac{pi}{4})+C
end{align}
Auxiliary angle formula $asin(x)+bcos(x)=sin(x+arctan(frac{b}{a}))$ has been used in second last line
answered Jan 23 at 13:10
DragunityMAXDragunityMAX
1618
$endgroup$
add a comment |
$begingroup$
When you dual with indefinite integral, you always assume implicitly that the function is on a valid interval. For example, in this question, you may assume the function is on $(frac{-3pi}{2},frac{pi}{2})$.
Your arguments about the condition $Sin(x)neq-1$ is correct, the final result obviously does not hold for $Sin(x)=-1$. You can get the complete solution as in Klaus answer. Here I give another solution which can avoid this situation:
begin{align}
int frac{sin(x)}{1-sin(x)}dx
&=int (-1+frac{1}{1-sin(x)})dx \
&=-x+int frac{1}{1-2 sin(frac{x}{2})cos(frac{x}{2})}dx \
&=-x+int frac{1}{(sin(frac{x}{2})-cos(frac{x}{2}))^2}dx \
&=-x+int frac{1}{2 sin^2(frac{x}{2}-frac{pi}{4})}dx \
&=-x- cot(frac{x}{2}-frac{pi}{4})+C
end{align}
Auxiliary angle formula $asin(x)+bcos(x)=sin(x+arctan(frac{b}{a}))$ has been used in second last line
answered Jan 23 at 13:10
DragunityMAXDragunityMAX
1618
$endgroup$
add a comment |
$begingroup$
When you dual with indefinite integral, you always assume implicitly that the function is on a valid interval. For example, in this question, you may assume the function is on $(frac{-3pi}{2},frac{pi}{2})$.
Your arguments about the condition $Sin(x)neq-1$ is correct, the final result obviously does not hold for $Sin(x)=-1$. You can get the complete solution as in Klaus answer. Here I give another solution which can avoid this situation:
begin{align}
int frac{sin(x)}{1-sin(x)}dx
&=int (-1+frac{1}{1-sin(x)})dx \
&=-x+int frac{1}{1-2 sin(frac{x}{2})cos(frac{x}{2})}dx \
&=-x+int frac{1}{(sin(frac{x}{2})-cos(frac{x}{2}))^2}dx \
&=-x+int frac{1}{2 sin^2(frac{x}{2}-frac{pi}{4})}dx \
&=-x- cot(frac{x}{2}-frac{pi}{4})+C
end{align}
Auxiliary angle formula $asin(x)+bcos(x)=sin(x+arctan(frac{b}{a}))$ has been used in second last line
answered Jan 23 at 13:10
DragunityMAXDragunityMAX
1618
$endgroup$
When you dual with indefinite integral, you always assume implicitly that the function is on a valid interval. For example, in this question, you may assume the function is on $(frac{-3pi}{2},frac{pi}{2})$.
Your arguments about the condition $Sin(x)neq-1$ is correct, the final result obviously does not hold for $Sin(x)=-1$. You can get the complete solution as in Klaus answer. Here I give another solution which can avoid this situation:
begin{align}
int frac{sin(x)}{1-sin(x)}dx
&=int (-1+frac{1}{1-sin(x)})dx \
&=-x+int frac{1}{1-2 sin(frac{x}{2})cos(frac{x}{2})}dx \
&=-x+int frac{1}{(sin(frac{x}{2})-cos(frac{x}{2}))^2}dx \
&=-x+int frac{1}{2 sin^2(frac{x}{2}-frac{pi}{4})}dx \
&=-x- cot(frac{x}{2}-frac{pi}{4})+C
end{align}
Auxiliary angle formula $asin(x)+bcos(x)=sin(x+arctan(frac{b}{a}))$ has been used in second last line
answered Jan 23 at 13:10
DragunityMAXDragunityMAX
1618
edited Jan 23 at 13:28
answered Jan 23 at 13:10
DragunityMAXDragunityMAX
1618
answered Jan 23 at 13:10
DragunityMAXDragunityMAX
1618
answered Jan 23 at 13:10
DragunityMAXDragunityMAX
1618
1618
add a comment |
add a comment |
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1
$begingroup$
It is not necessary to demand $cos xneq0$ apartly. The presence of term $tan x$ in the expression already indicates that.
$endgroup$
– drhab
Jan 23 at 9:52