Mixing
$ int bigl| asin(x)+bcos(x) bigr| {rm d}x = ? $
$begingroup$
My friend evaluated this to be
$$
int bigl| asin(x)+bcos(x) bigr| {rm d}x \
= sqrt{a^2+b^2} left(
sin(x-phi)text{sign}(cos(x-phi))
+frac{2}{pi} bigl(x-arctan(tan(x-phi)) bigr)
right) + C
$$
where $phi = arctan left(frac{a}{b} right)$.
My answer instead is much shorter, by simply looking for the intervals of $x$ for which I need to multiply by $-1$, I got
$$
(bsin(x)-acos(x)) cdot text{sign}(asin(x)+bcos(x)) + C
$$
Why does my friend's solution look so big and complex and how did she even come up with the idea to do that?
calculus integration indefinite-integrals absolute-value trigonometric-integrals
asked Jan 2 at 16:14
MintMint
5301417
$endgroup$
|
show 1 more comment
$begingroup$
My friend evaluated this to be
$$
int bigl| asin(x)+bcos(x) bigr| {rm d}x \
= sqrt{a^2+b^2} left(
sin(x-phi)text{sign}(cos(x-phi))
+frac{2}{pi} bigl(x-arctan(tan(x-phi)) bigr)
right) + C
$$
where $phi = arctan left(frac{a}{b} right)$.
My answer instead is much shorter, by simply looking for the intervals of $x$ for which I need to multiply by $-1$, I got
$$
(bsin(x)-acos(x)) cdot text{sign}(asin(x)+bcos(x)) + C
$$
Why does my friend's solution look so big and complex and how did she even come up with the idea to do that?
calculus integration indefinite-integrals absolute-value trigonometric-integrals
asked Jan 2 at 16:14
MintMint
5301417
$endgroup$
1
$begingroup$
Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
$endgroup$
– A.Γ.
Jan 2 at 16:23
2
$begingroup$
What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
$endgroup$
– John Doe
Jan 2 at 16:23
$begingroup$
@A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
$endgroup$
– John Doe
Jan 2 at 16:27
$begingroup$
@JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
$endgroup$
– A.Γ.
Jan 2 at 16:43
$begingroup$
@A.Γ. Ah yes, you're right.
$endgroup$
– John Doe
Jan 2 at 16:44
|
show 1 more comment
$begingroup$
My friend evaluated this to be
$$
int bigl| asin(x)+bcos(x) bigr| {rm d}x \
= sqrt{a^2+b^2} left(
sin(x-phi)text{sign}(cos(x-phi))
+frac{2}{pi} bigl(x-arctan(tan(x-phi)) bigr)
right) + C
$$
where $phi = arctan left(frac{a}{b} right)$.
My answer instead is much shorter, by simply looking for the intervals of $x$ for which I need to multiply by $-1$, I got
$$
(bsin(x)-acos(x)) cdot text{sign}(asin(x)+bcos(x)) + C
$$
Why does my friend's solution look so big and complex and how did she even come up with the idea to do that?
calculus integration indefinite-integrals absolute-value trigonometric-integrals
asked Jan 2 at 16:14
MintMint
5301417
$endgroup$
My friend evaluated this to be
$$
int bigl| asin(x)+bcos(x) bigr| {rm d}x \
= sqrt{a^2+b^2} left(
sin(x-phi)text{sign}(cos(x-phi))
+frac{2}{pi} bigl(x-arctan(tan(x-phi)) bigr)
right) + C
$$
where $phi = arctan left(frac{a}{b} right)$.
My answer instead is much shorter, by simply looking for the intervals of $x$ for which I need to multiply by $-1$, I got
$$
(bsin(x)-acos(x)) cdot text{sign}(asin(x)+bcos(x)) + C
$$
Why does my friend's solution look so big and complex and how did she even come up with the idea to do that?
calculus integration indefinite-integrals absolute-value trigonometric-integrals
calculus integration indefinite-integrals absolute-value trigonometric-integrals
asked Jan 2 at 16:14
MintMint
5301417
asked Jan 2 at 16:14
MintMint
5301417
edited Jan 4 at 14:29
Mint
asked Jan 2 at 16:14
MintMint
5301417
asked Jan 2 at 16:14
MintMint
5301417
asked Jan 2 at 16:14
MintMint
5301417
5301417
1
$begingroup$
Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
$endgroup$
– A.Γ.
Jan 2 at 16:23
2
$begingroup$
What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
$endgroup$
– John Doe
Jan 2 at 16:23
$begingroup$
@A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
$endgroup$
– John Doe
Jan 2 at 16:27
$begingroup$
@JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
$endgroup$
– A.Γ.
Jan 2 at 16:43
$begingroup$
@A.Γ. Ah yes, you're right.
$endgroup$
– John Doe
Jan 2 at 16:44
|
show 1 more comment
1
$begingroup$
Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
$endgroup$
– A.Γ.
Jan 2 at 16:23
2
$begingroup$
What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
$endgroup$
– John Doe
Jan 2 at 16:23
$begingroup$
@A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
$endgroup$
– John Doe
Jan 2 at 16:27
$begingroup$
@JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
$endgroup$
– A.Γ.
Jan 2 at 16:43
$begingroup$
@A.Γ. Ah yes, you're right.
$endgroup$
– John Doe
Jan 2 at 16:44
1
1
$begingroup$
Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
$endgroup$
– A.Γ.
Jan 2 at 16:23
$begingroup$
Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
$endgroup$
– A.Γ.
Jan 2 at 16:23
2
2
$begingroup$
What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
$endgroup$
– John Doe
Jan 2 at 16:23
$begingroup$
What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
$endgroup$
– John Doe
Jan 2 at 16:23
$begingroup$
@A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
$endgroup$
– John Doe
Jan 2 at 16:27
$begingroup$
@A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
$endgroup$
– John Doe
Jan 2 at 16:27
$begingroup$
@JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
$endgroup$
– A.Γ.
Jan 2 at 16:43
$begingroup$
@JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
$endgroup$
– A.Γ.
Jan 2 at 16:43
$begingroup$
@A.Γ. Ah yes, you're right.
$endgroup$
– John Doe
Jan 2 at 16:44
$begingroup$
@A.Γ. Ah yes, you're right.
$endgroup$
– John Doe
Jan 2 at 16:44
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
She is probably starting by replacing the $asin x + bcos x$ and rewriting in terms of a function with an amplitude and phase. One way to do this is to multiply by $1$ and draw a right triangle with angle $phi$, adjacent side length $b$, and opposite side length $a$. Then one has
$$begin{aligned} left|asin x + bcos xright| &= sqrt{a^{2}+b^{2}}left|frac{a}{sqrt{a^{2}+b^{2}}},sin x + frac{b}{sqrt{a^{2}+b^{2}}},cos xright| \
&= sqrt{a^{2}+b^{2}}left|sinphisin x + cosphicos xright| \
&= sqrt{a^{2}+b^{2}}left|cos(x-phi)right|.end{aligned}$$
This procedure is motivated in physics/engineering as a more elegant way of writing solutions to the harmonic oscillator equation, as amplitude and phase both have physical interpretations and are often easier to determine given initial conditions.
answered Jan 2 at 18:33
IninterrompueIninterrompue
63519
$endgroup$
add a comment |
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$begingroup$
She is probably starting by replacing the $asin x + bcos x$ and rewriting in terms of a function with an amplitude and phase. One way to do this is to multiply by $1$ and draw a right triangle with angle $phi$, adjacent side length $b$, and opposite side length $a$. Then one has
$$begin{aligned} left|asin x + bcos xright| &= sqrt{a^{2}+b^{2}}left|frac{a}{sqrt{a^{2}+b^{2}}},sin x + frac{b}{sqrt{a^{2}+b^{2}}},cos xright| \
&= sqrt{a^{2}+b^{2}}left|sinphisin x + cosphicos xright| \
&= sqrt{a^{2}+b^{2}}left|cos(x-phi)right|.end{aligned}$$
This procedure is motivated in physics/engineering as a more elegant way of writing solutions to the harmonic oscillator equation, as amplitude and phase both have physical interpretations and are often easier to determine given initial conditions.
answered Jan 2 at 18:33
IninterrompueIninterrompue
63519
$endgroup$
add a comment |
$begingroup$
She is probably starting by replacing the $asin x + bcos x$ and rewriting in terms of a function with an amplitude and phase. One way to do this is to multiply by $1$ and draw a right triangle with angle $phi$, adjacent side length $b$, and opposite side length $a$. Then one has
$$begin{aligned} left|asin x + bcos xright| &= sqrt{a^{2}+b^{2}}left|frac{a}{sqrt{a^{2}+b^{2}}},sin x + frac{b}{sqrt{a^{2}+b^{2}}},cos xright| \
&= sqrt{a^{2}+b^{2}}left|sinphisin x + cosphicos xright| \
&= sqrt{a^{2}+b^{2}}left|cos(x-phi)right|.end{aligned}$$
This procedure is motivated in physics/engineering as a more elegant way of writing solutions to the harmonic oscillator equation, as amplitude and phase both have physical interpretations and are often easier to determine given initial conditions.
answered Jan 2 at 18:33
IninterrompueIninterrompue
63519
$endgroup$
add a comment |
$begingroup$
She is probably starting by replacing the $asin x + bcos x$ and rewriting in terms of a function with an amplitude and phase. One way to do this is to multiply by $1$ and draw a right triangle with angle $phi$, adjacent side length $b$, and opposite side length $a$. Then one has
$$begin{aligned} left|asin x + bcos xright| &= sqrt{a^{2}+b^{2}}left|frac{a}{sqrt{a^{2}+b^{2}}},sin x + frac{b}{sqrt{a^{2}+b^{2}}},cos xright| \
&= sqrt{a^{2}+b^{2}}left|sinphisin x + cosphicos xright| \
&= sqrt{a^{2}+b^{2}}left|cos(x-phi)right|.end{aligned}$$
This procedure is motivated in physics/engineering as a more elegant way of writing solutions to the harmonic oscillator equation, as amplitude and phase both have physical interpretations and are often easier to determine given initial conditions.
answered Jan 2 at 18:33
IninterrompueIninterrompue
63519
$endgroup$
She is probably starting by replacing the $asin x + bcos x$ and rewriting in terms of a function with an amplitude and phase. One way to do this is to multiply by $1$ and draw a right triangle with angle $phi$, adjacent side length $b$, and opposite side length $a$. Then one has
$$begin{aligned} left|asin x + bcos xright| &= sqrt{a^{2}+b^{2}}left|frac{a}{sqrt{a^{2}+b^{2}}},sin x + frac{b}{sqrt{a^{2}+b^{2}}},cos xright| \
&= sqrt{a^{2}+b^{2}}left|sinphisin x + cosphicos xright| \
&= sqrt{a^{2}+b^{2}}left|cos(x-phi)right|.end{aligned}$$
This procedure is motivated in physics/engineering as a more elegant way of writing solutions to the harmonic oscillator equation, as amplitude and phase both have physical interpretations and are often easier to determine given initial conditions.
answered Jan 2 at 18:33
IninterrompueIninterrompue
63519
answered Jan 2 at 18:33
IninterrompueIninterrompue
63519
answered Jan 2 at 18:33
IninterrompueIninterrompue
63519
answered Jan 2 at 18:33
IninterrompueIninterrompue
63519
63519
add a comment |
add a comment |
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$begingroup$
Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
$endgroup$
– A.Γ.
Jan 2 at 16:23
2
$begingroup$
What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
$endgroup$
– John Doe
Jan 2 at 16:23
$begingroup$
@A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
$endgroup$
– John Doe
Jan 2 at 16:27
$begingroup$
@JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
$endgroup$
– A.Γ.
Jan 2 at 16:43
$begingroup$
@A.Γ. Ah yes, you're right.
$endgroup$
– John Doe
Jan 2 at 16:44