Show that $int_{0}^{2pi} cos^2(x) dx = int_{0}^{2pi} sin^2(x) dx$
$begingroup$
I'm trying to follow the argument in the image below, which aims to show that: $int_0^{2pi} cos^2(x) dx = int_0^{2pi} sin^2(x) dx$.
I believe this on an intuitive level, and I understand that it uses the periodicity of the sine and cosine functions to make the point. But, I'm stuck because the argument seems to only show that $int_{x=0}^{x=2pi} cos^2(x)dx = int_{u=0}^{u=2pi} sin^2(u) du$, and the variable $u$ is not the same as $x$.
Would you be able to clarify this argument?
calculus integration trigonometry
asked Nov 30 '18 at 4:15
lthompsonlthompson
1099
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying to follow the argument in the image below, which aims to show that: $int_0^{2pi} cos^2(x) dx = int_0^{2pi} sin^2(x) dx$.
I believe this on an intuitive level, and I understand that it uses the periodicity of the sine and cosine functions to make the point. But, I'm stuck because the argument seems to only show that $int_{x=0}^{x=2pi} cos^2(x)dx = int_{u=0}^{u=2pi} sin^2(u) du$, and the variable $u$ is not the same as $x$.
Would you be able to clarify this argument?
calculus integration trigonometry
asked Nov 30 '18 at 4:15
lthompsonlthompson
1099
$endgroup$
2
$begingroup$
Do you believe $int sin^2(x)dx = int sin^2(u)du$? What role does the variable play in the integral?
$endgroup$
– Ryan
Nov 30 '18 at 4:18
3
$begingroup$
For a definite integral, the name of the variable shouldn't matter.
$endgroup$
– Anurag A
Nov 30 '18 at 4:18
2
$begingroup$
$u$ is a dummy variable: it's like saying $sum_{k=1}^nf(x_k)$ is equal to $sum_{ell=1}^nf(x_ell)$, where here $k,ell$ are dummy indices. by the way, another way: $cos^2(x)=(1+sin(2x)]/2$, and $sin^2(x)=[1-sin 2x]/2$, and the doubly-periodic terms integrate to zero.
$endgroup$
– user254433
Nov 30 '18 at 4:19
3
$begingroup$
$$ int f( Calculus ) d Calculus = int f(x) dx = int f(u) du $$
$endgroup$
– Jimmy Sabater
Nov 30 '18 at 4:23
2
$begingroup$
@lthompson What is the difference between $int_a^b f(x),dx$ and $int_a^b f(t),dt$? There is no difference. The integration variable is a "dummy" and is not implicated in the answer.
$endgroup$
– Mark Viola
Nov 30 '18 at 4:29
|
show 1 more comment
$begingroup$
I'm trying to follow the argument in the image below, which aims to show that: $int_0^{2pi} cos^2(x) dx = int_0^{2pi} sin^2(x) dx$.
I believe this on an intuitive level, and I understand that it uses the periodicity of the sine and cosine functions to make the point. But, I'm stuck because the argument seems to only show that $int_{x=0}^{x=2pi} cos^2(x)dx = int_{u=0}^{u=2pi} sin^2(u) du$, and the variable $u$ is not the same as $x$.
Would you be able to clarify this argument?
calculus integration trigonometry
asked Nov 30 '18 at 4:15
lthompsonlthompson
1099
$endgroup$
I'm trying to follow the argument in the image below, which aims to show that: $int_0^{2pi} cos^2(x) dx = int_0^{2pi} sin^2(x) dx$.
I believe this on an intuitive level, and I understand that it uses the periodicity of the sine and cosine functions to make the point. But, I'm stuck because the argument seems to only show that $int_{x=0}^{x=2pi} cos^2(x)dx = int_{u=0}^{u=2pi} sin^2(u) du$, and the variable $u$ is not the same as $x$.
Would you be able to clarify this argument?
calculus integration trigonometry
calculus integration trigonometry
asked Nov 30 '18 at 4:15
lthompsonlthompson
1099
asked Nov 30 '18 at 4:15
lthompsonlthompson
1099
asked Nov 30 '18 at 4:15
lthompsonlthompson
1099
asked Nov 30 '18 at 4:15
lthompsonlthompson
1099
asked Nov 30 '18 at 4:15
lthompsonlthompson
1099
1099
2
$begingroup$
Do you believe $int sin^2(x)dx = int sin^2(u)du$? What role does the variable play in the integral?
$endgroup$
– Ryan
Nov 30 '18 at 4:18
3
$begingroup$
For a definite integral, the name of the variable shouldn't matter.
$endgroup$
– Anurag A
Nov 30 '18 at 4:18
2
$begingroup$
$u$ is a dummy variable: it's like saying $sum_{k=1}^nf(x_k)$ is equal to $sum_{ell=1}^nf(x_ell)$, where here $k,ell$ are dummy indices. by the way, another way: $cos^2(x)=(1+sin(2x)]/2$, and $sin^2(x)=[1-sin 2x]/2$, and the doubly-periodic terms integrate to zero.
$endgroup$
– user254433
Nov 30 '18 at 4:19
3
$begingroup$
$$ int f( Calculus ) d Calculus = int f(x) dx = int f(u) du $$
$endgroup$
– Jimmy Sabater
Nov 30 '18 at 4:23
2
$begingroup$
@lthompson What is the difference between $int_a^b f(x),dx$ and $int_a^b f(t),dt$? There is no difference. The integration variable is a "dummy" and is not implicated in the answer.
$endgroup$
– Mark Viola
Nov 30 '18 at 4:29
|
show 1 more comment
2
$begingroup$
Do you believe $int sin^2(x)dx = int sin^2(u)du$? What role does the variable play in the integral?
$endgroup$
– Ryan
Nov 30 '18 at 4:18
3
$begingroup$
For a definite integral, the name of the variable shouldn't matter.
$endgroup$
– Anurag A
Nov 30 '18 at 4:18
2
$begingroup$
$u$ is a dummy variable: it's like saying $sum_{k=1}^nf(x_k)$ is equal to $sum_{ell=1}^nf(x_ell)$, where here $k,ell$ are dummy indices. by the way, another way: $cos^2(x)=(1+sin(2x)]/2$, and $sin^2(x)=[1-sin 2x]/2$, and the doubly-periodic terms integrate to zero.
$endgroup$
– user254433
Nov 30 '18 at 4:19
3
$begingroup$
$$ int f( Calculus ) d Calculus = int f(x) dx = int f(u) du $$
$endgroup$
– Jimmy Sabater
Nov 30 '18 at 4:23
2
$begingroup$
@lthompson What is the difference between $int_a^b f(x),dx$ and $int_a^b f(t),dt$? There is no difference. The integration variable is a "dummy" and is not implicated in the answer.
$endgroup$
– Mark Viola
Nov 30 '18 at 4:29
2
2
$begingroup$
Do you believe $int sin^2(x)dx = int sin^2(u)du$? What role does the variable play in the integral?
$endgroup$
– Ryan
Nov 30 '18 at 4:18
$begingroup$
Do you believe $int sin^2(x)dx = int sin^2(u)du$? What role does the variable play in the integral?
$endgroup$
– Ryan
Nov 30 '18 at 4:18
3
3
$begingroup$
For a definite integral, the name of the variable shouldn't matter.
$endgroup$
– Anurag A
Nov 30 '18 at 4:18
$begingroup$
For a definite integral, the name of the variable shouldn't matter.
$endgroup$
– Anurag A
Nov 30 '18 at 4:18
2
2
$begingroup$
$u$ is a dummy variable: it's like saying $sum_{k=1}^nf(x_k)$ is equal to $sum_{ell=1}^nf(x_ell)$, where here $k,ell$ are dummy indices. by the way, another way: $cos^2(x)=(1+sin(2x)]/2$, and $sin^2(x)=[1-sin 2x]/2$, and the doubly-periodic terms integrate to zero.
$endgroup$
– user254433
Nov 30 '18 at 4:19
$begingroup$
$u$ is a dummy variable: it's like saying $sum_{k=1}^nf(x_k)$ is equal to $sum_{ell=1}^nf(x_ell)$, where here $k,ell$ are dummy indices. by the way, another way: $cos^2(x)=(1+sin(2x)]/2$, and $sin^2(x)=[1-sin 2x]/2$, and the doubly-periodic terms integrate to zero.
$endgroup$
– user254433
Nov 30 '18 at 4:19
3
3
$begingroup$
$$ int f( Calculus ) d Calculus = int f(x) dx = int f(u) du $$
$endgroup$
– Jimmy Sabater
Nov 30 '18 at 4:23
$begingroup$
$$ int f( Calculus ) d Calculus = int f(x) dx = int f(u) du $$
$endgroup$
– Jimmy Sabater
Nov 30 '18 at 4:23
2
2
$begingroup$
@lthompson What is the difference between $int_a^b f(x),dx$ and $int_a^b f(t),dt$? There is no difference. The integration variable is a "dummy" and is not implicated in the answer.
$endgroup$
– Mark Viola
Nov 30 '18 at 4:29
$begingroup$
@lthompson What is the difference between $int_a^b f(x),dx$ and $int_a^b f(t),dt$? There is no difference. The integration variable is a "dummy" and is not implicated in the answer.
$endgroup$
– Mark Viola
Nov 30 '18 at 4:29
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
What is $int_0^1x,dx$? It is $left.frac{1}{2}x^2right|_{x=0}^{x=1}$, which is $frac{1}{2}(1^2)-frac{1}{2}(0^2)$, which is $frac{1}{2}$.
What is $int_0^1u,du$? It is $left.frac{1}{2}u^2right|_{u=0}^{u=1}$, which is $frac{1}{2}(1^2)-frac{1}{2}(0^2)$, which is $frac{1}{2}$.
Does this help you see that $int_0^1x,dx=int_0^1u,du$? And that this generalizes to $int_a^bf(x),dx=int_a^bf(u),du$? In particular, $int_0^{2pi}sin^2(x),dx=int_0^{2pi}sin^2(u),du$
In a definite integral, the variable used is irrelevant. If you actually evaluate the definite integral, you will be replacing that variable with numbers at some point, and no trace of the variable will remain.
answered Nov 30 '18 at 4:30
alex.jordanalex.jordan
39.1k560121
$endgroup$
$begingroup$
That makes sense, thank you. But isn't there some notion of $x$ being fixed for the integral involving cosine, so that $x$ and $u$ are not just arbitrary variables with respect to each other in this particular case?
$endgroup$
– lthompson
Nov 30 '18 at 4:34
1
$begingroup$
$x$ and $u$ are arbitrary variables. And they aren't fixed. Each one is ranging from $0$ up to $2pi$ as the integral sum is assembled.
$endgroup$
– alex.jordan
Nov 30 '18 at 4:41
add a comment |
$begingroup$
As a more general fact (after you can convince yourself that dummy variables like $u$ or $x$ disappear after you do the integration), you can show that for any integers $n<m$,
$$
intlimits_{nfrac{pi}{2}}^{mfrac{pi}{2}} sin^2x; dx = intlimits_{nfrac{pi}{2}}^{mfrac{pi}{2}} cos^2x; dx = frac{1}{2}cdot(text{the length of the interval})
$$
You can convince yourself of the first equality by looking at the graphs and seeing that any of the "quarter-areas" under $|sin x|$ and $|cos x|$ are equal; and the second equality follows by adding the two integrals together and using $sin^2 x + cos^2 x = 1$.
answered Nov 30 '18 at 4:37
user25959user25959
1,573816
$endgroup$
add a comment |
$begingroup$
There are two ways to check the validity of the claim without changing the variable $x$:
1) Consider their difference:
$$int_0^{2pi} cos^2(x) dx - int_0^{2pi} sin^2(x) dx = int_0^{2pi} cos 2xdx=-frac12sin 2x|_0^{2pi}=0.$$
2) Use the half angle formula:
$$begin{align}int_0^{2pi} cos^2(x) dx &= int_0^{2pi} frac{1+cos 2x}{2} dx=\
&=int_0^{2pi} left(frac{1-cos 2x}{2}+cos 2xright) dx=\
&=int_0^{2pi} frac{1-cos 2x}{2}dx+underbrace{int_0^{2pi} cos 2xdx}_{=0}=\
&=int_0^{2pi} sin^2(x) dx. end{align}$$
answered Nov 30 '18 at 11:26
farruhotafarruhota
20.2k2738
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What is $int_0^1x,dx$? It is $left.frac{1}{2}x^2right|_{x=0}^{x=1}$, which is $frac{1}{2}(1^2)-frac{1}{2}(0^2)$, which is $frac{1}{2}$.
What is $int_0^1u,du$? It is $left.frac{1}{2}u^2right|_{u=0}^{u=1}$, which is $frac{1}{2}(1^2)-frac{1}{2}(0^2)$, which is $frac{1}{2}$.
Does this help you see that $int_0^1x,dx=int_0^1u,du$? And that this generalizes to $int_a^bf(x),dx=int_a^bf(u),du$? In particular, $int_0^{2pi}sin^2(x),dx=int_0^{2pi}sin^2(u),du$
In a definite integral, the variable used is irrelevant. If you actually evaluate the definite integral, you will be replacing that variable with numbers at some point, and no trace of the variable will remain.
answered Nov 30 '18 at 4:30
alex.jordanalex.jordan
39.1k560121
$endgroup$
$begingroup$
That makes sense, thank you. But isn't there some notion of $x$ being fixed for the integral involving cosine, so that $x$ and $u$ are not just arbitrary variables with respect to each other in this particular case?
$endgroup$
– lthompson
Nov 30 '18 at 4:34
1
$begingroup$
$x$ and $u$ are arbitrary variables. And they aren't fixed. Each one is ranging from $0$ up to $2pi$ as the integral sum is assembled.
$endgroup$
– alex.jordan
Nov 30 '18 at 4:41
add a comment |
$begingroup$
What is $int_0^1x,dx$? It is $left.frac{1}{2}x^2right|_{x=0}^{x=1}$, which is $frac{1}{2}(1^2)-frac{1}{2}(0^2)$, which is $frac{1}{2}$.
What is $int_0^1u,du$? It is $left.frac{1}{2}u^2right|_{u=0}^{u=1}$, which is $frac{1}{2}(1^2)-frac{1}{2}(0^2)$, which is $frac{1}{2}$.
Does this help you see that $int_0^1x,dx=int_0^1u,du$? And that this generalizes to $int_a^bf(x),dx=int_a^bf(u),du$? In particular, $int_0^{2pi}sin^2(x),dx=int_0^{2pi}sin^2(u),du$
In a definite integral, the variable used is irrelevant. If you actually evaluate the definite integral, you will be replacing that variable with numbers at some point, and no trace of the variable will remain.
answered Nov 30 '18 at 4:30
alex.jordanalex.jordan
39.1k560121
$endgroup$
$begingroup$
That makes sense, thank you. But isn't there some notion of $x$ being fixed for the integral involving cosine, so that $x$ and $u$ are not just arbitrary variables with respect to each other in this particular case?
$endgroup$
– lthompson
Nov 30 '18 at 4:34
1
$begingroup$
$x$ and $u$ are arbitrary variables. And they aren't fixed. Each one is ranging from $0$ up to $2pi$ as the integral sum is assembled.
$endgroup$
– alex.jordan
Nov 30 '18 at 4:41
add a comment |
$begingroup$
What is $int_0^1x,dx$? It is $left.frac{1}{2}x^2right|_{x=0}^{x=1}$, which is $frac{1}{2}(1^2)-frac{1}{2}(0^2)$, which is $frac{1}{2}$.
What is $int_0^1u,du$? It is $left.frac{1}{2}u^2right|_{u=0}^{u=1}$, which is $frac{1}{2}(1^2)-frac{1}{2}(0^2)$, which is $frac{1}{2}$.
Does this help you see that $int_0^1x,dx=int_0^1u,du$? And that this generalizes to $int_a^bf(x),dx=int_a^bf(u),du$? In particular, $int_0^{2pi}sin^2(x),dx=int_0^{2pi}sin^2(u),du$
In a definite integral, the variable used is irrelevant. If you actually evaluate the definite integral, you will be replacing that variable with numbers at some point, and no trace of the variable will remain.
answered Nov 30 '18 at 4:30
alex.jordanalex.jordan
39.1k560121
$endgroup$
What is $int_0^1x,dx$? It is $left.frac{1}{2}x^2right|_{x=0}^{x=1}$, which is $frac{1}{2}(1^2)-frac{1}{2}(0^2)$, which is $frac{1}{2}$.
What is $int_0^1u,du$? It is $left.frac{1}{2}u^2right|_{u=0}^{u=1}$, which is $frac{1}{2}(1^2)-frac{1}{2}(0^2)$, which is $frac{1}{2}$.
Does this help you see that $int_0^1x,dx=int_0^1u,du$? And that this generalizes to $int_a^bf(x),dx=int_a^bf(u),du$? In particular, $int_0^{2pi}sin^2(x),dx=int_0^{2pi}sin^2(u),du$
In a definite integral, the variable used is irrelevant. If you actually evaluate the definite integral, you will be replacing that variable with numbers at some point, and no trace of the variable will remain.
answered Nov 30 '18 at 4:30
alex.jordanalex.jordan
39.1k560121
edited Nov 30 '18 at 4:41
answered Nov 30 '18 at 4:30
alex.jordanalex.jordan
39.1k560121
answered Nov 30 '18 at 4:30
alex.jordanalex.jordan
39.1k560121
answered Nov 30 '18 at 4:30
alex.jordanalex.jordan
39.1k560121
39.1k560121
$begingroup$
That makes sense, thank you. But isn't there some notion of $x$ being fixed for the integral involving cosine, so that $x$ and $u$ are not just arbitrary variables with respect to each other in this particular case?
$endgroup$
– lthompson
Nov 30 '18 at 4:34
1
$begingroup$
$x$ and $u$ are arbitrary variables. And they aren't fixed. Each one is ranging from $0$ up to $2pi$ as the integral sum is assembled.
$endgroup$
– alex.jordan
Nov 30 '18 at 4:41
add a comment |
$begingroup$
That makes sense, thank you. But isn't there some notion of $x$ being fixed for the integral involving cosine, so that $x$ and $u$ are not just arbitrary variables with respect to each other in this particular case?
$endgroup$
– lthompson
Nov 30 '18 at 4:34
1
$begingroup$
$x$ and $u$ are arbitrary variables. And they aren't fixed. Each one is ranging from $0$ up to $2pi$ as the integral sum is assembled.
$endgroup$
– alex.jordan
Nov 30 '18 at 4:41
$begingroup$
That makes sense, thank you. But isn't there some notion of $x$ being fixed for the integral involving cosine, so that $x$ and $u$ are not just arbitrary variables with respect to each other in this particular case?
$endgroup$
– lthompson
Nov 30 '18 at 4:34
$begingroup$
That makes sense, thank you. But isn't there some notion of $x$ being fixed for the integral involving cosine, so that $x$ and $u$ are not just arbitrary variables with respect to each other in this particular case?
$endgroup$
– lthompson
Nov 30 '18 at 4:34
1
1
$begingroup$
$x$ and $u$ are arbitrary variables. And they aren't fixed. Each one is ranging from $0$ up to $2pi$ as the integral sum is assembled.
$endgroup$
– alex.jordan
Nov 30 '18 at 4:41
$begingroup$
$x$ and $u$ are arbitrary variables. And they aren't fixed. Each one is ranging from $0$ up to $2pi$ as the integral sum is assembled.
$endgroup$
– alex.jordan
Nov 30 '18 at 4:41
add a comment |
$begingroup$
As a more general fact (after you can convince yourself that dummy variables like $u$ or $x$ disappear after you do the integration), you can show that for any integers $n<m$,
$$
intlimits_{nfrac{pi}{2}}^{mfrac{pi}{2}} sin^2x; dx = intlimits_{nfrac{pi}{2}}^{mfrac{pi}{2}} cos^2x; dx = frac{1}{2}cdot(text{the length of the interval})
$$
You can convince yourself of the first equality by looking at the graphs and seeing that any of the "quarter-areas" under $|sin x|$ and $|cos x|$ are equal; and the second equality follows by adding the two integrals together and using $sin^2 x + cos^2 x = 1$.
answered Nov 30 '18 at 4:37
user25959user25959
1,573816
$endgroup$
add a comment |
$begingroup$
As a more general fact (after you can convince yourself that dummy variables like $u$ or $x$ disappear after you do the integration), you can show that for any integers $n<m$,
$$
intlimits_{nfrac{pi}{2}}^{mfrac{pi}{2}} sin^2x; dx = intlimits_{nfrac{pi}{2}}^{mfrac{pi}{2}} cos^2x; dx = frac{1}{2}cdot(text{the length of the interval})
$$
You can convince yourself of the first equality by looking at the graphs and seeing that any of the "quarter-areas" under $|sin x|$ and $|cos x|$ are equal; and the second equality follows by adding the two integrals together and using $sin^2 x + cos^2 x = 1$.
answered Nov 30 '18 at 4:37
user25959user25959
1,573816
$endgroup$
add a comment |
$begingroup$
As a more general fact (after you can convince yourself that dummy variables like $u$ or $x$ disappear after you do the integration), you can show that for any integers $n<m$,
$$
intlimits_{nfrac{pi}{2}}^{mfrac{pi}{2}} sin^2x; dx = intlimits_{nfrac{pi}{2}}^{mfrac{pi}{2}} cos^2x; dx = frac{1}{2}cdot(text{the length of the interval})
$$
You can convince yourself of the first equality by looking at the graphs and seeing that any of the "quarter-areas" under $|sin x|$ and $|cos x|$ are equal; and the second equality follows by adding the two integrals together and using $sin^2 x + cos^2 x = 1$.
answered Nov 30 '18 at 4:37
user25959user25959
1,573816
$endgroup$
As a more general fact (after you can convince yourself that dummy variables like $u$ or $x$ disappear after you do the integration), you can show that for any integers $n<m$,
$$
intlimits_{nfrac{pi}{2}}^{mfrac{pi}{2}} sin^2x; dx = intlimits_{nfrac{pi}{2}}^{mfrac{pi}{2}} cos^2x; dx = frac{1}{2}cdot(text{the length of the interval})
$$
You can convince yourself of the first equality by looking at the graphs and seeing that any of the "quarter-areas" under $|sin x|$ and $|cos x|$ are equal; and the second equality follows by adding the two integrals together and using $sin^2 x + cos^2 x = 1$.
answered Nov 30 '18 at 4:37
user25959user25959
1,573816
answered Nov 30 '18 at 4:37
user25959user25959
1,573816
answered Nov 30 '18 at 4:37
user25959user25959
1,573816
answered Nov 30 '18 at 4:37
user25959user25959
1,573816
1,573816
add a comment |
add a comment |
$begingroup$
There are two ways to check the validity of the claim without changing the variable $x$:
1) Consider their difference:
$$int_0^{2pi} cos^2(x) dx - int_0^{2pi} sin^2(x) dx = int_0^{2pi} cos 2xdx=-frac12sin 2x|_0^{2pi}=0.$$
2) Use the half angle formula:
$$begin{align}int_0^{2pi} cos^2(x) dx &= int_0^{2pi} frac{1+cos 2x}{2} dx=\
&=int_0^{2pi} left(frac{1-cos 2x}{2}+cos 2xright) dx=\
&=int_0^{2pi} frac{1-cos 2x}{2}dx+underbrace{int_0^{2pi} cos 2xdx}_{=0}=\
&=int_0^{2pi} sin^2(x) dx. end{align}$$
answered Nov 30 '18 at 11:26
farruhotafarruhota
20.2k2738
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add a comment |
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There are two ways to check the validity of the claim without changing the variable $x$:
1) Consider their difference:
$$int_0^{2pi} cos^2(x) dx - int_0^{2pi} sin^2(x) dx = int_0^{2pi} cos 2xdx=-frac12sin 2x|_0^{2pi}=0.$$
2) Use the half angle formula:
$$begin{align}int_0^{2pi} cos^2(x) dx &= int_0^{2pi} frac{1+cos 2x}{2} dx=\
&=int_0^{2pi} left(frac{1-cos 2x}{2}+cos 2xright) dx=\
&=int_0^{2pi} frac{1-cos 2x}{2}dx+underbrace{int_0^{2pi} cos 2xdx}_{=0}=\
&=int_0^{2pi} sin^2(x) dx. end{align}$$
answered Nov 30 '18 at 11:26
farruhotafarruhota
20.2k2738
$endgroup$
add a comment |
$begingroup$
There are two ways to check the validity of the claim without changing the variable $x$:
1) Consider their difference:
$$int_0^{2pi} cos^2(x) dx - int_0^{2pi} sin^2(x) dx = int_0^{2pi} cos 2xdx=-frac12sin 2x|_0^{2pi}=0.$$
2) Use the half angle formula:
$$begin{align}int_0^{2pi} cos^2(x) dx &= int_0^{2pi} frac{1+cos 2x}{2} dx=\
&=int_0^{2pi} left(frac{1-cos 2x}{2}+cos 2xright) dx=\
&=int_0^{2pi} frac{1-cos 2x}{2}dx+underbrace{int_0^{2pi} cos 2xdx}_{=0}=\
&=int_0^{2pi} sin^2(x) dx. end{align}$$
answered Nov 30 '18 at 11:26
farruhotafarruhota
20.2k2738
$endgroup$
There are two ways to check the validity of the claim without changing the variable $x$:
1) Consider their difference:
$$int_0^{2pi} cos^2(x) dx - int_0^{2pi} sin^2(x) dx = int_0^{2pi} cos 2xdx=-frac12sin 2x|_0^{2pi}=0.$$
2) Use the half angle formula:
$$begin{align}int_0^{2pi} cos^2(x) dx &= int_0^{2pi} frac{1+cos 2x}{2} dx=\
&=int_0^{2pi} left(frac{1-cos 2x}{2}+cos 2xright) dx=\
&=int_0^{2pi} frac{1-cos 2x}{2}dx+underbrace{int_0^{2pi} cos 2xdx}_{=0}=\
&=int_0^{2pi} sin^2(x) dx. end{align}$$
answered Nov 30 '18 at 11:26
farruhotafarruhota
20.2k2738
answered Nov 30 '18 at 11:26
farruhotafarruhota
20.2k2738
answered Nov 30 '18 at 11:26
farruhotafarruhota
20.2k2738
answered Nov 30 '18 at 11:26
farruhotafarruhota
20.2k2738
20.2k2738
add a comment |
add a comment |
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Do you believe $int sin^2(x)dx = int sin^2(u)du$? What role does the variable play in the integral?
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– Ryan
Nov 30 '18 at 4:18
3
$begingroup$
For a definite integral, the name of the variable shouldn't matter.
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– Anurag A
Nov 30 '18 at 4:18
2
$begingroup$
$u$ is a dummy variable: it's like saying $sum_{k=1}^nf(x_k)$ is equal to $sum_{ell=1}^nf(x_ell)$, where here $k,ell$ are dummy indices. by the way, another way: $cos^2(x)=(1+sin(2x)]/2$, and $sin^2(x)=[1-sin 2x]/2$, and the doubly-periodic terms integrate to zero.
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– user254433
Nov 30 '18 at 4:19
3
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$$ int f( Calculus ) d Calculus = int f(x) dx = int f(u) du $$
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– Jimmy Sabater
Nov 30 '18 at 4:23
2
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@lthompson What is the difference between $int_a^b f(x),dx$ and $int_a^b f(t),dt$? There is no difference. The integration variable is a "dummy" and is not implicated in the answer.
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– Mark Viola
Nov 30 '18 at 4:29