Why can we redefine the definition of a variable during substitution? Or let say assume it has two different...
This concept I have asked a few people, but none of them are able to help me understand, so hope that there's a hero can save me from this problem!!!
My question occurs during substitution process, for example, sometimes we let $x = π - u$. Then after some manipulation of numbers, we then let $x = u$ and integrate the guy that we want to integrate. That doesn't seem intuitive to me, isn't that we are changing the definition of x by omiting the rule of arithematic? $x = u implies x = π - x$. Why that operation doesn't affect the integration result?
Here is a concrete example illustrate my question
$$ int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx $$
then we let $x = π - u implies dx = -du$
$$= int^{π}_{0} frac{(π - u)sin(u)}{(1+cos^2(u))} du$$
$$= int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$$
And here (downward) is the part that I don't understand!!! (the x in _x_sin(x) in RHS)
we then let $x = u$
$$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx$$
move the rightmost guy to LHS and integrate RHS, solve the problem.
$$2int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du$$
Why can we let $x = u$? isn't that we have given it the value $pi - u$ in the beginning?
calculus integration definite-integrals substitution
add a comment |
This concept I have asked a few people, but none of them are able to help me understand, so hope that there's a hero can save me from this problem!!!
My question occurs during substitution process, for example, sometimes we let $x = π - u$. Then after some manipulation of numbers, we then let $x = u$ and integrate the guy that we want to integrate. That doesn't seem intuitive to me, isn't that we are changing the definition of x by omiting the rule of arithematic? $x = u implies x = π - x$. Why that operation doesn't affect the integration result?
Here is a concrete example illustrate my question
$$ int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx $$
then we let $x = π - u implies dx = -du$
$$= int^{π}_{0} frac{(π - u)sin(u)}{(1+cos^2(u))} du$$
$$= int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$$
And here (downward) is the part that I don't understand!!! (the x in _x_sin(x) in RHS)
we then let $x = u$
$$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx$$
move the rightmost guy to LHS and integrate RHS, solve the problem.
$$2int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du$$
Why can we let $x = u$? isn't that we have given it the value $pi - u$ in the beginning?
calculus integration definite-integrals substitution
1
I was told that this question is about dummy variable?
– Stephen Fong
Jan 4 at 3:13
Hi & welcome to MSE. Note the original $x$ is just a "dummy" variable, as your comment says, that you can replace with $y$ or anything else you like. It only has meaning within the integral, including with the limits. When you do a substitution, you are replacing it with another "dummy" variable. Later, you can reuse the same variable for another substitution, as your example indicates. However, I don't think that's a wise idea as it can cause confusion, like it did with you, so I normally, at least, would not do that.
– John Omielan
Jan 4 at 3:15
add a comment |
This concept I have asked a few people, but none of them are able to help me understand, so hope that there's a hero can save me from this problem!!!
My question occurs during substitution process, for example, sometimes we let $x = π - u$. Then after some manipulation of numbers, we then let $x = u$ and integrate the guy that we want to integrate. That doesn't seem intuitive to me, isn't that we are changing the definition of x by omiting the rule of arithematic? $x = u implies x = π - x$. Why that operation doesn't affect the integration result?
Here is a concrete example illustrate my question
$$ int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx $$
then we let $x = π - u implies dx = -du$
$$= int^{π}_{0} frac{(π - u)sin(u)}{(1+cos^2(u))} du$$
$$= int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$$
And here (downward) is the part that I don't understand!!! (the x in _x_sin(x) in RHS)
we then let $x = u$
$$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx$$
move the rightmost guy to LHS and integrate RHS, solve the problem.
$$2int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du$$
Why can we let $x = u$? isn't that we have given it the value $pi - u$ in the beginning?
calculus integration definite-integrals substitution
This concept I have asked a few people, but none of them are able to help me understand, so hope that there's a hero can save me from this problem!!!
My question occurs during substitution process, for example, sometimes we let $x = π - u$. Then after some manipulation of numbers, we then let $x = u$ and integrate the guy that we want to integrate. That doesn't seem intuitive to me, isn't that we are changing the definition of x by omiting the rule of arithematic? $x = u implies x = π - x$. Why that operation doesn't affect the integration result?
Here is a concrete example illustrate my question
$$ int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx $$
then we let $x = π - u implies dx = -du$
$$= int^{π}_{0} frac{(π - u)sin(u)}{(1+cos^2(u))} du$$
$$= int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$$
And here (downward) is the part that I don't understand!!! (the x in _x_sin(x) in RHS)
we then let $x = u$
$$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx$$
move the rightmost guy to LHS and integrate RHS, solve the problem.
$$2int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du$$
Why can we let $x = u$? isn't that we have given it the value $pi - u$ in the beginning?
calculus integration definite-integrals substitution
calculus integration definite-integrals substitution
edited Jan 4 at 4:26
Mint
5301417
5301417
asked Jan 4 at 3:05
Stephen FongStephen Fong
64
64
1
I was told that this question is about dummy variable?
– Stephen Fong
Jan 4 at 3:13
Hi & welcome to MSE. Note the original $x$ is just a "dummy" variable, as your comment says, that you can replace with $y$ or anything else you like. It only has meaning within the integral, including with the limits. When you do a substitution, you are replacing it with another "dummy" variable. Later, you can reuse the same variable for another substitution, as your example indicates. However, I don't think that's a wise idea as it can cause confusion, like it did with you, so I normally, at least, would not do that.
– John Omielan
Jan 4 at 3:15
add a comment |
1
I was told that this question is about dummy variable?
– Stephen Fong
Jan 4 at 3:13
Hi & welcome to MSE. Note the original $x$ is just a "dummy" variable, as your comment says, that you can replace with $y$ or anything else you like. It only has meaning within the integral, including with the limits. When you do a substitution, you are replacing it with another "dummy" variable. Later, you can reuse the same variable for another substitution, as your example indicates. However, I don't think that's a wise idea as it can cause confusion, like it did with you, so I normally, at least, would not do that.
– John Omielan
Jan 4 at 3:15
1
1
I was told that this question is about dummy variable?
– Stephen Fong
Jan 4 at 3:13
I was told that this question is about dummy variable?
– Stephen Fong
Jan 4 at 3:13
Hi & welcome to MSE. Note the original $x$ is just a "dummy" variable, as your comment says, that you can replace with $y$ or anything else you like. It only has meaning within the integral, including with the limits. When you do a substitution, you are replacing it with another "dummy" variable. Later, you can reuse the same variable for another substitution, as your example indicates. However, I don't think that's a wise idea as it can cause confusion, like it did with you, so I normally, at least, would not do that.
– John Omielan
Jan 4 at 3:15
Hi & welcome to MSE. Note the original $x$ is just a "dummy" variable, as your comment says, that you can replace with $y$ or anything else you like. It only has meaning within the integral, including with the limits. When you do a substitution, you are replacing it with another "dummy" variable. Later, you can reuse the same variable for another substitution, as your example indicates. However, I don't think that's a wise idea as it can cause confusion, like it did with you, so I normally, at least, would not do that.
– John Omielan
Jan 4 at 3:15
add a comment |
2 Answers
2
active
oldest
votes
This is called, not in the most humble manner, the "abuse" of notations. Let us go step by step in understanding what happened.
First, for the integral $intlimits_{0}^{pi} dfrac{x sin x}{1 + cos^2 x} dx$, we put $x = pi - u$, where since $x$ varies from $0$ to $pi$, $u$ varies from $pi$ to $0$. And from the exact differential term $dx = -du$, we can reverse the limits of integration so that it becomes $intlimits_{0}^{pi} dfrac{left( pi - u right) sin u}{1 + cos^2 u} du$ (Here, some trigonometric formulae are also used). After this we split the numerator to obtain the two integrals.
Well, this was pretty easy! Now comes the step where problem arises. One of the two integrals is $intlimits_{0}^{pi} dfrac{u sin u}{1 + cos^2 u} du$. Now, let us look what we got in this integral. First, the limits are from $0$ to $pi$, just as the original integral. Next, the integrand is the same as the original integral except that we now have the variable $u$ instead of $x$.
But, when it comes to integration, it does not really matter what the variable's "name" is. What matters is that integrand and the limits of integrations. So, the integral will be the same if we call the variable as $u$ or $x$, or for that matter any other name!
Hence, we get $intlimits_{0}^{pi} dfrac{u sin u}{1 + cos^2 u} du = intlimits_{0}^{pi} dfrac{x sin x}{1 + cos^2 x} dx$.
In other words, we DO NOT "let $x = u$" in that step. Rather, we use facts from calculus to conclude the above mentioned equality. Since these two integrals are the same, it does not matter what variable we use and we rather replace $u$ by $x$ in the second integral to get the answer.
Not so familiar with the terminology ~ what does it mean by "u varies from π to x"?
– Stephen Fong
Jan 4 at 3:28
I changed it. It varies from $pi$ to $0$ as $x$ goes from $0$ to $pi$.
– Aniruddha Deshmukh
Jan 4 at 3:29
@John Omielan I find combining you twos opinions make the explanation more clear. Can I think of it in this way? Whenever I do substitution, it's like I draw the same graph (with same size) on the another axis and try to integrate about the new axis (and so don't care the former definition)? So after the substitution, it's like I'm integrating with respect to a new axis. And here comes Aniruddha's idea, since the two graphs have same size, integrating them gives the same result. So for the about example, I can just replace x by u assuming they are the same.
– Stephen Fong
Jan 4 at 3:53
Let me end up with another example to see if my understanding is fine, suppose we have int u+1 du = int x dx + int x + 1 dx, then we can say 2int u+1 du = int x dx
– Stephen Fong
Jan 4 at 3:54
add a comment |
We have:
$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$
Rather than say "let $u = x$" I think it would be cleaner to say
$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx =
int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$
And then bring it over to the other side
or
$I = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - I\
2I = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du\
I = frac 12 int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du$
add a comment |
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2 Answers
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2 Answers
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This is called, not in the most humble manner, the "abuse" of notations. Let us go step by step in understanding what happened.
First, for the integral $intlimits_{0}^{pi} dfrac{x sin x}{1 + cos^2 x} dx$, we put $x = pi - u$, where since $x$ varies from $0$ to $pi$, $u$ varies from $pi$ to $0$. And from the exact differential term $dx = -du$, we can reverse the limits of integration so that it becomes $intlimits_{0}^{pi} dfrac{left( pi - u right) sin u}{1 + cos^2 u} du$ (Here, some trigonometric formulae are also used). After this we split the numerator to obtain the two integrals.
Well, this was pretty easy! Now comes the step where problem arises. One of the two integrals is $intlimits_{0}^{pi} dfrac{u sin u}{1 + cos^2 u} du$. Now, let us look what we got in this integral. First, the limits are from $0$ to $pi$, just as the original integral. Next, the integrand is the same as the original integral except that we now have the variable $u$ instead of $x$.
But, when it comes to integration, it does not really matter what the variable's "name" is. What matters is that integrand and the limits of integrations. So, the integral will be the same if we call the variable as $u$ or $x$, or for that matter any other name!
Hence, we get $intlimits_{0}^{pi} dfrac{u sin u}{1 + cos^2 u} du = intlimits_{0}^{pi} dfrac{x sin x}{1 + cos^2 x} dx$.
In other words, we DO NOT "let $x = u$" in that step. Rather, we use facts from calculus to conclude the above mentioned equality. Since these two integrals are the same, it does not matter what variable we use and we rather replace $u$ by $x$ in the second integral to get the answer.
Not so familiar with the terminology ~ what does it mean by "u varies from π to x"?
– Stephen Fong
Jan 4 at 3:28
I changed it. It varies from $pi$ to $0$ as $x$ goes from $0$ to $pi$.
– Aniruddha Deshmukh
Jan 4 at 3:29
@John Omielan I find combining you twos opinions make the explanation more clear. Can I think of it in this way? Whenever I do substitution, it's like I draw the same graph (with same size) on the another axis and try to integrate about the new axis (and so don't care the former definition)? So after the substitution, it's like I'm integrating with respect to a new axis. And here comes Aniruddha's idea, since the two graphs have same size, integrating them gives the same result. So for the about example, I can just replace x by u assuming they are the same.
– Stephen Fong
Jan 4 at 3:53
Let me end up with another example to see if my understanding is fine, suppose we have int u+1 du = int x dx + int x + 1 dx, then we can say 2int u+1 du = int x dx
– Stephen Fong
Jan 4 at 3:54
add a comment |
This is called, not in the most humble manner, the "abuse" of notations. Let us go step by step in understanding what happened.
First, for the integral $intlimits_{0}^{pi} dfrac{x sin x}{1 + cos^2 x} dx$, we put $x = pi - u$, where since $x$ varies from $0$ to $pi$, $u$ varies from $pi$ to $0$. And from the exact differential term $dx = -du$, we can reverse the limits of integration so that it becomes $intlimits_{0}^{pi} dfrac{left( pi - u right) sin u}{1 + cos^2 u} du$ (Here, some trigonometric formulae are also used). After this we split the numerator to obtain the two integrals.
Well, this was pretty easy! Now comes the step where problem arises. One of the two integrals is $intlimits_{0}^{pi} dfrac{u sin u}{1 + cos^2 u} du$. Now, let us look what we got in this integral. First, the limits are from $0$ to $pi$, just as the original integral. Next, the integrand is the same as the original integral except that we now have the variable $u$ instead of $x$.
But, when it comes to integration, it does not really matter what the variable's "name" is. What matters is that integrand and the limits of integrations. So, the integral will be the same if we call the variable as $u$ or $x$, or for that matter any other name!
Hence, we get $intlimits_{0}^{pi} dfrac{u sin u}{1 + cos^2 u} du = intlimits_{0}^{pi} dfrac{x sin x}{1 + cos^2 x} dx$.
In other words, we DO NOT "let $x = u$" in that step. Rather, we use facts from calculus to conclude the above mentioned equality. Since these two integrals are the same, it does not matter what variable we use and we rather replace $u$ by $x$ in the second integral to get the answer.
Not so familiar with the terminology ~ what does it mean by "u varies from π to x"?
– Stephen Fong
Jan 4 at 3:28
I changed it. It varies from $pi$ to $0$ as $x$ goes from $0$ to $pi$.
– Aniruddha Deshmukh
Jan 4 at 3:29
@John Omielan I find combining you twos opinions make the explanation more clear. Can I think of it in this way? Whenever I do substitution, it's like I draw the same graph (with same size) on the another axis and try to integrate about the new axis (and so don't care the former definition)? So after the substitution, it's like I'm integrating with respect to a new axis. And here comes Aniruddha's idea, since the two graphs have same size, integrating them gives the same result. So for the about example, I can just replace x by u assuming they are the same.
– Stephen Fong
Jan 4 at 3:53
Let me end up with another example to see if my understanding is fine, suppose we have int u+1 du = int x dx + int x + 1 dx, then we can say 2int u+1 du = int x dx
– Stephen Fong
Jan 4 at 3:54
add a comment |
This is called, not in the most humble manner, the "abuse" of notations. Let us go step by step in understanding what happened.
First, for the integral $intlimits_{0}^{pi} dfrac{x sin x}{1 + cos^2 x} dx$, we put $x = pi - u$, where since $x$ varies from $0$ to $pi$, $u$ varies from $pi$ to $0$. And from the exact differential term $dx = -du$, we can reverse the limits of integration so that it becomes $intlimits_{0}^{pi} dfrac{left( pi - u right) sin u}{1 + cos^2 u} du$ (Here, some trigonometric formulae are also used). After this we split the numerator to obtain the two integrals.
Well, this was pretty easy! Now comes the step where problem arises. One of the two integrals is $intlimits_{0}^{pi} dfrac{u sin u}{1 + cos^2 u} du$. Now, let us look what we got in this integral. First, the limits are from $0$ to $pi$, just as the original integral. Next, the integrand is the same as the original integral except that we now have the variable $u$ instead of $x$.
But, when it comes to integration, it does not really matter what the variable's "name" is. What matters is that integrand and the limits of integrations. So, the integral will be the same if we call the variable as $u$ or $x$, or for that matter any other name!
Hence, we get $intlimits_{0}^{pi} dfrac{u sin u}{1 + cos^2 u} du = intlimits_{0}^{pi} dfrac{x sin x}{1 + cos^2 x} dx$.
In other words, we DO NOT "let $x = u$" in that step. Rather, we use facts from calculus to conclude the above mentioned equality. Since these two integrals are the same, it does not matter what variable we use and we rather replace $u$ by $x$ in the second integral to get the answer.
This is called, not in the most humble manner, the "abuse" of notations. Let us go step by step in understanding what happened.
First, for the integral $intlimits_{0}^{pi} dfrac{x sin x}{1 + cos^2 x} dx$, we put $x = pi - u$, where since $x$ varies from $0$ to $pi$, $u$ varies from $pi$ to $0$. And from the exact differential term $dx = -du$, we can reverse the limits of integration so that it becomes $intlimits_{0}^{pi} dfrac{left( pi - u right) sin u}{1 + cos^2 u} du$ (Here, some trigonometric formulae are also used). After this we split the numerator to obtain the two integrals.
Well, this was pretty easy! Now comes the step where problem arises. One of the two integrals is $intlimits_{0}^{pi} dfrac{u sin u}{1 + cos^2 u} du$. Now, let us look what we got in this integral. First, the limits are from $0$ to $pi$, just as the original integral. Next, the integrand is the same as the original integral except that we now have the variable $u$ instead of $x$.
But, when it comes to integration, it does not really matter what the variable's "name" is. What matters is that integrand and the limits of integrations. So, the integral will be the same if we call the variable as $u$ or $x$, or for that matter any other name!
Hence, we get $intlimits_{0}^{pi} dfrac{u sin u}{1 + cos^2 u} du = intlimits_{0}^{pi} dfrac{x sin x}{1 + cos^2 x} dx$.
In other words, we DO NOT "let $x = u$" in that step. Rather, we use facts from calculus to conclude the above mentioned equality. Since these two integrals are the same, it does not matter what variable we use and we rather replace $u$ by $x$ in the second integral to get the answer.
edited Jan 4 at 3:29
answered Jan 4 at 3:18
Aniruddha DeshmukhAniruddha Deshmukh
934418
934418
Not so familiar with the terminology ~ what does it mean by "u varies from π to x"?
– Stephen Fong
Jan 4 at 3:28
I changed it. It varies from $pi$ to $0$ as $x$ goes from $0$ to $pi$.
– Aniruddha Deshmukh
Jan 4 at 3:29
@John Omielan I find combining you twos opinions make the explanation more clear. Can I think of it in this way? Whenever I do substitution, it's like I draw the same graph (with same size) on the another axis and try to integrate about the new axis (and so don't care the former definition)? So after the substitution, it's like I'm integrating with respect to a new axis. And here comes Aniruddha's idea, since the two graphs have same size, integrating them gives the same result. So for the about example, I can just replace x by u assuming they are the same.
– Stephen Fong
Jan 4 at 3:53
Let me end up with another example to see if my understanding is fine, suppose we have int u+1 du = int x dx + int x + 1 dx, then we can say 2int u+1 du = int x dx
– Stephen Fong
Jan 4 at 3:54
add a comment |
Not so familiar with the terminology ~ what does it mean by "u varies from π to x"?
– Stephen Fong
Jan 4 at 3:28
I changed it. It varies from $pi$ to $0$ as $x$ goes from $0$ to $pi$.
– Aniruddha Deshmukh
Jan 4 at 3:29
@John Omielan I find combining you twos opinions make the explanation more clear. Can I think of it in this way? Whenever I do substitution, it's like I draw the same graph (with same size) on the another axis and try to integrate about the new axis (and so don't care the former definition)? So after the substitution, it's like I'm integrating with respect to a new axis. And here comes Aniruddha's idea, since the two graphs have same size, integrating them gives the same result. So for the about example, I can just replace x by u assuming they are the same.
– Stephen Fong
Jan 4 at 3:53
Let me end up with another example to see if my understanding is fine, suppose we have int u+1 du = int x dx + int x + 1 dx, then we can say 2int u+1 du = int x dx
– Stephen Fong
Jan 4 at 3:54
Not so familiar with the terminology ~ what does it mean by "u varies from π to x"?
– Stephen Fong
Jan 4 at 3:28
Not so familiar with the terminology ~ what does it mean by "u varies from π to x"?
– Stephen Fong
Jan 4 at 3:28
I changed it. It varies from $pi$ to $0$ as $x$ goes from $0$ to $pi$.
– Aniruddha Deshmukh
Jan 4 at 3:29
I changed it. It varies from $pi$ to $0$ as $x$ goes from $0$ to $pi$.
– Aniruddha Deshmukh
Jan 4 at 3:29
@John Omielan I find combining you twos opinions make the explanation more clear. Can I think of it in this way? Whenever I do substitution, it's like I draw the same graph (with same size) on the another axis and try to integrate about the new axis (and so don't care the former definition)? So after the substitution, it's like I'm integrating with respect to a new axis. And here comes Aniruddha's idea, since the two graphs have same size, integrating them gives the same result. So for the about example, I can just replace x by u assuming they are the same.
– Stephen Fong
Jan 4 at 3:53
@John Omielan I find combining you twos opinions make the explanation more clear. Can I think of it in this way? Whenever I do substitution, it's like I draw the same graph (with same size) on the another axis and try to integrate about the new axis (and so don't care the former definition)? So after the substitution, it's like I'm integrating with respect to a new axis. And here comes Aniruddha's idea, since the two graphs have same size, integrating them gives the same result. So for the about example, I can just replace x by u assuming they are the same.
– Stephen Fong
Jan 4 at 3:53
Let me end up with another example to see if my understanding is fine, suppose we have int u+1 du = int x dx + int x + 1 dx, then we can say 2int u+1 du = int x dx
– Stephen Fong
Jan 4 at 3:54
Let me end up with another example to see if my understanding is fine, suppose we have int u+1 du = int x dx + int x + 1 dx, then we can say 2int u+1 du = int x dx
– Stephen Fong
Jan 4 at 3:54
add a comment |
We have:
$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$
Rather than say "let $u = x$" I think it would be cleaner to say
$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx =
int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$
And then bring it over to the other side
or
$I = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - I\
2I = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du\
I = frac 12 int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du$
add a comment |
We have:
$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$
Rather than say "let $u = x$" I think it would be cleaner to say
$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx =
int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$
And then bring it over to the other side
or
$I = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - I\
2I = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du\
I = frac 12 int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du$
add a comment |
We have:
$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$
Rather than say "let $u = x$" I think it would be cleaner to say
$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx =
int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$
And then bring it over to the other side
or
$I = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - I\
2I = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du\
I = frac 12 int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du$
We have:
$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$
Rather than say "let $u = x$" I think it would be cleaner to say
$int^{π}_{0} frac{xsin(x)}{(1+cos^2(x))} dx =
int^{π}_{0} frac{usin(u)}{(1+cos^2(u))} du$
And then bring it over to the other side
or
$I = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du - I\
2I = int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du\
I = frac 12 int^{π}_{0} frac{πsin(u)}{(1+cos^2(u))} du$
answered Jan 4 at 4:41
Doug MDoug M
44.2k31854
44.2k31854
add a comment |
add a comment |
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I was told that this question is about dummy variable?
– Stephen Fong
Jan 4 at 3:13
Hi & welcome to MSE. Note the original $x$ is just a "dummy" variable, as your comment says, that you can replace with $y$ or anything else you like. It only has meaning within the integral, including with the limits. When you do a substitution, you are replacing it with another "dummy" variable. Later, you can reuse the same variable for another substitution, as your example indicates. However, I don't think that's a wise idea as it can cause confusion, like it did with you, so I normally, at least, would not do that.
– John Omielan
Jan 4 at 3:15