Question

how to solve the following integral.

Answer 1

\[\Large \int\limits_{0}^{\pi}\frac{xsin(x)}{1+\cos^{2}(x)}\]

Answer 2

such hard. wow. much complicate.

Answer 3

First we need to find the antiderivative of \[\frac{ x \times \sin(x) }{ 1+\cos ^{2}(x) }\]

Answer 4

Why not brainstorm regarding possible methods of integrating this?

I see right away that if we let u = cos x, the derivative of that is (du/dx) = sin x, and that we have sin x dx in the numerator. What does that tell you?

Answer 5

We do not because of the following
http://www.wolframalpha.com/input/?i=integrate+%28x+sin%28x%29%29%2F%281+%2B+cos%28x%29^2%29

Answer 6

Excuse me. loser66 has kindly pointed out to me that the derivative of cos x is -sin x, not sin x (as I previously typed).

Answer 7

did yoy try using the property of definite integral states that.
\[\Large \int\limits_{0}^{a} f(x)dx=\int\limits_{0}^{a}f(a-x)dx\]

Answer 8

@eliassaab : Interesting wolframalpha.com result. Are you saying, in effect, that integration by parts will not succeed here?

Answer 9

@sami-21: would you mind telling us more about this so-called property? what's it's name? When would one apply it?

Answer 10

If you believe that it can be done, show it to me and inform wolframaplpha about it.

Answer 11

lets give it a try.
Let
\[\Large I=\int\limits_{0}^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\cos^{2}(\pi-x)}dx\]
since
sin(pi-x+sin(x)
and
cos^2(pi-x)=cos^2(x) so original integral repeates here we can write as
\[\Large 2I=\int\limits_{0}^{\pi}\frac{(\pi)\sin(x)}{1+\cos^{2}(x)}dx\]

now simply substitue t=cos(x) then divide the result by 2.

Answer 12

typo
sin(pi-x)=sin(x)

Answer 13

@mathmale you can find this in any standard calculus book. here its given on no 11.
http://www.pinkmonkey.com/studyguides/subjects/calc/chap7/c0707501.asp

Answer 14

@sami-21: That's new to me. Thank you for sharing this perspective. Would you mind providing the name by which this property of integrals is called, and what it was about the problem posted here that led you to consider applying this property?

Answer 15

\[\ t=\cos(x)\]
\[dt=\sin(x)dx\]
\[\pi \int\limits_{0}^{\pi} \frac{1}{1+t^2}dt\]
\[2I=(\pi)\tan^{-1} {\pi}\]

Answer 16

somewhere long ago have solved similiar kind of problem in my problem sheet.

Answer 17

@mathmale the limits were corresponding to the property tried and it worked.

Answer 18

The answer is \(\Large \frac{\pi^2 }4 =2.4674\)
See
http://www.wolframalpha.com/input/?i=integral+%28x+sin%28x%29%29%2F%281+%2B+cos%28x%29^2%29+from+x%3D0+to+x%3Dpi

Answer 19

yes i am also getting the same answer as eliassaab.

@saadi you need to chage the limits of integrations as well when trying the substitution. it should be from 1 to -1 bcz
when x=0 then t=cos(0)=1
when x=pi then t=cos(pi)=-1

Answer 20

thank you all for your help.

Answer 21

Here is a complete solution
\[
I=\int_0^{\pi } \frac{x \sin (x)}{\cos ^2(x)+1} \, dx=\int_0^{\pi } \frac{(\pi -x) \sin (\pi -x)}{\cos ^2(\pi -x)+1} \, dx=\\
\int_0^{\pi } \frac{(\pi -x) \sin (x)}{\cos ^2(x)+1} \,
dx=\pi \int_0^{\pi } \frac{\sin (x)}{\cos ^2(x)+1} \, dx-\int_0^{\pi } \frac{x \sin (x)}{\cos ^2(x)+1} \, dx=\\
\frac{\pi \pi }{2}-\text{I}=\frac{\pi ^2}{2}-\text{I}\\
2 I = \frac{\pi^2}2\\
I = \frac{\pi^2}4\\


\]

Answer 22

The only integral we have to compute is
\[
\int_0^{\pi } \frac{\sin (x)}{\cos ^2(x)+1} \, dx=\frac{\pi }{2}
\]

Answer 23

http://www.wolframalpha.com/input/?i=integral+%28sin%28x%29%29%2F%281+%2B+cos%28x%29^2%29+from+x%3D0+to+x%3Dpi

Answer 24

Wow you guys explained that in good detail.