1. ## Definite Integration.

Evaluate

$I = \int_{0}^{\pi} \frac{x sin x}{1+cos^2 x} dx$

2. Hello,
Originally Posted by varunnayudu
Evaluate

$I = \int_{0}^{\pi} \frac{x sin x}{1+cos^2 x} dx$
Let $t=\pi-x$ :

$I=-\int_\pi^0 \frac{(\pi-t) \sin(\pi-t)}{1+\cos^2(\pi-t)} ~ dt=\int_0^\pi \frac{(\pi-t) \sin(\pi-t)}{1+\cos^2(\pi-t)} ~ dt$

But $\sin(\pi-t)=\sin(t)$ and $\cos(\pi-t)=-\cos(t)$
Hence :

$I=\int_0^\pi \frac{(\pi-t) \sin(t)}{1+\cos^2(t)}~dt$

$I=\pi \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~dt-\underbrace{\int_0^\pi \frac{t \sin(t)}{1+\cos^2(t)} ~ dt}_{=I}$

$2I=\pi \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~ dt$

$I=\frac{\pi}{2} \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~ dt$

Now substitute $\varphi=\cos(t)$ and it'll be easy.

3. ## hw did u do that

hw did u get that

4. Originally Posted by zorro
hw did u get that
You will need to be more specific. What part of Moo's reply don't you understand?

5. ## Please tell me if its correct.

Originally Posted by Moo
Hello,

Let $t=\pi-x$ :

$I=-\int_\pi^0 \frac{(\pi-t) \sin(\pi-t)}{1+\cos^2(\pi-t)} ~ dt=\int_0^\pi \frac{(\pi-t) \sin(\pi-t)}{1+\cos^2(\pi-t)} ~ dt$

But $\sin(\pi-t)=\sin(t)$ and $\cos(\pi-t)=-\cos(t)$
Hence :

$I=\int_0^\pi \frac{(\pi-t) \sin(t)}{1+\cos^2(t)}~dt$

$I=\pi \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~dt-\underbrace{\int_0^\pi \frac{t \sin(t)}{1+\cos^2(t)} ~ dt}_{=I}$

$2I=\pi \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~ dt$

$I=\frac{\pi}{2} \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~ dt$

Now substitute $\varphi=\cos(t)$ and it'll be easy.
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$cos(t) = \varphi$
therefore
$- sin t dt= d\varphi$

$
- \frac{\pi}{2} \int_0^\pi \frac{1}{1+\varphi^2}d\varphi
$

what to do next ................

6. Originally Posted by zorro
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$cos(t) = \varphi$
therefore
$- sin t dt= d\varphi$

$
- \frac{\pi}{2} \int_0^\pi \frac{1}{1+\varphi^2}d\varphi
$

what to do next ................
$\int \frac{1}{1+\varphi^2} \, d\varphi$ is a standard form and is equal to $\tan^{-1} \varphi$ (I have omitted the arbitrary cnostant of integration).

7. Originally Posted by mr fantastic
$\int \frac{1}{1+\varphi^2} \, d\varphi$ is a standard form and is equal to $\tan^{-1} \varphi$ (I have omitted the arbitrary cnostant of integration).

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= - $\frac{\pi}{2} (tan^{-1} \pi - tan^{-1}0)$

= - $\frac{\pi}{2} (0-0)$

= - $\frac{\pi}{2} .0$

= $0$

is this right .........

8. You forgot to change the limits.

9. The straight dashed line diff's / anti-diff's with respect to the dashed balloon expression.

Don't integrate - balloontegrate! http://www.ballooncalculus.org/examples]

10. ## hw did t conert to pi

Originally Posted by tom@ballooncalculus

The straight dashed line diff's / anti-diff's with respect to the dashed balloon expression.

Don't integrate - balloontegrate! http://www.ballooncalculus.org/examples]

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Here hw did t= $\pi$ turn into 1 and -1.

11. Originally Posted by zorro
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Here hw did t= $\pi$ turn into 1 and -1.
Recall that there was a change of variable: $\varphi = \cos t$.

$t = 0 \Rightarrow \varphi = \cos 0 = 1$.

$t = \pi \Rightarrow \varphi = \cos \pi = -1$.

12. Originally Posted by zorro
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Here hw did t= $\pi$ turn into 1 and -1.
they are the limits for the new variable