Definite Integration.

**varunnayudu** · December 23rd 2008, 01:39 AM

Evaluate

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

**Moo** · December 23rd 2008, 02:07 AM

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.

**zorro** · December 30th 2008, 09:02 PM

hw did u get that

**mr fantastic** · December 30th 2008, 11:08 PM

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?

**zorro** · January 1st 2009, 01:15 AM

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.

--------------------------------------------------------------------

$cos(t) = \varphi$
therefore
$- sin t dt= d\varphi$

$<br /> - \frac{\pi}{2} \int_0^\pi \frac{1}{1+\varphi^2}d\varphi<br />$

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

**mr fantastic** · January 1st 2009, 01:49 AM

Originally Posted by zorro

--------------------------------------------------------------------

$cos(t) = \varphi$
therefore
$- sin t dt= d\varphi$

$<br /> - \frac{\pi}{2} \int_0^\pi \frac{1}{1+\varphi^2}d\varphi<br />$

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).

**zorro** · January 1st 2009, 02:08 AM

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).

--------------------------------

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

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

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

= $0$

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

**Chop Suey** · January 1st 2009, 02:10 AM

You forgot to change the limits.

**tom@ballooncalculus** · January 1st 2009, 05:00 AM

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]

**zorro** · January 4th 2009, 06:39 PM

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]

-------------------------------------------------------------------

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

**mr fantastic** · January 4th 2009, 06:56 PM

Originally Posted by zorro

-------------------------------------------------------------------

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

Recall that there was a change of variable: $\varphi = \cos t$ .

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

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

**qpmathelp** · January 4th 2009, 07:10 PM

Originally Posted by zorro

-------------------------------------------------------------------

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

they are the limits for the new variable

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Definite Integration.

hw did u do that

Please tell me if its correct.

hw did t conert to pi

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