Integration Problem

**varunnayudu** · December 19th 2008, 04:05 AM

Evaluate
$\int\limits_{0}^{\pi} \frac{xsinx}{1+sinx}dx.$

**PaulRS** · December 19th 2008, 04:13 AM

Set $u=\pi-x$ then $ I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( {\pi - x} \right)}} {{1 + \sin \left( {\pi - x} \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} $

So: $ 2 \cdot I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} + \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} = \int_0^\pi {\left( {\tfrac{{x \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}} + \tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}} \right)dx} $

We get $ 2 \cdot I = \pi \cdot \int_0^\pi {\left( {\tfrac{{\sin \left( x \right)}} {{1 + \sin \left( x \right)}}} \right)dx} $ and this can be solved by standard methods

**varunnayudu** · December 28th 2008, 03:24 AM

I am unable to understand hw u carried out the steps.................

**mr fantastic** · December 28th 2008, 03:28 AM

Originally Posted by varunnayudu

I am unable to understand hw u carried out the steps.................

Read the reply again - carefully. Now try and be more specific - what part or parts don't you understand?

**ADARSH** · December 28th 2008, 03:46 AM

And kindly show your try

**Prove It** · December 28th 2008, 03:54 AM

I'm also having trouble following.

Where did u come into it? i.e. how did x become $\pi - x$ ?

**mr fantastic** · December 28th 2008, 04:03 AM

Originally Posted by Prove It

I'm also having trouble following.

Where did u come into it? i.e. how did x become $\pi - x$ ?

The substitution $u = \pi - x$ is made and then the dummy variable is switched back to x:

$I = \int_0^\pi {\frac{{x \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}} \, dx}$

Substitute $u = \pi - x$ :

$= \int_\pi^0 {\frac{{\left( {\pi - u} \right) \cdot \sin \left( {\pi - u} \right)}}{{1 + \sin \left( {\pi - u} \right)}} \, (-du) }$

$= \int_0^\pi {\frac{{\left( {\pi - u} \right) \cdot \sin \left( {\pi - u} \right)}}{{1 + \sin \left( {\pi - u} \right)}} \, du }$

Now switch the dummy variable back to x:

$= \int_0^\pi {\frac{{\left( {\pi - x} \right) \cdot \sin \left( {\pi - x} \right)}} {{1 + \sin \left( {\pi - x} \right)}} \, dx}$

etc.

This is a standard technique.

**varunnayudu** · December 28th 2008, 04:36 AM

Now switch the dummy variable back to x:

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

Hw can one substitute $(\pi - u)$ into $(\pi - x)$

and secondly even after doing the sub the sum is not evaluated it is still the same ...........only instead of $x$ u have $(\pi - x)$

$ I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( {\pi - x} \right)}} {{1 + \sin \left( {\pi - x} \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} $

$ 2 \cdot I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} + \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} $

tell me hw u got this conversion ..............

**ADARSH** · December 28th 2008, 08:14 AM

Hi Varun,
$ I=\int^{a}_{b}{f(x)}$

$I=\int^{a}_{b}{f(a+b-x)} $

I hope you know the above property

Here $ f(x)=\frac{xsinx}{1-sin(x)} $
Hence
the two equations are
added

$I+I= 2I$
which is given

----------------------------------------------------------------
to prove that property
Put t=a+b-x
dt=-dx
When x=a ,t=b
and when x=b,t=a
Now
$ \int^{a}_{b}{f(x)dx}=-\int^{b}_{a}{f(a+b-t)dt} $ $ =\int^{a}_{b}{f(a+b-t)dt} $
As we know
$\int^{x}_{y}{g(\alpha)d{\alpha }} =\int^{x}_{y}{g(\beta)d{\beta }} $
So $ =\int^{a}_{b}{f(a+b-x)dx}=\int^{a}_{b}{f(x)dx} $
And that's the end of that show..

**Danny** · December 28th 2008, 08:17 AM

Originally Posted by varunnayudu

Now switch the dummy variable back to x:

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

Hw can one substitute $(\pi - u)$ into $(\pi - x)$

and secondly even after doing the sub the sum is not evaluated it is still the same ...........only instead of $x$ u have $(\pi - x)$

$ I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( {\pi - x} \right)}} {{1 + \sin \left( {\pi - x} \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} $

$ 2 \cdot I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} + \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}} {{1 + \sin \left( x \right)}}dx} $

tell me hw u got this conversion ..............

Maybe I can help. After you make the substitution $x = \pi - u$ your integral, as Mr. F says is

$I = \int_{\pi}^0 \frac{(\pi-u) \sin (\pi-u)}{1 + \sin(\pi-u)}(-du)$

using the fact that

$\sin(\pi-u) = \sin u$

then (I used the negative to switch the limits of integration)

$I = \int_0^{\pi} \frac{(\pi-u) \sin (u)}{1 + \sin(u)}\,du$

or, upon expanding,

$I = \int_0^{\pi} \frac{\pi \sin (u)}{1 + \sin(u)}\,du - \int_0^{\pi} \frac{u \sin (u)}{1 + \sin(u)}\,du$

If you look at your original integral and the second one of the right these are the same except for the variable $u$ . As this is only a dummy variable (it is used and never appears in the answer) then we can replace $u$ with $x$ giving

$I = \int_0^{\pi} \frac{\pi \sin (u)}{1 + \sin(u)}\,du - I$

or

$2I = \int_0^{\pi} \frac{\pi \sin (u)}{1 + \sin(u)}\,du$

as others have said. Others can correct me if I'm wrong but I believe this question appears in Stewart's Calculus book.

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