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Integral Calculus

Hot goIITian

 Joined: 18 Jan 2012 Post: 147
21 Feb 2012 12:06:34 IST
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Integration Challenge-7!!
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Integral Calculus

Integration Challenge-7!!

The final Integration Challenge:(Press View post if not visible)

(A)π/2

(B)π

(C)2π

(D) None of these

Blazing goIITian

Joined: 19 Jan 2008
Posts: 1089
21 Feb 2012 16:50:08 IST
1 people liked this

$\hspace{-16}Let \mathbf{f(\phi)=\int_{0}^{\pi}e^{\phi.\cos \theta}.\cos(\phi.\sin \theta)d\theta...............(1)}\\\\\\ Rembember Here We Calculate \mathbf{f(1)}\\\\\\ Diff. both side w.r.to \mathbf{\phi}, We Get\\\\\\ \mathbf{\frac{df}{d\phi}=\int_{0}^{\pi}e^{\phi.\cos \theta}\left(\cos \theta.cos(\phi.\sin \theta)-\sin \theta.\sin (\phi.\cos \theta)\right)d\theta}\\\\\\ \mathbf{=\int_{0}^{\pi}\frac{1}{\phi}.\frac{d}{d\theta}\left(e^{\phi.\cos \theta}.\sin (\phi.\sin \theta)\right)d\theta}\\\\\\ \mathbf{=\frac{1}{\phi}\int_{0}^{2\pi}d\left(e^{\phi.\cos \theta}.\sin (\phi.\sin \theta)\right)}\\\\\\ \mathbf{=\frac{1}{\pi}.\left[e^{\phi.\cos \theta}.\sin (\phi.\sin \theta\right]_{0}^{\pi}=0}\\\\\\ Now From \mathbf{f(0)=\pi}\\\\\\ So \mathbf{\int_{0}^{1}\frac{df}{d\phi}.d\phi=\int_{0}^{1}0.d\phi=0}\\\\\\ So \mathbf{f(1)-f(0)=0\Leftrightarrow f(1)-\pi=0}\\\\\\ So \boxed{\boxed{\mathbf{f(1)=\int_{0}^{\pi}e^{\cos \theta}.\cos(\sin \theta)d\theta=\pi}}}$

Blazing goIITian

Joined: 19 Jan 2008
Posts: 1089
21 Feb 2012 16:51:46 IST
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Some Silly Mistakes in Answer So Dont Mind , Thanks Thanks Visal for Such a Nice Challenging Question in Integrationagain Thanks bhai.

Hot goIITian

Joined: 18 Jan 2012
Posts: 147
21 Feb 2012 18:50:27 IST
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My solution was different...but gave me the answer..

Putting cosθ+i.sinθ=z

We get cosθ-i.sinθ=1/z

and (-sinθ+i.cosθ)dθ=dz

or i(z)dθ=dz

or dθ=dz/iz

Limits for z: 1 to -1

Now using

and using ln(-1)=iπ, you'll find that

finishes it neatly.

Cool goIITian

Joined: 19 Feb 2011
Posts: 68
30 Mar 2012 14:44:04 IST
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here is a different approach tell me it is true or not i wrote

so integral becomes

Hot goIITian

Joined: 18 Jan 2012
Posts: 147
31 Mar 2012 17:24:05 IST
0 people liked this

It is wrong owing to your poor understanding of exponent basics..(Sorry I'm rude).

How can you write

???

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