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Integral Calculus

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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

(D) None of these

Comments (5)

jagdish singh

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21 Feb 2012 16:50:08 IST

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$\hspace{-16}$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}}}$$

jagdish singh

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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.

Vishal U. Karve

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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.

I really liked your solution.

Mayank Mittal

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Joined: 19 Feb 2011

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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

Vishal U. Karve

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Joined: 18 Jan 2012

Posts: 147

31 Mar 2012 17:24:05 IST

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

How can you write

???

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