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Integral Calculus

Hot goIITian

 Joined: 18 Jan 2012 Post: 147
19 Feb 2012 19:18:48 IST
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Integration Challenge-2!!
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Integral Calculus

Integration Challenge-2!!

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Integrate:

Blazing goIITian

Joined: 19 Jan 2008
Posts: 1142
19 Feb 2012 19:57:37 IST
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$\hspace{-16}\mathbf{\int \frac{x\cos x-\sin x}{x^2+\sin^2 x}dx}\\\\\\ \mathbf{\int \frac{\frac{x\cos x-\sin x}{x^2}}{1+\left(\frac{\sin x}{x}\right)^2}dx}\\\\\\ Now Put \mathbf{\frac{\sin x}{x}=t\Leftrightarrow \left(\frac{x\cos x-\sin x}{x^2}\right)dx=dt}\\\\\\ So \mathbf{\int\frac{1}{1+t^2}dt=\tan^{-1}(t)+C}\\\\\\ \mathbf{\int \frac{x\cos x-\sin x}{x^2+\sin^2 x}dx=\tan^{-1}\left(\frac{\sin x}{x}\right)+C}$

Hot goIITian

Joined: 18 Jan 2012
Posts: 147
19 Feb 2012 20:06:13 IST
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That was quick!....challenge 3 tomorrow...

Blazing goIITian

Joined: 19 Jan 2008
Posts: 1142
19 Feb 2012 20:09:55 IST
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Thanks man.

New kid on the Block

Joined: 19 Feb 2012
Posts: 17
19 Feb 2012 21:39:22 IST
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the answer to this  is (tan- sinx / x)+c

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