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Integral Calculus

Hot goIITian

 Joined: 18 Jan 2012 Post: 147
20 Feb 2012 15:31:52 IST
3 People liked this
4
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Integration Challenge-3!!
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Integral Calculus

Integration Challenge-3!!

I'm posting SEVEN challenges...i know i said six earlier but the 7th one was the best question I've ever encountered (and solved).     I'll post challenges 3 and 4 today and 5,6 & 7 tomorrow. Without wasting any more time,

Integrate:

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Blazing goIITian

Joined: 19 Jan 2008
Posts: 1139
20 Feb 2012 17:13:57 IST
3 people liked this

$\hspace{-16}\mathbf{\int\frac{x^2}{(x\cos x-\sin x)^2}dx}\\\\\\ Now Simplifying\\\\\\ \mathbf{(x\cos x-\sin x)=\sqrt{1+x^2}\left\{\frac{x}{\sqrt{1+x^2}}.cos x-\frac{1}{\sqrt{1+x^2}}.\sin x\right\}}\\\\\\ \mathbf{=\sqrt{1+x^2}\left\{\\sin \alpha.\cos x-\cos \alpha.\sin x \right\}=\sqrt{1+x^2}.\sin (\alpha-x)}\\\\\\ Where \mathbf{\sin \alpha=\frac{x}{\sqrt{1+x^2}}} and \mathbf{\cos \alpha =\frac{1}{\sqrt{1+x^2}}}\\\\\\ and \mathbf{\tan\alpha=x\Leftrightarrow \alpha =\tan^{-1}(x)}\\\\\\ So \mathbf{\int\frac{x^2}{(x\cos x-\sin x)^2}dx=\int\csc^2(\alpha-x)\frac{x^2}{(1+x^2)}dx}\\\\\\ Now Put \mathbf{(\alpha-x)=t\Leftrightarrow \tan^{-1}(x)-x=t\Leftrightarrow \left(\frac{x^2}{1+x^2}\right)=-dt}\\\\\\ So \mathbf{\int \csc^2(t)dt=-\cot (t)+C}\\\\\\ \mathbf{=-\cot(\alpha-x)+C=-\cot \left(\tan^{-1}(x)-x\right)+C}\\\\\\ After Simplification\\\\\\ \mathbf{\int\frac{x^2}{(x\cos x-\sin x)^2}dx=\left(\frac{x\sin x+\cos x}{\sin x-x\cos x}\right)+C}$

Blazing goIITian

Joined: 19 Jan 2008
Posts: 1139
20 Feb 2012 17:47:48 IST
0 people liked this

I think mistake of (-) sign

Sorry for That (bcz there is no option of editing)

Thanks

Scorching goIITian

Joined: 18 Feb 2012
Posts: 264
20 Feb 2012 17:51:29 IST
0 people liked this

a good question and aawesum answer.....

Hot goIITian

Joined: 15 Feb 2012
Posts: 180
20 Feb 2012 17:56:35 IST
0 people liked this

how to give votes to singh... iam a new user i dont know.. please tell me..but i applouse to him

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