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Discussion Response Post to: integrate

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®µD®A (2710)

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$I=\int\frac{x^2}{(x\sin x+\cos x)^2}\ dx$

$\frac d{dx}\left(\frac{1}{x\sin x+\cos x}\right)=\frac{-\{\sin x+\cos x-\sin x\}}{(x\sin x+\cos x)^2}$

We know that;

$\frac d{dx}\left(\frac{1}{x\sin x+\cos x}\right)=\frac{-x\cos x}{(x\sin x+\cos x)^2}$

$\therefore\quad\int\frac{-x\cos x}{(x\sin x+\cos x)^2}\ dx=\frac 1{x\sin x+\cos x}$

$\Rightarrow\quad\int\underbrace{\frac{-x\cos x}{(x\sin x+\cos x)^2}}_{\text{II}}\underbrace{\cdot\frac {-x}{\cos x}}_{\text{I}}\ dx$

$=\left(\frac{-x}{\cos x}\right)\cdot\frac 1{x\sin x+\cos x}-\int\frac{-\cos x-x\sin x}{\cos^2x}\cdot\frac 1{x\sin x+\cos x}\cdot dx$

$=\frac{-x}{\cos x(x\sin x+\cos x)}+\int\sec^2x\ dx$

$I=\frac{-x}{\cos x(x\sin x+\cos x)}+\tan x+C$

[IMG]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/6/3/0638cc54ee34d87aa803d78fe3ec072aa8749bf5.gif[/IMG] :)

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