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Question: \[\int_{}^{}{\frac{x^{2}dx}{(x\sin x + \cos x)^{2}} =}\]...

∫x2dx(xsin⁡x+cos⁡x)2=\int_{}^{}{\frac{x^{2}dx}{(x\sin x + \cos x)^{2}} =}

A

sin⁡x+cos⁡xxsin⁡x+cos⁡x\frac{\sin x + \cos x}{x\sin x + \cos x}

B

xsin⁡x−cos⁡xxsin⁡x+cos⁡x\frac{x\sin x - \cos x}{x\sin x + \cos x}

C

sin⁡x−xcos⁡xxsin⁡x+cos⁡x\frac{\sin x - x\cos x}{x\sin x + \cos x}

D

None of these

Answer

sin⁡x−xcos⁡xxsin⁡x+cos⁡x\frac{\sin x - x\cos x}{x\sin x + \cos x}

Explanation

Solution

Differentiation of xsin⁡x+cos⁡xx\sin x + \cos x is xcos⁡x.x\cos x. Then,

I=∫x2dx(xsin⁡x+cos⁡x)2=∫xcos⁡x(xsin⁡x+cos⁡x)2.xcos⁡xdxI = \int_{}^{}\frac{x^{2}dx}{(x\sin x + \cos x)^{2}} = \int_{}^{}\frac{x\cos x}{(x\sin x + \cos x)^{2}}.\frac{x}{\cos x}dx

Integrate by parts, [∵∫1t2dt=−1t]\left\lbrack \because\int_{}^{}{\frac{1}{t^{2}}dt = - \frac{1}{t}} \right\rbrack

∴I=−1(xsin⁡x+cos⁡x).(xcos⁡x)+∫1(xsin⁡x+cos⁡x).cos⁡x.1−x(−sin⁡x)cos⁡2xdx\therefore I = \frac{- 1}{(x\sin x + \cos x)}.\left( \frac{x}{\cos x} \right) + \int_{}^{}{\frac{1}{(x\sin x + \cos x)}.\frac{\cos x.1 - x( - \sin x)}{\cos^{2}x}dx}

⇒I=−1xsin⁡x+cos⁡x.xcos⁡x+∫sec⁡2xdx\Rightarrow I = \frac{- 1}{x\sin x + \cos x}.\frac{x}{\cos x} + \int_{}^{}{\sec^{2}xdx}

I=−1xsin⁡x+cos⁡x.xcos⁡x+sin⁡xcos⁡xI = \frac{- 1}{x\sin x + \cos x}.\frac{x}{\cos x} + \frac{\sin x}{\cos x}

⇒\Rightarrow I=−x+xsin⁡2x+sin⁡xcos⁡x(xsin⁡x+cos⁡x)cos⁡xI = \frac{- x + x\sin^{2}x + \sin x\cos x}{(x\sin x + \cos x)\cos x}

⇒I=sin⁡xcos⁡x−x(1−sin⁡2x)(xsin⁡x+cos⁡x)cos⁡x=sin⁡x−xcos⁡xxsin⁡x+cos⁡x\Rightarrow I = \frac{\sin x\cos x - x(1 - \sin^{2}x)}{(x\sin x + \cos x)\cos x} = \frac{\sin x - x\cos x}{x\sin x + \cos x}