Integral Calculus Question 456
Question: $ \int_{{}}^{{}}{\frac{x^{2}}{{{(x\sin x+\cos x)}^{2}}}\ dx=} $
[MNR 1989; RPET 2000]
Options:
A) $ \frac{\sin x+\cos x}{x\sin x+\cos x} $
B) $ \frac{x\sin x-\cos x}{x\sin x+\cos x} $
C) $ \frac{\sin x-x\cos x}{x\sin x+\cos x} $
D) None of these
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Answer:
Correct Answer: C
Solution:
Differentiation of $ x\sin x+\cos x $ is $ x\cos x, $ then
$ I=\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 $ [ \int_{{}}^{{}}{\frac{1}{t^{2}},dt=-\frac{1}{t}} ] $
$ \therefore ,I=\frac{-1}{(x\sin x+\cos x)}.\frac{x}{\cos x} $
$ +\int_{{}}^{{}}{\frac{1}{(x\sin x+\cos x)}}.\frac{\cos x,.,1-x(-\sin x)}{{{\cos }^{2}}x},dx $
$ =-\frac{1}{x\sin x+\cos x}.\frac{x}{\cos x}+\int_{{}}^{{}}{{{\sec }^{2}}x,dx} $
$ =-\frac{1}{x\sin x+\cos x}.\frac{x}{\cos x}+\frac{\sin x}{\cos x} $
$ =\frac{-x+x{{\sin }^{2}}x+\sin x\cos x}{(x\sin x+\cos x)\cos x} $
$ =\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} $ .
BETA
