_ ^ x^2 (x x + x) ^2 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{x^2}}}{{{{(x\sin x + \cos x)}^2}}}\;dx = } $

A
$\frac{{\sin x + \cos x}}{{x\sin x + \cos x}}$
B
$\frac{{x\sin x - \cos x}}{{x\sin x + \cos x}}$
✓
$\frac{{\sin x - x\cos x}}{{x\sin x + \cos x}}$
D
None of these

✓

Answer

Correct option: C.

$\frac{{\sin x - x\cos x}}{{x\sin x + \cos x}}$

c
(c) 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 $\left[ {\int_{}^{} {\frac{1}{{{t^2}}}\,dt = - \frac{1}{t}} } \right]$
$\therefore \,\,\,I = \frac{{ - 1}}{{(x\sin x + \cos x)}}.\frac{x}{{\cos x}}$
$ + \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}}$.

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