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

MCQ

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

A
$\frac{1}{b}(a + b\cos x) + c$
✓
$\frac{1}{{b(a + b\cos x)}} + c$
C
$\frac{1}{b}\log (a + b\cos x) + c$
D
None of these

✓

Answer

Correct option: B.

$\frac{1}{{b(a + b\cos x)}} + c$

b
(b) Put $a + b\cos x = t \Rightarrow dx = - \frac{{dt}}{{b\sin x}},$ then
$\int_{}^{} {\frac{{\sin x}}{{{{(a + b\cos x)}^2}}}\,dx = - \frac{1}{b}\int_{}^{} {\frac{1}{{{t^2}}}\,dt} = \frac{1}{b}\frac{1}{t} + c} $
$ = \frac{1}{{b(a + b\cos x)}} + c.$

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