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

MCQ

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

✓
$\frac{1}{6}\log (1 - \cos x) + \frac{1}{2}\log (1 + \cos x) - \frac{2}{3}\log (1 + 2\cos x)$
B
$6\log (1 - \cos x) + 2\log (1 + \cos x) - \frac{2}{3}\log (1 + 2\cos x)$
C
$6\log (1 - \cos x) + \frac{1}{2}\log (1 + \cos x) + \frac{2}{3}\log (1 + 2\cos x)$
D
None of these

✓

Answer

Correct option: A.

$\frac{1}{6}\log (1 - \cos x) + \frac{1}{2}\log (1 + \cos x) - \frac{2}{3}\log (1 + 2\cos x)$

a
(a)$I = \int_{}^{} {\frac{{dx}}{{\sin x(1 + 2\cos x)}}} = \int_{}^{} {\frac{{\sin x\,dx}}{{{{\sin }^2}x(1 + 2\cos x)}}} $
$ = \int_{}^{} {\frac{{\sin x\,dx}}{{(1 - \cos x)(1 + \cos x)(1 + 2\cos x)}}} $
Now differential coefficient of $\cos x$ is $ - \sin x$ which is given in numerator and hence we make the substitution $\cos x = t \Rightarrow - \sin x\,dx = dt$
$\therefore \,\,\,I = - \int_{}^{} {\frac{{dt}}{{(1 - t)(1 + t)(1 + 2t)}}} $
We split the integrand into partial fractions
$I = - \int {\left[ {\frac{1}{{6(1 - t)}} - \frac{1}{{2(1 + t)}} + \frac{4}{{3(1 + 2t)}}} \right]} \,dt$ etc.

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