_ ^ 2x 5x 3x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{\sin 2x}}{{\sin 5x\sin 3x}}} \;dx = $

A
$\log \sin 3x - \log \sin 5x + c$
B
$\frac{1}{3}\log \sin 3x + \frac{1}{5}\log \sin 5x + c$
✓
$\frac{1}{3}\log \sin 3x - \frac{1}{5}\log \sin 5x + c$
D
$3\log \sin 3x - 5\log \sin 5x + c$

✓

Answer

Correct option: C.

$\frac{1}{3}\log \sin 3x - \frac{1}{5}\log \sin 5x + c$

c
(c)$\int_{}^{} {\frac{{\sin 2x}}{{\sin 5x\sin 3x}}\,dx} = \int_{}^{} {\frac{{\sin (5x - 3x)}}{{\sin 5x\sin 3x}}\,dx} $
$ = \int_{}^{} {\frac{{\sin 5x\cos 3x - \cos 5x\sin 3x}}{{\sin 5x\sin 3x}}\,dx} $
$ = \frac{1}{3}\log \sin 3x - \frac{1}{5}\log \sin 5x + c.$

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