Question

$\int \frac{\sin 6\theta +\sin 2\theta +\sin 4\theta +\sin 8\theta }{4\sin 5\theta \cos 2\theta \cos \theta }d\theta =$
(where $c$ is constant of integration)

• A. $\cos \theta +c$
• B. $-\cos \theta +c$
• C. $\sin \theta +c$
• D. $\theta +c$

Answer 1

The correct option is D $\theta +c$

$\int \frac{\sin 6\theta +\sin 2\theta +\sin 4\theta +\sin 8\theta }{4\sin 5\theta \cos 2\theta \cos \theta }d\theta$
$=\int \frac{(\sin 6\theta +\sin 4\theta )+(\sin 2\theta +\sin 8\theta )}{4\sin 5\theta \cos 2\theta \cos \theta }d\theta$
$=\int \frac{2\sin 5\theta \cos \theta +2\sin 5\theta \cos 3\theta }{4\sin 5\theta \cos 2\theta \cos \theta }d\theta$
$=\int \frac{2\sin 5\theta (\cos \theta +\cos 3\theta )}{4\sin 5\theta \cos 2\theta \cos \theta }d\theta$
$=\int \frac{2\sin 5\theta \left(2\cos 2\theta \cos \theta \right)}{4\sin 5\theta \cos 2\theta \cos \theta }d\theta$
$=\int \frac{4\sin 5\theta \cos 2\theta \cos \theta }{4\sin 5\theta \cos 2\theta \cos \theta }d\theta$
$=\int d\theta$
$=\theta +c$