First slide

Methods of integration

Question

If the primitive of $f (x) = \frac{1}{3 \sin x + \sin^{3} x}$
is equal to $\frac{1}{6} \log |\frac{t - 1}{t + 1}| + \frac{1}{12} \log |\frac{2 + t}{2 - t}| + C$ then

Moderate

A

$t = \cos x$
B

$t = \tan x / 2$
C

$t = 2 \cos x$
D

$t = \sin x$

Solution

$\begin{array}{l} \int \frac{d x}{3 \sin x + \sin^{3} x} = \int \frac{\sin x d x}{(3 + \sin^{2} x) \sin^{2} x} \\ = \int \frac{\sin x d x}{(4 - \cos^{2} x) (1 - \cos^{2} x)} \end{array}$
$\begin{aligned} = - \int \frac{d t}{(4 - t^{2}) (1 - t^{2})} \\ = - \frac{1}{3} \int (\frac{1}{t^{2} - 4} - \frac{1}{t^{2} - 1}) (t = \cos x) \\ = \frac{1}{6} \log |\frac{t - 1}{t + 1}| + \frac{1}{12} \log |\frac{2 + t}{2 - t}| + C \end{aligned}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App