Question

The value of $2 \int \sin x \cdot \text{cosec} 4 x d x$ is equal to

• A. $\frac{1}{2 \sqrt{2}} \ln \left|\frac{1+\sqrt{2} \sin x}{1-\sqrt{2} \sin x}\right|-\frac{1}{4} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C$
• B. $\frac{1}{2 \sqrt{2}} \ln \left|\frac{1+\sqrt{2} \sin x}{1-\sqrt{2} \sin x}\right|+\frac{1}{4} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C$
• C. $\frac{1}{2 \sqrt{2}}$ en $\left|\frac{1-\sqrt{2} \sin x}{1+\sqrt{2} \sin x}\right|-\frac{1}{4}$ en $\left|\frac{1+\sin x}{1-\sin x}\right|+C$
• D. none of these

Answer 1

Answer: (A)

$ I =2 \int \sin x \cdot \text{cosec} 4 x d x=2 \int \frac{\sin x d x}{4 \sin x \cos x \cos 2 x}$
$ =\frac{1}{2} \int \frac{d t}{\left(1-t^2\right)\left(1-2 t^2\right)} $
$ \text { Put } \sin x=t, \text { then } \cos x d x=d t$
$\Rightarrow =\frac{1}{2} \int \frac{d t}{\left(1-t^2\right)\left(1-2 t^2\right)}$
$=\frac{1}{2}\left[-\int \frac{1}{1-t^2} d t+\int \frac{2}{1-2 t^2} d t\right]$[By partial fraction]
$=\frac{1}{2}\left[-\frac{1}{2} \ln \left|\frac{1+t}{1-t}\right|+\frac{2}{2 \sqrt{2}} \ln \mid\left(\frac{1+\sqrt{2} t}{1-\sqrt{2} t}\right)\right]$
$=\frac{1}{2 \sqrt{2}} \ln \left|\left(\frac{1+\sqrt{2} \sin x}{1-\sqrt{2} \sin x}\right)\right|-\frac{1}{4} \ln \left|\left(\frac{1+\sin x}{1-\sin x}\right)\right|+C$