Question

The value of $2∫\sin xâ‹\dots \text{cosec}4xdx$ is equal to

• A. $\frac{1}{2\sqrt{2}}\ln â\frac{1+\sqrt{2}\sin x}{1−\sqrt{2}\sin x}â−\frac{1}{4}\ln â\frac{1+\sin x}{1−\sin x}â+C$
• B. $\frac{1}{2\sqrt{2}}\ln â\frac{1+\sqrt{2}\sin x}{1−\sqrt{2}\sin x}â+\frac{1}{4}\ln â\frac{1+\sin x}{1−\sin x}â+C$
• C. $\frac{1}{2\sqrt{2}}$ en $â\frac{1−\sqrt{2}\sin x}{1+\sqrt{2}\sin x}â−\frac{1}{4}$ en $â\frac{1+\sin x}{1−\sin x}â+C$
• D. none of these

Answer 1

Answer: (A)

$I=2∫\sin xâ‹\dots \text{cosec}4xdx=2∫\frac{\sin xdx}{4\sin x\cos x\cos 2x}$
$=\frac{1}{2}∫\frac{dt}{\left(1−t^2\right)\left(1−2t^2\right)}$
$\text{ Put }\sin x=t,\text{ then }\cos xdx=dt$
$⇒=\frac{1}{2}∫\frac{dt}{\left(1−t^2\right)\left(1−2t^2\right)}$
$=\frac{1}{2}\left[−∫\frac{1}{1−t^2}dt+∫\frac{2}{1−2t^2}dt\right]$[By partial fraction]
$=\frac{1}{2}\left[−\frac{1}{2}\ln â\frac{1+t}{1−t}â+\frac{2}{2\sqrt{2}}\ln âˆ\pounds \left(\frac{1+\sqrt{2}t}{1−\sqrt{2}t}\right)\right]$
$=\frac{1}{2\sqrt{2}}\ln â\left(\frac{1+\sqrt{2}\sin x}{1−\sqrt{2}\sin x}\right)â−\frac{1}{4}\ln â\left(\frac{1+\sin x}{1−\sin x}\right)â+C$