Question

$\int \frac{\left(\sin x + \cos x\right)\left(2 - \sin 2x\right)}{\sin^{2} 2x}dx =$

• A. $\frac{\sin x + \cos x}{\sin2x } + c$
• B. $\frac{\sin x - \cos x}{\sin2x} +c$
• C. $\frac{\sin \, x}{\sin x + \cos x} +c$
• D. $\frac{\sin \, x}{\sin x - \cos x} + c$

Answer 1

Answer: (B)

$\frac{\sin x - \cos x}{\sin2x} +c$

Answer 2

We have,
$I=\int \frac{(\sin x+\cos x)(2-\sin 2 x)}{\sin ^{2} 2 x} d x$
Put $\sin x-\cos x=t \Rightarrow(\sin x+\cos x) d x=d t$
and $(\sin x-\cos x)^{2}=t^{2} \Rightarrow 1-\sin 2 x=t^{2}$
$\Rightarrow \, \sin 2 x=1-t^{2}$
$\therefore \, I=\int \frac{\left(2-\left(1-t^{2}\right)\right) d t}{\left(1-t^{2}\right)^{2}}$
$\Rightarrow \, I=\int \frac{\left(1+t^{2}\right) d t}{\left(1-t^{2}\right)^{2}}$
$\Rightarrow \, I-\int \frac{1+t^{2}}{1-2 t^{2}+t^{4}} d t$
$\Rightarrow \, I=\int \frac{1+1 / t^{2}}{\frac{1}{t^{2}}+t^{2}-2} d t$
$\Rightarrow \, I=\int \frac{1+\frac{1}{t^{2}}}{\left(t-\frac{1}{t}\right)^{2}} d t$
Put $t-\frac{1}{t}-y \Rightarrow \left(1+\frac{1}{t^{2}}\right) d t-d y$
$\therefore \, I=\int \frac{d y}{y^{2}}=-\frac{1}{y}+C$
$\Rightarrow \, I=\frac{-1}{t-\frac{1}{t}}+C$
$\Rightarrow \, I=\frac{t}{1-t^{2}}+C$
$\Rightarrow \, I=\frac{\sin x-\cos x}{\sin 2 x}+C$