Question

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

• A. $\frac{\sin x+\cos x}{\sin 2x}+c$
• B. $\frac{\sin x-\cos x}{\sin 2x}+c$
• C. $\frac{\sin x}{\sin x+\cos x}+c$
• D. $\frac{\sin x}{\sin x-\cos x}+c$
• E. $\frac{\sin x-\cos x}{\sin x+\cos x}+c$

Answer 1

Answer: (B)

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