Question

$\int \frac{1}{{\sin }^{2}x+\sin 2x}dx.$

• A. $\frac{1}{2}\log \frac{\tan x}{\tan x+2}$
• B. $-\frac{1}{2}\log \frac{\tan x}{\tan x+2}$
• C. $\frac{1}{2}\log \frac{\tan x}{\tan x-2}$
• D. $\log \frac{\tan x}{\tan x+2}$

Answer 1

The correct option is A $\frac{1}{2}\log \frac{\tan x}{\tan x+2}$

Let $I=\int \frac{1}{{\sin }^{2}x+\sin 2x}dx=\int \frac{dx}{{\sin }^{2}x+2\sin x\cos x}$
Dividing numerator and denominator by ${\cos }^{2}x$, we get
$I=\int \frac{{\sec }^{2}xdx}{{\tan }^{2}x+2\tan x}$
Put $tanx=t\Rightarrow {\sec }^{2}xdx=dt$
Therefore
$I=\int \frac{dt}{t^2+2t}=\int \frac{dt}{t(t+2)}=\frac{1}{2}\int (\frac{1}{2}-\frac{1}{t+2})dt$
$=\frac{1}{2}[\log t-\log (t+2)]=\frac{1}{2}\log \frac{\tan x}{\tan x+2}$
Hence, option 'A' is correct.