For $x ∈ (0,\frac{\pi}{2}),\int\frac{1}{\sin^2x+\sin 2x} dx$ is equal to

$\frac{1}{2}\log\left|\frac{\tan x}{\tan x+2}\right|+C$, where C is constant of integration

The correct answer is Option (2) → $\frac{1}{2}\log\left|\frac{\tan x}{\tan x+2}\right|+C$, where C is constant of integration

Given: $\displaystyle \int \frac{1}{\sin^2 x + \sin 2x} \, dx$ for $x \in \left(0, \frac{\pi}{2}\right)$

Try substitution $u = \tan x$

Then:

• $\sin x = \frac{u}{\sqrt{1 + u^2}}$
• $\cos x = \frac{1}{\sqrt{1 + u^2}}$
• $\sin 2x = \frac{2u}{1 + u^2}$

So:

$\sin^2 x + \sin 2x = \frac{u^2}{1 + u^2} + \frac{2u}{1 + u^2} = \frac{u^2 + 2u}{1 + u^2}$

$dx = \frac{1}{1 + u^2} du$

So the integral becomes:

$\int \frac{1}{\frac{u^2 + 2u}{1 + u^2}} \cdot \frac{1}{1 + u^2} du = \int \frac{1 + u^2}{u^2 + 2u} \cdot \frac{1}{1 + u^2} du = \int \frac{1}{u^2 + 2u} du$

Now simplify and integrate:

$\int \frac{1}{u(u + 2)} du = \int \left( \frac{1}{2u} - \frac{1}{2(u + 2)} \right) du$

$= \frac{1}{2} \ln|u| - \frac{1}{2} \ln|u + 2| + C = \frac{1}{2} \ln \left| \frac{u}{u + 2} \right| + C$

Recall $u = \tan x$

Final Answer: $\frac{1}{2} \ln \left| \frac{\tan x}{\tan x + 2} \right| + C$