For \(x \in \left(0,\displaystyle\frac{\pi}{2}\right)\), \(\int\displaystyle\frac{1}{\sin^2 x+\sin 2x}\, dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\log\left|\frac{\tan x}{\tan x+2}\right|+C\), where C is constant of integration
Since sin 2x = 2 sin x cos x, dividing the numerator and the denominator by \(\cos^2 x\) gives
\(\displaystyle I = \int \frac{\sec^2 x}{\tan^2 x + 2\tan x}\,dx\)
Let t = tan x, so dt = sec2x dx:
\(\displaystyle I = \int \frac{dt}{t(t + 2)}\)
By partial fractions:
\(\displaystyle \frac{1}{t(t + 2)} = \frac{1}{2}\left(\frac{1}{t} - \frac{1}{t + 2}\right)\)
\(\displaystyle I = \frac{1}{2}\left(\log|t| - \log|t + 2|\right) + C\)
\(\displaystyle = \frac{1}{2}\log\left|\frac{\tan x}{\tan x + 2}\right| + C\)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{2}\log\left|\frac{\tan x+2}{\tan x-2}\right|+C\), where C is constant of integration — Incorrect. The denominator factors as t(t + 2); there is no factor t − 2.
Option (3) \(\displaystyle\frac{1}{2}\log\left|\frac{\tan x+2}{\tan x}\right|+C\), where C is constant of integration — Incorrect. This is the negative of the answer: it comes from taking the partial fractions as \(\frac{1}{t + 2} - \frac{1}{t}\), a sign slip.
Option (4) \(\displaystyle\frac{1}{2}\log\left|\frac{\tan x+1}{\tan x+2}\right|+C\), where C is constant of integration — Incorrect. The denominator factors as t(t + 2); there is no factor t + 1.