Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
$I = \int \frac{dx}{(\sin x - 2\cos x)(2\cos x + \sin x)}$ is equal to
$\log_e \sqrt{\frac{\tan x - 2}{\tan x + 2}} + c$
$\frac{1}{4}\log\left|\frac{\sin x - 2\cos x}{\sin x + 2\cos x}\right| + c$
$-\frac{1}{4}\log\left|\frac{2\sin x + \cos x}{\sin x + 2\cos x}\right| + c$
None of these
Step-by-Step Solution
Key Concept: Converting to a rational function via $t = \tan x$ enables the use of partial fractions.
Rewrite the integral as $I = \int \frac{\sec^2 x}{\tan^2 x - 4} dx$. Substitute $\tan x = t$ to get $I = \int \frac{dt}{t^2 - 4}$. Using partial fractions: $\frac{1}{t^2-4} = \frac{1}{4}\left(\frac{1}{t-2} - \frac{1}{t+2}\right)$, yielding $I = \frac{1}{4}\ln\left|\frac{\tan x - 2}{\tan x + 2}\right| + C$.
Correct Answer: 1,2