Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
$I = \int \sec^3(2\theta)d\theta$ is equal to:
\frac{1}{2}(\sec \theta \tan \theta) + \log_e \sqrt{\sec \theta + \tan \theta} + c
\frac{1}{4}(\sec 2\theta \tan 2\theta) + \frac{1}{2}\log_e \sqrt{\sec 2\theta + \tan 2\theta} + c
\frac{1}{4}(\sec 2\theta \tan 2\theta) + \log_e \sqrt{\sec 2\theta + \tan 2\theta} + c
None of these
Step-by-Step Solution
Key Concept: Apply the substitution $u = 2\theta$ to convert $\int \sec^3(2\theta)d\theta$ into a standard form using the reduction formula for $\sec^3(u)$.
To find $I = \int \sec^3(2\theta)d\theta$, we use the standard reduction formula for $\int \sec^n(x)dx$. For $\sec^3(x)$, the formula is $\int \sec^3(x)dx = \frac{1}{2}(\sec x \tan x + \ln|\sec x + \tan x|) + c$. Substituting $u = 2\theta$ gives $du = 2d\theta$, so $I = \frac{1}{2}\int \sec^3(u)du = \frac{1}{2} \cdot \frac{1}{2}(\sec u \tan u + \ln|\sec u + \tan u|) + c = \frac{1}{4}(\sec 2\theta \tan 2\theta) + \frac{1}{4}\ln|\sec 2\theta + \tan 2\theta| + c$. Since $\frac{1}{4}\ln|\sec 2\theta + \tan 2\theta| = \frac{1}{2}\ln\sqrt{|\sec 2\theta + \tan 2\theta|}$, both options 2 and 3 are equivalent forms of the correct answer.
Correct Answer: 2,3