Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade None
Question:
The integral $\int \frac{\sec^{3/2}\theta - \sec^{1/2}\theta}{2 + \tan^2\theta} \tan\theta d\theta$ is:
$\sqrt{2}\tan^{-1}\left(\frac{\sec\theta + 1}{\sqrt{2}\sec\theta}\right) + C$
$\frac{1}{\sqrt{2}}\log_e\left|\frac{\sec\theta - \sqrt{2}\sec\theta + 1}{\sec\theta + \sqrt{2}\sec\theta + 1}\right| + C$
$\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sec\theta + 1}{\sqrt{2}\sec\theta}\right) + C$
$\sqrt{2}\log_e\left|\frac{\sec\theta - \sqrt{2}\sec\theta + 1}{\sec\theta + \sqrt{2}\sec\theta + 1}\right| + C$
Step-by-Step Solution
Key Concept: The substitution $\sqrt{\sec\theta} = t$ linearizes the radical and simplifies the denominator into a form manageable by further substitution.
The integral $I = \int \frac{(\sec\theta - 1)\sqrt{\sec\theta}}{1+\sec^2\theta} \tan\theta d\theta$ is simplified by substituting $\sqrt{\sec\theta} = t$, giving $\sec\theta \tan\theta d\theta = dt/2$. This transforms the integral to $I = \int \frac{(t^2-1)}{1+t^4} 2dt$, which is solved via $u = t + 1/t$ to obtain $I = \frac{1}{\sqrt{2}} \log\left|\frac{t + 1/t - \sqrt{2}}{t + 1/t + \sqrt{2}}\right| + C$.
Correct Answer: 2