Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
If $\int\left[\frac{1}{1-x^8}\left\{\cos^{-1}\left(\frac{2x}{1+x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right)\right\}\right]dx$ has $p\tan^{-1}f(x)$ & $q\tan^{-1}g(x)$ terms and $x \in (-1, 1)$, then: (where $p$ and $q$ are constant)
f(x).g(x) = \frac{x^2-1}{\sqrt{2}}
p.q = \frac{\pi^2}{64\sqrt{2}}
p.q = \frac{\pi^2}{\sqrt{2}}
None of these
Step-by-Step Solution
Key Concept: Decompose inverse trigonometric expressions into recognizable forms and use partial fractions with the factorization $1-x^8=(1-x^4)(1+x^4)$.
The integral $\int \frac{1}{1-x^8}\left[\cos^{-1}\left(\frac{2x}{1+x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right)\right]dx$ is evaluated by recognizing that $\cos^{-1}\left(\frac{2x}{1+x^2}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ and $\tan^{-1}\left(\frac{2x}{1-x^2}\right)$ relates to $2\tan^{-1}(x)$. After simplification using partial fractions and trigonometric identities, the result involves logarithmic and inverse tangent terms combined with radicals.
Correct Answer: 1,2