Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $I(n) = \int_{1}^{e} x^3(\log x)^n dx$, where $n$ is a whole number.
MATCH THE FOLLOWING:
(A) $\frac{64}{5e^4 - 1} = I(2)$
(B) $\frac{4I(n) + nI(n-1)}{e^4}$, for $n \geq 1 =$
(C) The least value of $n$, for which $I(n) < \frac{e^4 - 4}{4}$, is
(D) $\lim_{n \to \infty} I(n) =$
Step-by-Step Solution
Key Concept: Use the substitution $\sqrt{x} = \tan\theta$ to convert the complicated inverse tangent expression into a manageable trigonometric form.
Substitute $\sqrt{x} = \tan\theta$, so $\frac{1}{2\sqrt{x}}dx = 2\tan\theta\sec^2\theta d\theta$ and $dx = 4\tan\theta\sec^2\theta d\theta$. The integral becomes $\int\tan^2\theta\tan(2\tan^{-1}(\sqrt{1+\tan^2\theta+1}-\sqrt{1+\tan^2\theta-1})) \cdot 4\tan\theta\sec^2\theta d\theta$. Simplify using $\sqrt{1+\tan^2\theta} = \sec\theta$ to get $\int 4\tan^3\theta\sec^2\theta\tan(\frac{\pi}{4}-\frac{\theta}{2})d\theta = \int 4\tan^3\theta\sec^2\theta d\theta$. This evaluates to $\frac{4}{5}\tan^5\theta + c = \frac{4}{5}x^{5/4} + c$.
Correct Answer: [A-r] [B-p] [C-q] [D-s]