Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade None
Question:
If $\int \frac{\cos^2 x + \sin 2x}{(2\cos x - \sin x)^2} dx = \frac{\cos x}{2\cos x - \sin x} + ax + b\ln|2\cos x - \sin x| + c$, then:
$a = \frac{1}{5}, b = \frac{2}{5}$
$a = \frac{1}{5}, b = -\frac{2}{5}$
$a = -\frac{1}{5}, b = \frac{2}{5}$
$a = -\frac{1}{5}, b = -\frac{2}{5}$
Step-by-Step Solution
Key Concept: Successive differentiation of a parameter in an integral formula generates higher power terms in the denominator.
Given $\int_0^{\pi} \frac{dx}{a-\cos x} = \frac{\pi}{\sqrt{a^2-1}}$. Differentiating both sides with respect to $a$ yields $-\int_0^{\pi} \frac{dx}{(a-\cos x)^2} = -\frac{a}{(a^2-1)^{3/2}}$. Differentiating again: $2\int_0^{\pi} \frac{dx}{(a-\cos x)^3} = \frac{\pi(1+2a^2)}{(a^2-1)^{5/2}}$. For $a = \sqrt{10}$, we substitute to get $\int_0^{\pi} \frac{dx}{(\sqrt{10}-\cos x)^3} = \frac{7\pi}{81}$.
Correct Answer: 4