Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
If $x$ satisfies the equation $x^2\left[\int_0^x \frac{dt}{t^2 + 2t\cos\alpha + 1} - x\left(\int_{-3}^x \frac{t^2\sin 2t}{t^2+1}dt\right)\right] - 2 = 0$ $(0 < \alpha < \pi)$, then the value of $x$ is:
$2\sqrt{\frac{\sin\alpha}{\alpha}}$
$-2\sqrt{\frac{\sin\alpha}{\alpha}}$
$4\sqrt{\frac{\sin\alpha}{\alpha}}$
$-4\sqrt{\frac{\sin\alpha}{\alpha}}$
Step-by-Step Solution
Key Concept: Strategic substitution transforms a complex rational function into a polynomial integrand that can be expanded and integrated term by term.
We use the substitution $1 - \frac{6}{x^7} = p$, which gives $\frac{42}{x^8}dx = dp$ and $x^7 = \frac{6}{1-p}$. The integral becomes $I = \frac{1}{42}\int \frac{(1-p)^7}{(6)^7}dp$, which after expansion and integration yields $I = \frac{1}{54432}[\ln p^6 + 9p^2 - 2p^3 - 18p] + c$.
Correct Answer: 3,2