Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
$$\int \frac{(ax^2 - b)dx}{\sqrt{c^2x^2 - (ax^2 + b)^2}} =$$
$$\sin^{-1}\left(\frac{ax + bx^2}{c}\right) + k$$
$$\tan^{-1}\left(\frac{a + bx^2}{cx}\right) + k$$
$$\sin^{-1}\left(\frac{ax^2 + b}{cx}\right) + k$$
$$\tan^{-1}\left(ax^2 + bx + c\right) + k$$
Step-by-Step Solution
Key Concept: The substitution $ax + \frac{b}{x} = c\sin\theta$ converts the complicated radical expression into a trigonometric integral.
Rewrite $I = \int \frac{(ax^2-b)dx}{x\sqrt{c^2x^2-(ax^2+b)^2}}$ and substitute $ax + \frac{b}{x} = c\sin\theta$. After trigonometric simplification, $I = \int \frac{c\cos\theta d\theta}{c\cos\theta} = \theta + k = \sin^{-1}\left(\frac{ax^2+b}{cx}\right) + k$.
Correct Answer: 3