Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade None
Question:
If $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx = P\cos x + Q\ln |f(x)| + R$, then:
P = 1/2, Q = -\frac{3}{4\sqrt{2}}$, $f(x) = \frac{\sqrt{2}\cos x + 1}{\sqrt{2}\cos x - 1}$
P = 1/4, Q = -\frac{1}{\sqrt{2}}$, $f(x) = \frac{\sqrt{2}\cos x - 1}{\sqrt{2}\cos x + 1}$
P = 1/2, Q = -\frac{3}{4\sqrt{2}}$, $f(x) = \frac{\sqrt{2}\cos x - 1}{\sqrt{2}\cos x + 1}$
P = -1/2, Q = -\frac{3}{4\sqrt{2}}$, $f(x) = \frac{\sqrt{2}\cos x + 1}{\sqrt{2}\cos x - 1}$
Step-by-Step Solution
Key Concept: Substitution $t = \cos x$ converts a trigonometric integral into a rational function solvable by partial fractions.
The integral $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx$ is simplified by rewriting the denominator as $\cos 2x = 2\cos^2 x - 1$ and the numerator as $\sin x(1 + \sin^2 x)$. Using substitution $t = \cos x$ (so $dt = -\sin x dx$), the integral transforms to $I = \int \frac{2 - t^2}{(1-2t^2)} dt$. This is decomposed and integrated using partial fractions, yielding $I = \frac{\cos x}{2} - \frac{3}{4\sqrt{2}} \ln\left|\frac{t - \frac{1}{\sqrt{2}}}{t + \frac{1}{\sqrt{2}}}\right| + c$, which simplifies to $I = \frac{\cos x}{2} - \frac{3}{4\sqrt{2}} \ln\left|\frac{\sqrt{2}\cos x - 1}{\sqrt{2}\cos x + 1}\right| + c$.
Correct Answer: 1,3