Question

If $I=\int\frac{\sin x+\sin^{3}x}{\cos 2x}dx = P\cos x+Q\log\left|\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}\right|+c$ (where c is a constant of integration), then values of P and Q are respectively

• A. $\frac{1}{2}, \frac{3}{4\sqrt{2$
• B. $\frac{1}{2}, \frac{-3}{4\sqrt{2$
• C. $1, \frac{3}{2\sqrt{2$
• D. $1, \frac{-3}{2\sqrt{2$

Hint:

Logic Tip: When integrating rational functions of $t$ where the numerator degree equals the denominator degree (e.g. $\frac{t^2-2}{2t^2-1}$), you MUST perform polynomial long division or algebraic manipulation before integrating.

Answer 1

The Correct Option is B

Concept:
Transform the integrand so that it is entirely in terms of $\cos x$, except for a single $\sin x$ factor in the numerator. This allows for the substitution $t = \cos x$. Then, use partial fractions or algebraic division to integrate the resulting rational function.
Step 1: Rewrite the integrand in terms of $\cos x$.
Given integral: $$I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx$$ Factor out $\sin x$ in the numerator and use the double angle formula $\cos 2x = 2\cos^2 x - 1$: $$I = \int \frac{\sin x(1 + \sin^2 x)}{2\cos^2 x - 1} dx$$ Use the Pythagorean identity $\sin^2 x = 1 - \cos^2 x$: $$I = \int \frac{\sin x(1 + 1 - \cos^2 x)}{2\cos^2 x - 1} dx = \int \frac{\sin x(2 - \cos^2 x)}{2\cos^2 x - 1} dx$$
Step 2: Apply substitution method.
Let $t = \cos x$. Then $dt = -\sin x \, dx$, which means $\sin x \, dx = -dt$. Substitute this into the integral: $$I = \int \frac{2 - t^2}{2t^2 - 1} (-dt) = \int \frac{t^2 - 2}{2t^2 - 1} dt$$
Step 3: Perform polynomial division and integrate.
Divide the numerator by the denominator. To make this easier, multiply and divide the integral by 2: $$I = \frac{1}{2} \int \frac{2t^2 - 4}{2t^2 - 1} dt$$ Rewrite the numerator to match the denominator: $$I = \frac{1}{2} \int \frac{(2t^2 - 1) - 3}{2t^2 - 1} dt$$ $$I = \frac{1}{2} \int \left( 1 - \frac{3}{2t^2 - 1} \right) dt$$ Now, integrate term by term. Use the standard formula $\int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \log\left|\frac{x-a}{x+a}\right|$ where we write $2t^2-1$ as $(\sqrt{2}t)^2 - 1^2$: $$I = \frac{1}{2}t - \frac{3}{2} \int \frac{1}{(\sqrt{2}t)^2 - 1^2} dt$$ $$I = \frac{1}{2}t - \frac{3}{2} \cdot \left(\frac{1}{\sqrt{2\right) \cdot \frac{1}{2(1)} \log\left|\frac{\sqrt{2}t - 1}{\sqrt{2}t + 1}\right| + c$$ $$I = \frac{1}{2}t - \frac{3}{4\sqrt{2 \log\left|\frac{\sqrt{2}t - 1}{\sqrt{2}t + 1}\right| + c$$
Step 4: Substitute t back and compare with the given form.
Substitute $t = \cos x$: $$I = \frac{1}{2}\cos x - \frac{3}{4\sqrt{2 \log\left|\frac{\sqrt{2}\cos x - 1}{\sqrt{2}\cos x + 1}\right| + c$$ Comparing this with $P\cos x+Q\log\left|\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}\right|+c$, we get: $$P = \frac{1}{2}, \quad Q = \frac{-3}{4\sqrt{2$$