Question

If $I = \int \frac{\sin x + \sin^3 x}{\cos 2x}\,dx = P \cos x + Q \log \left| \frac{\sqrt{2}\cos x - 1}{\sqrt{2}\cos x + 1} \right| + C$, then the values of $P$ and $Q$ are respectively:

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

Hint:

For integrals with mixed trigonometric powers, the substitution $u = \cos x$ often converts the integrand into a rational function that is easier to handle.

Answer 1

The Correct Option is B

Step 1: Concept
Use the identity $\cos 2x = 2\cos^2 x - 1$ and factor the numerator using $\sin^3 x = \sin x(1-\cos^2 x)$.

Step 2: Simplification
The numerator becomes $\sin x + \sin x(1-\cos^2 x) = \sin x(2-\cos^2 x)$. Thus: \[I = \int \frac{\sin x(2-\cos^2 x)}{2\cos^2 x - 1}\,dx.\]

Step 3: Analysis
Substitute $u = \cos x$, $du = -\sin x\,dx$: \[I = -\int \frac{2-u^2}{2u^2-1}\,du = \int \frac{u^2-2}{2u^2-1}\,du = \int\!\left(\frac{1}{2} - \frac{3/2}{2u^2-1}\right)du.\] \[= \frac{1}{2}u - \frac{3}{2}\cdot\frac{1}{2\sqrt{2}}\log\left|\frac{\sqrt{2}u-1}{\sqrt{2}u+1}\right| + C = \frac{1}{2}\cos x - \frac{3}{4\sqrt{2}}\log\left|\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}\right| + C.\]

Step 4: Conclusion
Comparing with the given form: $P = \dfrac{1}{2}$ and $Q = \dfrac{-3}{4\sqrt{2}}$. Final Answer: (B)