Question

If \( \int \frac{\sin x}{\sin^3 x + \cos^3 x} dx = \alpha \log_e |1 + \tan x| + \beta \log_e |1 - \tan x + \tan^2 x| + \gamma \tan^{-1} \left( \frac{2 \tan x - 1}{\sqrt{3}} \right) + C \), when C is a constant of integration, then the value of \( 18(\alpha + \beta + \gamma^2) \) is _________.

Hint:

For integrals like \( \int \frac{x}{(x+a)(x^2+bx+c)} dx \), partial fractions is the standard route. Always check if a simple substitution like \( t = x^2 \) or \( t = x+a \) simplifies the quadratic part first.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
To integrate a function involving powers of trigonometric terms, we often divide the numerator and denominator by a suitable power of \(\cos x\) to transform the expression into a function of \(\tan x\).
Step 2: Detailed Explanation:
Divide numerator and denominator by \(\cos^3 x\): \[ I = \int \frac{\tan x \sec^2 x}{\tan^3 x + 1} dx \] Let \( u = \tan x \), then \( du = \sec^2 x dx \). \[ I = \int \frac{u}{u^3 + 1} du = \int \frac{u}{(u+1)(u^2-u+1)} du \] Using partial fractions: \[ \frac{u}{(u+1)(u^2-u+1)} = \frac{A}{u+1} + \frac{Bu+C}{u^2-u+1} \] Solving gives \( A = -1/3, B = 1/3, C = 1/3 \). \[ I = -\frac{1}{3} \int \frac{1}{u+1} du + \frac{1}{3} \int \frac{u+1}{u^2-u+1} du \] \[ I = -\frac{1}{3} \log|u+1| + \frac{1}{6} \int \frac{2u+2}{u^2-u+1} du \] \[ I = -\frac{1}{3} \log|u+1| + \frac{1}{6} \log|u^2-u+1| + \frac{1}{2} \int \frac{1}{(u-1/2)^2 + (\sqrt{3}/2)^2} du \] \[ I = -\frac{1}{3} \log|1+\tan x| + \frac{1}{6} \log|1-\tan x+\tan^2 x| + \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{2\tan x-1}{\sqrt{3}} \right) + C \] Comparing with the given form: \( \alpha = -1/3, \beta = 1/6, \gamma = 1/\sqrt{3} \).
Calculate \( 18(\alpha + \beta + \gamma^2) \): \[ 18(-1/3 + 1/6 + 1/3) = 18(1/6) = 3 \]
Step 3: Final Answer:
The value is 3.