JEE Main 2021 (Online) 31st August Evening ShiftQuestion 1
If $\int\frac{\sin x}{\sin^{3}x+\cos^{3}x}dx=\alpha\log_e|1+\tan x|+\beta\log_e\left|1-\tan x+\tan^{2}x\right|+\gamma\tan^{-1}\left(\frac{2\tan x-1}{\sqrt{3}}\right)+C$, when $C$ is constant of integration, then the value of $18(\alpha+\beta+\gamma^{2})$ is $\_\_\_\_\_\_$ .
Indefinite Integration
Answer$3$
Solution
Concept. Divide by $\cos^3x$: with $t=\tan x$ the integral is $\int\frac{t\,dt}{1+t^3}$; use partial fractions. Formulas used. (1) $\dfrac{t}{(1+t)(t^2-t+1)}=-\dfrac13\cdot\dfrac1{1+t}+\dfrac13\cdot\dfrac{t+1}{t^2-t+1}$(2) $t+1=\frac12(2t-1)+\frac32$ and $t^2-t+1=\big(t-\frac12\big)^2+\frac34$Step 1 - Substitute. $\frac{\sin x}{\cos^3x}=\tan x\sec^2x$, so $I=\int\frac{t\,dt}{1+t^3}$. Step 2 - Integrate (1) using (2). $$I=-\frac13\ln|1+t|+\frac16\ln(t^2-t+1)+\frac12\cdot\frac2{\sqrt3}\tan^{-1}\frac{2t-1}{\sqrt3}+C.$$ So $\alpha=-\frac13$, $\beta=\frac16$, $\gamma=\frac1{\sqrt3}$. Step 3 - Value. $18\Big(-\frac13+\frac16+\frac13\Big)=18\cdot\frac16=3$. Answer: 3