JEE Main 2020 (Online) 8th January Morning ShiftQuestion 1
If $\int\frac{\cos x\,dx}{\sin^3x\left(1+\sin^6x\right)^{2/3}}=f(x)\left(1+\sin^6x\right)^{1/\lambda}+c$ where $c$ is a constant of integration, then $\lambda f\left(\frac\pi3\right)$ is equal to :
Indefinite Integration
Solution
Concept. Put $u=\sin x$, then take $u^6$ out of the bracket; the rest is $w=1+u^{-6}$. Formulas used. (1) $w=1+u^{-6}\Rightarrow dw=-6u^{-7}\,du$Step 1 - Put $u=\sin x$. $du=\cos x\,dx$ and $$\int\frac{du}{u^3(1+u^6)^{2/3}}=\int\frac{du}{u^3\cdot u^4(1+u^{-6})^{2/3}}=\int\frac{u^{-7}\,du}{(1+u^{-6})^{2/3}}.$$ Step 2 - Use (1). $$-\frac16\int w^{-2/3}\,dw=-\frac12\,w^{1/3}+c=-\frac{1}{2\sin^2x}\left(1+\sin^6x\right)^{1/3}+c,$$ since $w^{1/3}=\dfrac{(1+u^6)^{1/3}}{u^2}$. Step 3 - Read off. $\lambda=3$ and $f(x)=-\dfrac{1}{2\sin^2x}$, so $f\left(\frac\pi3\right)=-\dfrac{1}{2\cdot\frac34}=-\dfrac23$. Step 4 - Value. $\lambda f\left(\frac\pi3\right)=-2$. Answer: (D)