JEE Main 2020 (Online) 3rd September Evening ShiftQuestion 1
If $\int\sin^{-1}\left(\sqrt{\frac x{1+x}}\right)dx=A(x)\tan^{-1}(\sqrt x)+B(x)+C$, where $C$ is a constant of integration, then the ordered pair $(A(x),B(x))$ can be :
Indefinite Integration
Solution
Concept. $\sin^{-1}\sqrt{\frac{x}{1+x}}=\tan^{-1}\sqrt x$; integrate by parts. Formulas used. (1) If $\tan\theta=\sqrt x$, then $\sin\theta=\sqrt{\dfrac{x}{1+x}}$(2) $\int u\,dv=uv-\int v\,du$Step 1 - Rewrite. By (1), the integrand is $\tan^{-1}\sqrt x$. Step 2 - By parts. With $u=\tan^{-1}\sqrt x$, $dv=dx$: $du=\dfrac{1}{1+x}\cdot\dfrac{1}{2\sqrt x}\,dx$, so $$\int\tan^{-1}\sqrt x\,dx=x\tan^{-1}\sqrt x-\int\frac{\sqrt x}{2(1+x)}\,dx.$$ Step 3 - The last integral. Put $x=s^2$: $\displaystyle\int\frac{s\cdot2s}{2(1+s^2)}\,ds=\int\left(1-\frac{1}{1+s^2}\right)ds=\sqrt x-\tan^{-1}\sqrt x$. Step 4 - Combine. $$x\tan^{-1}\sqrt x-\sqrt x+\tan^{-1}\sqrt x=(x+1)\tan^{-1}\sqrt x-\sqrt x+C.$$ So $(A(x),B(x))=(x+1,-\sqrt x)$. Answer: (A)