JEE Main 2013 — Mathematics Calculus

Consider $$\int_0^{\sin ^2 x} \sin ^{-1}(\sqrt{t}) d t+\int_0^{\cos ^2 x} \cos ^{-1}(\sqrt{t}) d t$$∫0sin2x​sin−1(t​)dt+∫0cos2x​cos−1(t​)dt Let $\mathrm{I}=f(x)$I=f(x) after integrating and putting the limits. $$\begin{aligned} &f^{\prime}(x)=\sin ^{-1} \sqrt{\sin ^2 x}(2 \sin x \cos x)-0 \\ &\quad+\cos ^{-1} \sqrt{\cos ^2 x}(-2 \cos x \sin x)-0 \\ &\therefore f^{\prime}(x)=0 \Rightarrow f(x)=\mathrm{C} \quad \text { (constant) } \end{aligned}$$​f′(x)=sin−1sin2x​(2sinxcosx)−0+cos−1cos2x​(−2cosxsinx)−0∴f′(x)=0⇒f(x)=C (constant) ​ Now, we find $f(x)$f(x) at $x=\frac{\pi}{4}$x=4π​ $$\begin{aligned} \therefore \mathrm{I} &=\int_0^{1 / 2} \sin ^{-1} \sqrt{t} d t+\int_0^{1 / 2} \cos ^{-1} \sqrt{t} d t \\ &=\int_0^{1 / 2}\left(\sin ^{-1} \sqrt{t}+\cos ^{-1} \sqrt{t}\right) d t \\ &=\int_0^{1 / 2} \frac{\pi}{2} d t=\frac{\pi}{4}=\mathrm{C} \\ \therefore \quad f(x)=\frac{\pi}{4} \\ \therefore \text { Required integration }=\frac{\pi}{4} \end{aligned}$$∴I∴f(x)=4π​∴ Required integration =4π​​=∫01/2​sin−1t​dt+∫01/2​cos−1t​dt=∫01/2​(sin−1t​+cos−1t​)dt=∫01/2​2π​dt=4π​=C​