Question

The value of $[\stackrel{{\sin }^2\theta }{\underset{0}{\int }}{\sin }^{-1}\sqrt{\varphi }d\varphi +\stackrel{{\cos }^2\theta }{\underset{0}{\int }}{\cos }^{-1}\sqrt{\varphi }d\varphi ]$ is equal to.

• A. $\pi$
• B. $\pi /2$
• C. $\pi /3$
• D. $\pi /4$

Answer 1

Answer: (D)

$\pi /4$

Given : $I={\int }_0^{{\sin }^2\theta }{\sin }^{-1}\sqrt{\theta }d\theta +{\int }_0^{{\cos }^2\theta }{\cos }^{-1}\sqrt{\theta }d\theta$

Let :

$\varphi =t^2\varphi \to {\sin }^2\theta t\to \sin \theta \varphi \to {\cos }^2\theta t\to \cos \theta d\varphi =2tdt\varphi \to 0t\to 0$

$I={\int }_0^{\sin \theta }2t{\sin }^{-1}tdt+{\int }_0^{\cos \theta }2t{\cos }^{-1}\theta dt$ Using by parts

$I={\int }_0^{\sin \theta }2t{\sin }^{-1}tdt+{\int }_0^{\cos \theta }2t{\cos }^{-1}\theta dtI=2[{\left[\right({\sin }^{-1}t)\frac{t^2}{2}-\int \frac{t^{2}dt}{2\sqrt{1-t^2}}]}_0^{\sin \theta }+{\left[\right({\cos }^{-1}t)\frac{t^2}{2}-\int \frac{t^{2}dt}{2\sqrt{1-t^2}}]}_0^{\cos \theta }]I=[\left({\sin }^{-1}t\right)t^2+\int {[\sqrt{1-t^2}-\frac{1}{\sqrt{1-t^2}}dt]}_0^{\sin \theta }+\left({\cos }^{-1}t\right)t^2+\int {[\sqrt{1-t^2}-\frac{1}{\sqrt{1-t^2}}dt]}_0^{\cos \theta }]I={\left[\left({\sin }^{-1}t\right)t^2\right]}_0^{\sin \theta }+{\left[\left({\cos }^{-1}t\right)t^2\right]}_0^{\cos \theta }+{[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t]}_0^{\sin \theta }-{[-{\sin }^{-1}\left(t\right)]}_0^{\sin \theta }+{[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t]}_0^{\cos \theta }+{\left[{\sin }^{-1}\left(t\right)\right]}_0^{\cos \theta }I=\left[\right(\theta {\sin }^2\theta +\theta {\cos }^2\theta )+\frac{\sin \theta \cos \theta }{2}+\frac{\theta }{2}-\theta -\frac{\cos \theta \cos \theta }{2}-\frac{1}{2}(\frac{\pi }{2}-0)+(\frac{\pi }{2}-0)]I=[\theta -\theta +\frac{\theta }{2}+\frac{\theta }{2}-\theta +\frac{\pi }{2}-\frac{\pi }{4}]I=\frac{\pi }{4}$

Hence the correct answer is $\frac{\pi }{4}$