Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Definite Integration

Question:

The value of $\int\limits_0^{\sin ^2 x} \sin ^{-1} \sqrt{t} d t+\int\limits_0^{\cos ^2 x} \cos ^{-1} \sqrt{t} d t$, is

Options:

$\pi$

$\frac{\pi}{2}$

$\frac{\pi}{4}$

none of these

Correct Answer:

$\frac{\pi}{4}$

Explanation:

Let $\phi(x)=\int\limits_0^{\sin ^2 x} \sin ^{-1} \sqrt{t} d t+\int\limits_0^{\cos ^2 x} \cos ^{-1} \sqrt{t} d t$

Then,

$\frac{d \phi}{d x}=\int\limits_0^{\sin ^2 x} 0 d t+\left\{\frac{d}{d x}\left(\sin ^2 x\right)\right\} \times \sin ^{-1}\left(\sqrt{\sin ^2 x}\right)-0 + \int\limits_0^{\cos ^2 x} 0 d t+\left\{\frac{d}{d x}\left(\cos ^2 x\right)\right\} \times \cos ^{-1}\left(\sqrt{\cos ^2 x}\right)-0$

$\Rightarrow \frac{d \phi}{d x} =(2 \sin x \cos x) x-(2 \sin x \cos x) x$

$\Rightarrow \frac{d \phi}{d x}=0$ for all $x$

∴  $\phi(x)=$ Constant for all $x$

Let $\phi(x)=k$, for all $x$           .......(i)

$\Rightarrow \phi\left(\frac{\pi}{4}\right)=k$

$\Rightarrow \int\limits_0^{1 / 2} \sin ^{-1} \sqrt{t} d t+\int\limits_0^{1 / 2} \cos ^{-1} \sqrt{t} d t=k$

$\Rightarrow \int\limits_0^{1 / 2}\left(\sin ^{-1} \sqrt{t}+\cos ^{-1} \sqrt{t}\right) d t=k$

$\Rightarrow \int\limits_0^{1 / 2} \frac{\pi}{2} d t=k \Rightarrow k=\frac{\pi}{4}$

Putting $k=\pi / 4$ in (i), we get $\phi(x)=\pi / 4$ for all $x$.