Show that

Question:

$\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}, x \in[0,1]$

Solution:

Let $I=\int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} x+\cos ^{-1} x} d x$

It is known that, $\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}=\frac{\pi}{2}$

$\Rightarrow I=\int \frac{\left(\frac{\pi}{2}-\cos ^{-1} \sqrt{x}\right)-\cos ^{-1} \sqrt{x}}{\frac{\pi}{2}} d x$

$=\frac{2}{\pi} \int\left(\frac{\pi}{2}-2 \cos ^{-1} \sqrt{x}\right) d x$

$=\frac{2}{\pi} \cdot \frac{\pi}{2} \int 1 . d x-\frac{4}{\pi} \int \cos ^{-1} \sqrt{x} d x$

$=x-\frac{4}{\pi} \int \cos ^{-1} \sqrt{x} d x$                     ...(1)

Let $I_{1}=\int \cos ^{-1} \sqrt{x} d x$

Also, let $\sqrt{x}=t \Rightarrow d x=2 t d t$

$\Rightarrow I_{1}=2 \int \cos ^{-1} t \cdot t d t$

$=2\left[\cos ^{-1} t \cdot \frac{t^{2}}{2}-\int \frac{-1}{\sqrt{1-t^{2}}} \cdot \frac{t^{2}}{2} d t\right]$

$=t^{2} \cos ^{-1} t+\int \frac{t^{2}}{\sqrt{1-t^{2}}} d t$

$=t^{2} \cos ^{-1} t-\int \frac{1-t^{2}-1}{\sqrt{1-t^{2}}} d t$

$=t^{2} \cos ^{-1} t-\int \sqrt{1-t^{2}} d t+\int \frac{1}{\sqrt{1-t^{2}}} d t$

$=t^{2} \cos ^{-1} t-\frac{t}{2} \sqrt{1-t^{2}}-\frac{1}{2} \sin ^{-1} t+\sin ^{-1} t$

$=t^{2} \cos ^{-1} t-\frac{t}{2} \sqrt{1-t^{2}}+\frac{1}{2} \sin ^{-1} t$

From equation (1), we obtain

$I=x-\frac{4}{\pi}\left[t^{2} \cos ^{-1} t-\frac{t}{2} \sqrt{1-t^{2}}+\frac{1}{2} \sin ^{-1} t\right]$

$=x-\frac{4}{\pi}\left[x \cos ^{-1} \sqrt{x}-\frac{\sqrt{x}}{2} \sqrt{1-x}+\frac{1}{2} \sin ^{-1} \sqrt{x}\right]$

$=x-\frac{4}{\pi}\left[x\left(\frac{\pi}{2}-\sin ^{-1}(\sqrt{x})\right)-\frac{\sqrt{x-x^{2}}}{2}+\frac{1}{2} \sin ^{-1}(\sqrt{x})\right]$

$=x-2 x+\frac{4 x}{\pi} \sin ^{-1} \sqrt{x}+\frac{2}{\pi} \sqrt{x-x^{2}}-\frac{2}{\pi} \sin ^{-1} \sqrt{x}$

$=-x+\frac{2}{\pi}\left[(2 x-1) \sin ^{-1} \sqrt{x}\right]+\frac{2}{\pi} \sqrt{x-x^{2}}+\mathrm{C}$

$=\frac{2(2 x-1)}{\pi} \sin ^{-1} \sqrt{x}+\frac{2}{\pi} \sqrt{x-x^{2}}-x+\mathrm{C}$

Leave a comment

None
Free Study Material