# Show that

Question:

$\frac{d y}{d x}=\sin ^{-1} x$

Solution:

The given differential equation is:

$\frac{d y}{d x}=\sin ^{-1} x$

$\Rightarrow d y=\sin ^{-1} x d x$

Integrating both sides, we get:

$\int d y=\int \sin ^{-1} x d x$

$\Rightarrow y=\int\left(\sin ^{-1} x \cdot 1\right) d x$

$\Rightarrow y=\sin ^{-1} x \cdot \int(1) d x-\int\left[\left(\frac{d}{d x}\left(\sin ^{-1} x\right) \cdot \int(1) d x\right)\right] d x$

$\Rightarrow y=\sin ^{-1} x \cdot x-\int\left(\frac{1}{\sqrt{1-x^{2}}} \cdot x\right) d x$

$\Rightarrow y=x \sin ^{-1} x+\int \frac{-x}{\sqrt{1-x^{2}}} d x$               ...(1)

Let $1-x^{2}=t$.

$\Rightarrow \frac{d}{d x}\left(1-x^{2}\right)=\frac{d t}{d x}$

$\Rightarrow-2 x=\frac{d t}{d x}$

$\Rightarrow x d x=-\frac{1}{2} d t$

Substituting this value in equation (1), we get:

$y=x \sin ^{-1} x+\int \frac{1}{2 \sqrt{t}} d t$

$\Rightarrow y=x \sin ^{-1} x+\frac{1}{2} \cdot \int(t)^{-\frac{1}{2}} d t$

$\Rightarrow y=x \sin ^{-1} x+\frac{1}{2} \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\mathrm{C}$

$\Rightarrow y=x \sin ^{-1} x+\sqrt{t}+\mathrm{C}$

$\Rightarrow y=x \sin ^{-1} x+\sqrt{1-x^{2}}+\mathrm{C}$

This is the required general solution of the given differential equation.