# Find the general solution $\large\frac{dy}{dx}=$$\sin^{-1}x \begin{array}{1 1}y = x\sin^{-1}x + \sqrt { 1 - x^2} + C \\ y = x\cos^{-1}x - \sqrt { 1 - x^2} + C \\y = x\tan^{-1}x + \sqrt { 1 + x^2} + C \\ y = x\cos^{-1}x + \sqrt { 1 + x^2} + C\end{array} ## 1 Answer Toolbox: • Integration by parts is \int udv = uv - \int vdu Step 1: Given:\large\frac{ dy}{dx}$$= \sin^{-1}x$
Seperating the variables we get
$dy = \sin^{-1}x\; dx$
Integrating on both sides we get
$\int dy = \int\sin^{-1}xdx$
Integrating the RHS by parts using the information in the tool box we get,
Let $u = \sin^{-1}x\; du =\large\frac{ 1}{\sqrt{1-x^2}}$ and $dv = dx$ hence $x = v$
Step 2:
substituting in the formula we get
$\int y = x\sin^{-1}x -\int\large\frac{ xdx}{\sqrt{1-x^2 }}$
applying the substitution method we get,
Let $1-x^2 = t$, hence $-2xdx = dt$ or $xdx = \large\frac{-dt}{2}$
substituting this
$y = x\sin^{-1}x -\int \large\frac{(-1/2)dt}{\sqrt { t}}$
$y = x\sin^{-1}x +\large\frac{\Large\frac{ 1}{2 \sqrt t}}{\Large\frac{1}{2}}$
$y = x\sin^{-1}x + \sqrt { 1 - x^2} + C$