Simple integration just learning this application I want to integrate a function as $f(x)=\sin^{-1}x$. What should be the proper method of doing it?
 A: Notice, use integration by parts as follows $$\int \sin^{-1}(x)\ dx=\int \sin^{-1}(x)\cdot 1\ dx$$
$$=\sin^{-1}(x)\int 1\ dx-\int \left(\frac{d}{dx}(\sin^{-1}(x))\cdot \int 1\ dx\right)\ dx$$
$$=\sin^{-1}(x)\cdot(x)-\int \frac{1}{\sqrt{1-x^2}}\cdot (x)\ dx$$
$$=x\sin^{-1}(x)+\frac 12\int \frac{(-2x)}{\sqrt{1-x^2}}\ dx$$
$$=x\sin^{-1}(x)+\frac 12\int (1-x^2)^{-1/2}d(1-x^2)$$
$$=x\sin^{-1}(x)+\frac 12\cdot \frac{(1-x^2)^{1/2}}{1/2}+C$$
$$=\color{red}{x\sin^{-1}(x)+\sqrt{1-x^2}+C}$$
A: We can do it integration by parts method as taking f(x)= sin and f(x)2 = 1.
Also we can take x=sin(t) as second method.
My maths is this level i can solve this sum like this only..but i want a method that we can use directly.
A: Another thing you could try is substituting $x=\sin t$, so that $\mathrm{d}x=\cos t\,\mathrm{d}t$.
$$\int\sin^{-1}x\,\mathrm{d}x=\int t\cos t\,\mathrm{d}t$$
If you can find a way to integrate this without using integration by parts, then more power to you.
A: Integration by parts is the method for this one, but here is another way of thinking (I don't say it is really different from integration by parts, and I don't suggest to use this instead of integration by parts).
Let us think of $0<x<1$. Draw the graph of $y=\arcsin x$. 

From the fundamental theorem of calculus we know that $x\mapsto \int_0^x\arcsin t\,dt$ is a primitive of $x\mapsto\arcsin x$. From the area interpretation of integrals and from the figure, we find that (both sides correspond to the area of the rectangle)
$$
\int_0^{\arcsin x}\sin t\,dt+\int_0^x \arcsin t\,dt=x\arcsin x.
$$
Thus, using the fundamental theorem (the version with insertion of limits) on the integral with integrand $\sin t$ (which is easy to find a primitive to), we get 
$$
\begin{aligned}
\int_0^x\arcsin t\,dt&=x\arcsin x-\bigl[-\cos t\bigr]_0^{\arcsin x}\\
&=x\arcsin x+\cos(\arcsin x)-1\\
&=x\arcsin x+\sqrt{1-x^2}-1.
\end{aligned}
$$
Here we have used the fact that (in the given domain) $\cos(\arcsin x)=\sqrt{1-\sin^2(\arcsin x)}=\sqrt{1-x^2}$. We conclude that (invoking the $-1$ into the arbitrary constant $C$)
$$
\int \arcsin x\,dx=x\arcsin x+\sqrt{1-x^2}+C.
$$
A: HINT:
When integrating by parts in the middle a stage comes to integrate
$$ \int \frac{x\, dx}{\sqrt{1-x^2}} $$
$$= \dfrac12 \int \frac{2 x\, dx}{\sqrt{1-x^2}}= \dfrac12 \int \frac{d\,(x^2) }{\sqrt{1-x^2}}$$ 
which  can be integrated with $ x^2 = u$ substitution or directly. 
