Integration by parts of inverse sine, a solved exercise, some doubts...

[mcastillo356]

TL;DR Summary

There are steps I would like to understand, or, better said, share, check out.

Hi, PF, here goes an easy integral, meant to be an example of integration by parts.

Use integration by parts to evaluate
##\int \sin^{-1}x \, dx##

Let ##U=\sin^{-1}x,\quad{dV=dx}##
Then ##dU=dx/\sqrt{1-x^2},\quad{V=x}##

##=x\sin^{-1}x-\int \frac{x}{\sqrt{1-x^2} \, dx}##

Let ##u=1-x^2##,
##du=-2xdx##

##=x\sin^{-1}x+\frac{1}{2}\int u^{-1/2} \, du##
##=x\sin^{-1}x+u^{1/2}+C##
##=x\sin^{-1}x+\sqrt{1-x^2}+C##
Last substitution steps are where I need some clue: specifically understanding how, passed the first step, i.e, integration by parts, second integral is solved.

Attempt:
Second integral, ##\int \frac{x}{\sqrt{1-x^2}}\,dx##, is not integration by parts, but the substitution method. If so, I will outline it:

##\int \frac{1}{\sqrt{1-x^2}}\cdot x##, which suits the circumstances to evaluate by the substitution method they way it does.

Greetings!

 

[mathhabibi]

After letting ##u=1-x^2##, we have that ##\frac{du}{-2x}=dx##. Substituting this into the second integral, we get that it is equivalent to $$-\frac12\int u^{-1/2}du$$. Using power rule and replacing the definition of ##u##, we get the result.

 

[mcastillo356]

After letting ##u=1-x^2##, we have that ##\frac{du}{-2x}=dx##. Substituting this into the second integral, we get that it is equivalent to ##-\frac12\int u^{-1/2}du##. Using power rule and replacing the definition of ##u##, we get the result.

Hi, @mathhabibi, you mean power rule for integration?.
By the way, how is it to display a bigger and elegant indefinite integal?
Thanks!

 

[Mark44]

Hi, @mathhabibi, you mean power rule for integration?.

Yes.

By the way, how is it to display a bigger and elegant indefinite integal?

He used standalone (i.e. $$) tex delimiters rather than inline (##).

 

[mcastillo356]

Hi, PF

After letting ##u=1-x^2##, we have that ##\frac{du}{-2x}=dx##. Substituting this into the second integral, we get that it is equivalent to $$-\frac12\int u^{-1/2}du$$.

$$-\int {\frac{x}{\sqrt{1-x^2}}\,dx}\Leftrightarrow{\frac{1}{2}\int u^{-1/2}\,du}$$

Using power rule and replacing the definition of ##u##, we get the result.

$$\frac{1}{2}\int u^{-1/2}\,du$$
Applying the Power Rule for integration
$$=\frac{1}{2}\int {u^{-1/2}\,du}=\frac{1}{2}\cdot{\frac{u^{1/2}}{1/2}}+C=u^{1/2}+C$$
Replacing,
$$u=1-x^2\Rightarrow{u^{1/2}=\sqrt{1-x^2}}$$

Greetings!