Trigonometric substitution isn’t typically taught in the AP AB class, but it can be very useful. The techniques are a more generic form of the techniques that we used here in the Derivatives and Integrals of Inverse Trigonometric Functions section and uses Right Triangle Trigonometry. Notice U-Substitution Integration will typically be used in the more complicated problems.
Basically, Trigonometric Substitutions are techniques to evaluate integrals with radical functions, where the radicals are replaced by trigonometric expressions before the integration is done. Then, at the end, the trig functions are replaced back by the algebraic expressions! They can get a lot more complicated and can be solved by completing the square, but I’ll only address some “simplistic” problems. Notice also that not every integral with radical expressions can be solved by trig substitution!
Here are the inverse trig integration rules we learned about in the Derivatives and Integrals of Inverse Trigonometric Functions:
Integrals Involving the Inverse Trig Functions
$ \displaystyle \begin{align}\int{{\frac{{du}}{{\sqrt{{1-{{u}^{2}}}}}}}}\,=\arcsin \,u+C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int{{\frac{{du}}{{\sqrt{{{{a}^{2}}-{{u}^{2}}}}}}}}\,=\arcsin \,\frac{u}{a}+C\\\int{{\frac{{du}}{{1+{{u}^{2}}}}}}\,=\arctan \,u+C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int{{\frac{{du}}{{{{a}^{2}}+{{u}^{2}}}}}}\,=\frac{1}{a}\arctan \,\frac{u}{a}+C\\\int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-1}}}}}}\,=\text{arcsec}\,\left| u \right|+C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-{{a}^{2}}}}}}}}\,=\frac{1}{a}\text{arcsec}\,\frac{{\left| u \right|}}{a}+C\end{align}$
Also, here are the Pythagorean Identities we learned about in the Trigonometric Identities section:
$ {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \,\,\,\,\,\,\,\,\,\,\,\,{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \,\,\,\,\,\,\,\,\,\,\,\,{{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$
Putting this altogether, we can come up with the following. You don’t really need to memorize this; I just like to just set up Pythagorean Triangles to get the same results, as in the problems below.
Here are some problems:
Problem: $ \displaystyle \int{{\frac{{\sqrt{{4-{{x}^{2}}}}}}{{{{x}^{2}}}}\,\,dx}}$
Solution: We have $ \sqrt{{{{a}^{2}}-{{x}^{2}}}}=\sqrt{{{{2}^{2}}-{{x}^{2}}}}$, so fill in triangle so that $ \displaystyle \sin \theta =\frac{x}{a}=\frac{x}{2}$. Thus, $ x=2\sin \theta $, and $ dx=2\cos \theta \,d\theta $. From the triangle, $ \displaystyle \cos \theta =\frac{{\sqrt{{4-{{x}^{2}}}}}}{2};\,\,\,\,\sqrt{{4-{{x}^{2}}}}=2\cos \theta $. 
Now perform substitution so that trig functions are gone:
$ \displaystyle \begin{align}\int{{\frac{{\sqrt{{4-{{x}^{2}}}}}}{{{{x}^{2}}}}\,dx}}&=\int{{\frac{{2\cos \theta }}{{{{{\left( {2\sin \theta } \right)}}^{2}}}}\,\cdot 2\cos \theta d\theta }}\\&=\int{{\frac{{4{{{\cos }}^{2}}\theta }}{{4{{{\sin }}^{2}}\theta }}\,d\theta }}=\int{{{{{\cot }}^{2}}\theta \,d\theta }}\\&=\int{{\left( {{{{\csc }}^{2}}\theta -1} \right)\,d\theta }}\\&=-\cot \theta -\theta \\&=-\frac{{\sqrt{{4-{{x}^{2}}}}}}{x}-{{\sin }^{{-1}}}\left( {\frac{x}{2}} \right)+C\end{align}$
Problem: $ \displaystyle \int{{\frac{1}{{\sqrt{{16+{{x}^{2}}}}}}dx}}$
Solution: We have $ \sqrt{{{{a}^{2}}+{{x}^{2}}}}=\sqrt{{{{4}^{2}}+{{x}^{2}}}}$, so fill in triangle so that $ \displaystyle \tan \theta =\frac{x}{a}=\frac{x}{4}$. Thus, $ \displaystyle x=4\tan \theta $, and $ dx=4\,{{\sec }^{2}}\theta \,d\theta $. From the triangle, $ \displaystyle \sec \theta =\frac{{\sqrt{{16+{{x}^{2}}}}}}{4};\,\,\,\,\sqrt{{16+{{x}^{2}}}}=4\sec \theta $. 
Now perform substitution so that trig functions are gone:
$ \displaystyle \begin{align}\int{{\frac{x}{{\sqrt{{16+{{x}^{2}}}}}}\,dx}}&=\int{{\frac{1}{{4\sec \theta }}\,\cdot 4\,{{{\sec }}^{2}}\theta d\theta }}\\&=\int{{\sec \theta \,d\theta }}\\&=\ln \left| {\sec \theta +\tan \theta } \right|+C\\&=\ln \left| {\frac{{\sqrt{{16+{{x}^{2}}}}}}{4}+\frac{x}{4}} \right|+C\end{align}$