**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**:

Also, here are the Pythagorean Identities we learned about in the **Trigonometric Identities** section:

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}$

**For Practice**: Use the **Mathway** widget below to try a problem. Click on **Submit** (the blue arrow to the right of the problem) and click on **Examples**, **Integrals, Substitution Rule** to try a problem.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

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