Integration by Trigonometric Substitution
Depending on the function we need to integrate, we can use this trigonometric expression as substitution to simplify the integral:
1. When $ \displaystyle \sqrt{{{{a}^{2}}-{{b}^{2}}{{x}^{2}}}}$ then substitute $ \displaystyle x=\frac{a}{b}\sin \theta $ and the helpful trigonometric identities is $ \displaystyle {{\sin }^{2}}x=1-{{\cos }^{2}}x$
2. When $ \displaystyle \sqrt{{{{a}^{2}}+{{b}^{2}}{{x}^{2}}}}$ then substitute $ \displaystyle x=\frac{a}{b}\tan \theta $ and the helpful trigonometric identities is $ \displaystyle {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $
3. When $ \displaystyle \sqrt{{{{b}^{2}}{{x}^{2}}-{{a}^{2}}}}$ then substitute $ \displaystyle x=\frac{a}{b}\sec \theta $ and the helpful trigonometric identities is $ \displaystyle {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$
Example 1: Find the integral $ \displaystyle \int{{\frac{{dx}}{{\sqrt{{{{x}^{2}}+16}}}}}}$?
If we rewrite the expression inside the integral as $ \displaystyle \sqrt{{{{x}^{2}}+16}}=\sqrt{{{{{(4)}}^{2}}+{{x}^{2}}}}$ then we see that we have the second form above that is:
$ \displaystyle \sqrt{{{{a}^{2}}+{{b}^{2}}{{x}^{2}}}}$, with a=4 and b=1
So we will substitute $ \displaystyle x=4\tan \theta $ and have $ \displaystyle dx=4{{\sec }^{2}}\theta d\theta $
The helpful trigonometric identities we will use is $ \displaystyle {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $
Then we will have $ \displaystyle \sqrt{{{{{(4)}}^{2}}+{{x}^{2}}}}=\sqrt{{{{{(4)}}^{2}}+{{{(4\tan \theta )}}^{2}}}}=$
$ \displaystyle \sqrt{{16+16{{{\tan }}^{2}}\theta }}=\sqrt{{16(1+{{{\tan }}^{2}}\theta )}}=$
$ \displaystyle \sqrt{{16{{{\sec }}^{2}}\theta }}=4\sec \theta $
Substituting $ \displaystyle 4\sec \theta $ and $ \displaystyle dx=4{{\sec }^{2}}\theta d\theta $ into the given integral:
$ \displaystyle \int{{\frac{{dx}}{{\sqrt{{{{x}^{2}}+16}}}}}}=\int{{\frac{{4{{{\sec }}^{2}}\theta d\theta }}{{4\sec \theta }}}}=$
$ \displaystyle \int{{\sec \theta =\ln \left| {\tan \theta +\sec \theta } \right|}}+C$
Now we write the answer in terms of x.
$ \displaystyle \tan \theta =\frac{x}{4}$
$ \displaystyle \sec \theta =\sqrt{{1+{{{\tan }}^{2}}\theta }}=$
$ \displaystyle \sqrt{{1+{{{\left( {\frac{x}{4}} \right)}}^{2}}}}=\sqrt{{1+\frac{{{{x}^{2}}}}{{16}}}}$
Substituting, we get the final answer:
$ \displaystyle \int{{\frac{{dx}}{{\sqrt{{{{x}^{2}}+16}}}}}}=\ln \left| {\frac{x}{4}+\sqrt{{1+\frac{{{{x}^{2}}}}{{16}}}}} \right|+C$
Example 2: Find the integral $ \displaystyle \int{{\sqrt{{4{{x}^{2}}-9}}}}dx$?
If we rewrite the expression inside the integral as $ \displaystyle \sqrt{{4{{x}^{2}}-9}}=\sqrt{{{{{(2x)}}^{2}}-{{{(3)}}^{2}}}}$ then we see that we have the third form above that is:
$ \displaystyle \sqrt{{{{b}^{2}}{{x}^{2}}-{{a}^{2}}}}$, with a=3 and b=2
So we will$ \displaystyle x=\frac{3}{2}\sec \theta $ and $ \displaystyle dx=\frac{3}{2}\sec \theta \tan \theta d\theta $
The helpful trigonometric identities we will use$ \displaystyle {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$
Then we will have $ \displaystyle \sqrt{{4{{x}^{2}}-9}}=\sqrt{{4{{{(\frac{3}{2}\sec \theta )}}^{2}}-9}}$
$ \displaystyle \sqrt{{9{{{\sec }}^{2}}\theta -9}}=\sqrt{{9({{{\sec }}^{2}}\theta -1)}}$
$ \displaystyle \sqrt{{{{{\tan }}^{2}}\theta }}=\tan \theta $
Substituting $ \displaystyle x=\frac{3}{2}\sec \theta $ and $ \displaystyle dx=\frac{3}{2}\sec \theta \tan \theta d\theta $ into the given integral:
$ \displaystyle \int{{\sqrt{{4{{x}^{2}}-9}}}}dx=\int{{\tan \theta }}\frac{3}{2}\sec \theta \tan \theta d\theta =$
$ \displaystyle \frac{3}{2}\int{{{{{\tan }}^{2}}}}\theta \sec \theta d\theta =\frac{3}{2}\int{{({{{\sec }}^{2}}\theta -1)}}\sec \theta d\theta =$
$ \displaystyle \frac{3}{2}\int{{({{{\sec }}^{3}}\theta -\sec \theta }})d\theta =\frac{3}{2}\int{{{{{\sec }}^{3}}}}\theta d\theta -\frac{3}{2}\int{{\sec \theta d\theta }}=$
$ \displaystyle \frac{3}{2}(\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}\ln \left| {\sec \theta +\tan \theta } \right|)-\frac{3}{2}\ln \left| {\sec \theta +\tan \theta } \right|+C=$
$ \displaystyle \frac{3}{4}(\sec \theta \tan \theta -\ln (\sec \theta +\tan \theta )+C$
Now we write the answer in terms of x.
$ \displaystyle \sec \theta =\frac{{2x}}{3}$ or ($ \displaystyle \cos \theta =\frac{3}{{2x}}$)
Using Pythagorean Theorem we find $ \displaystyle \tan \theta $.
$ \displaystyle \tan \theta =\frac{{\sqrt{{9+4{{x}^{2}}}}}}{3}$
Substituting, we get the final answer:
$ \displaystyle \int{{\sqrt{{4{{x}^{2}}-9}}}}dx=$
$ \displaystyle \frac{3}{4}(\frac{{2x}}{3}\cdot \frac{{\sqrt{{9+4{{x}^{2}}}}}}{3}-\ln \left| {\frac{{2x}}{3}+\frac{{\sqrt{{9+4{{x}^{2}}}}}}{3}} \right|)+C$
Example 3: Find the integral $ \displaystyle \int{{\sqrt{{36-{{x}^{2}}}}}}dx$?
If we rewrite the expression inside the integral as $ \displaystyle \sqrt{{36-{{x}^{2}}}}=\sqrt{{{{6}^{2}}-{{x}^{2}}}}$ then we see that we have the first form above that is:
$ \displaystyle \sqrt{{{{a}^{2}}-{{b}^{2}}{{x}^{2}}}}$, with a=6 and b=1.
So we will $ \displaystyle x=6\sin \theta $ than $ \displaystyle dx=6\cos \theta d\theta $
The helpful trigonometric identities we will use is $ \displaystyle {{\sin }^{2}}x=1-{{\cos }^{2}}x$
Then we will have $ \displaystyle \sqrt{{36-{{x}^{2}}}}=\sqrt{{36-{{{(6\sin \theta )}}^{2}}}}=$
$ \displaystyle \sqrt{{36-36{{{\sin }}^{2}}\theta }}=\sqrt{{36(1-{{{\sin }}^{2}}\theta )}}=$
$ \displaystyle \sqrt{{36{{{\cos }}^{2}}\theta }}=6\cos \theta $
Substituting $ \displaystyle x=6\sin \theta $ and $ \displaystyle dx=6\cos \theta d\theta $ into the given integral and using power reducing or half angles identities $ \displaystyle \int{{\sqrt{{36-{{x}^{2}}}}}}dx=\int{{6\cos \theta \cdot }}6\cos \theta d\theta =$
$ \displaystyle 36\int{{{{{\cos }}^{2}}\theta d\theta =}}36\int{{\frac{{1+\cos 2\theta }}{2}}}d\theta =$
$ \displaystyle 18\int{{(1+\cos 2\theta )d\theta =18\theta }}+9\sin 2\theta +C$
Now we write the answer in terms of x.
$ \displaystyle \sin \theta =\frac{x}{6}$
$ \displaystyle \theta =\arcsin \frac{x}{6}$
Substituting, we get the final answer:
$ \displaystyle \int{{\sqrt{{36-{{x}^{2}}}}}}dx=18\arcsin \frac{x}{6}+9\cdot \frac{x}{6}+C$
Example 4: Find the integral $\displaystyle \int{{\frac{{5dx}}{{x\sqrt{{9-{{x}^{2}}}}}}}}$
This integral contains an expression of the first form above $ \displaystyle \sqrt{{{{a}^{2}}-{{b}^{2}}{{x}^{2}}}}$ with a=3 and b=1.
So we will substitute $ \displaystyle x=3\sin \theta $ then $ \displaystyle dx=3\cos \theta d\theta $
The helpful trigonometric identities we will use is $ \displaystyle {{\sin }^{2}}x=1-{{\cos }^{2}}x$
Then we will have $ \displaystyle \sqrt{{9-{{x}^{2}}}}=\sqrt{{9-{{{(3\sin \theta )}}^{2}}}}=$
$ \displaystyle \sqrt{{9-9{{{\sin }}^{2}}\theta }}=\sqrt{{9(1-{{{\sin }}^{2}}\theta )}}=$
$ \displaystyle \sqrt{{9{{{\cos }}^{2}}\theta }}=3\cos \theta $
Substituting $ \displaystyle x=3\sin \theta $ and $ \displaystyle dx=3\cos \theta d\theta $
Into the given integral:
$ \displaystyle \int{{\frac{{5dx}}{{x\sqrt{{9-{{x}^{2}}}}}}}}=\int{{\frac{{5\cdot 3\cos \theta d\theta }}{{3\sin \theta \cdot 3\cos \theta }}}}=$
$ \displaystyle \int{{\frac{{15\cos \theta d\theta }}{{9\sin \theta \cos \theta }}}}=\frac{5}{3}\int{{\frac{1}{{\sin \theta }}}}d\theta =$
$ \displaystyle \frac{5}{3}\int{{\cos ecd\theta }}=\frac{5}{3}\ln \left| {\cos ec\theta -\cot \theta } \right|+C$
Now we write the answer in terms of x.
$ \displaystyle \sin \theta =\frac{x}{3}$
$ \displaystyle \cos ec\theta =\frac{3}{x}$
$ \displaystyle \cot \theta =\frac{{\sqrt{{9-{{x}^{2}}}}}}{3}$
Substituting, we get the final answer:
$ \displaystyle \int{{\frac{{5dx}}{{x\sqrt{{9-{{x}^{2}}}}}}}}=\frac{5}{3}\ln \left| {\frac{3}{x}-\frac{{\sqrt{{9-{{x}^{2}}}}}}{3}} \right|+C$
Example 5: Find the integral $ \displaystyle \int{{\frac{x}{{\sqrt{{2{{x}^{2}}-8x-8}}}}}}dx$?
On a first look the expression under the integral doesn’t fit in any of those categories above, but
If we complete the square of the quadratic equation we can make it look like one of those forms above:
$ \displaystyle 2{{x}^{2}}-8x-8=2({{x}^{2}}-4x-4)=2{{(x-2)}^{2}}-16$
So the roots becomes $ \displaystyle \sqrt{{2{{x}^{2}}-8x-8}}=\sqrt{{2{{{(x-2)}}^{2}}-16}}$
Now this looks like the third form $ \displaystyle \sqrt{{{{b}^{2}}{{x}^{2}}-{{a}^{2}}}}$ with the difference that we have x-2 except x, but we will do the same substitution $ \displaystyle x-2=\frac{a}{b}\sec \theta $
So we will substitute $ \displaystyle x-2=\frac{4}{{\sqrt{2}}}\sec \theta $ then $ \displaystyle x=\frac{4}{{\sqrt{2}}}\sec \theta +2$ and $ \displaystyle dx=\frac{4}{{\sqrt{2}}}\sec \theta \tan \theta d\theta $
The helpful trigonometric identities we will use is $ \displaystyle {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$
Then we will have $ \displaystyle \sqrt{{2{{{(x-2)}}^{2}}-16}}=\sqrt{{2(\frac{4}{{\sqrt{2}}}\sec \theta +2-2)-16}}=$
$ \displaystyle \sqrt{{16{{{\sec }}^{2}}\theta -16}}=\sqrt{{16({{{\sec }}^{2}}\theta -1)}}=$
$ \displaystyle \sqrt{{16{{{\tan }}^{2}}\theta }}=4\tan \theta $
Substituting $ \displaystyle x=\frac{4}{{\sqrt{2}}}\sec \theta +2$ and $ \displaystyle dx=\frac{4}{{\sqrt{2}}}\sec \theta \tan \theta d\theta $
Into the given integral:
$ \displaystyle \int{{\frac{x}{{\sqrt{{2{{x}^{2}}-8x-8}}}}}}dx=\int{{\frac{{\frac{4}{{\sqrt{2}}}\sec \theta +2}}{{4\tan \theta }}}}\frac{4}{{\sqrt{2}}}\sec \theta \tan \theta d\theta =$
$ \displaystyle \int{{(\frac{{4{{{\sec }}^{2}}\theta }}{2}+\frac{{2\sec \theta }}{{\sqrt{2}}}}})d\theta =\int{{(2{{{\sec }}^{2}}\theta +}}\frac{{2\sec \theta }}{{\sqrt{2}}})d\theta =$
$ \displaystyle 2\tan \theta +\frac{2}{{\sqrt{2}}}\ln \left| {\sec \theta +\tan \theta } \right|+C$
Now we write the answer in terms of x.
$ \displaystyle \sec \theta =\frac{{\sqrt{2}(x-2)}}{4}$
$ \displaystyle \cos \theta =\frac{4}{{\sqrt{2}(x-2)}}$
Using Pythagorean Theorem we find:
$ \displaystyle \tan \theta =\frac{{\sqrt{{2{{x}^{2}}-8x-8}}}}{4}$
Substituting, we get the final answer:
A summary of the trigonometric substitution you should make in each case:
