Question

What is the integral of `tan^3xsecxdx`?

Answer 1

The answer can be written as: `\frac{1}{3}\sec^{3}(x)-\sec(x)+C`.

Use the fact that `\tan^{2}(x)=\sec^{2}(x)-1` to write `\tan^{3}(x)\sec(x)` as `(sec^{2}(x)-1)\tan(x)\sec(x)`. Note now that if `u=\sec(x)` then `du=\sec(x)\tan(x)dx`. Hence

`\int\tan^{3}(x)\sec(x)dx=\int(u^{2}-1)du=u^{3}/3-u+C=\frac{1}{3}\sec^{3}(x)-\sec(x)+C`