Question

What is the indefinite integral of tan^5xdx?

Answer 1

`\displaystyle\int\tan^5x\,\mathrm dx=\int\tan^4x\tan x`

`=\displaystyle\int(\sec^2x-1)^2\tan x\,\mathrm dx`

`=\displaystyle\int(\sec^4x-2\sec^2x+1)\tan x\,\mathrm dx`

`=\displaystyle\int\sec^3x\sec x\tan x\,\mathrm dx-2\int\sec^2x\tan x\,\mathrm dx+\int\tan x\,\mathrm dx`

For the first integral, let
`a=\sec x` so that
`\mathrm da=\sec x\tan x\,\mathrm dx`. For the second,
`b=\tan x` so that
`\mathrm db=\sec^2x\,\mathrm dx`. The last integral is standard, but if you're not familiar with it, you could write
`\tan x=(\sin x)/(\cos x)` and set
`c=\cos x` so that
`\mathrm dc=-\sin x\,\mathrm dx`. Now


`=\displaystyle\int a^3\,\mathrm da-2\int b\,\mathrm db-\int\frac{\mathrm dc}c`

`=\displaystyle\frac{a^4}4-b^2-\ln|c|+C`

`=\displaystyle\frac{\sec^4x}4-\tan^2x-\ln|\cos x|+C`