`int tan^5(x) dx` Evaluate the integral

`int tan^5 (x) dx`

To solve, apply the Pythagorean identity `tan^2(x) = sec^2(x) - 1` until the integrand is in the form `int u^n du` .

`= int tan^3(x) tan^2(x) dx`

`= int tan^3(x)(sec^2x-1)dx`

`= int [tan^3(x) sec^2(x) - tan^3(x)]dx`

`= int [tan^3(x)sec^2(x) - tan(x) tan^2(x)]dx`

`= int [tan^3(x) sec^2(x) - tan(x) (sec^2x-1)]dx`

`= int [tan^3(x) sec^2(x) - tan(x)sec^2(x) +tan(x)]dx`

`= int tan^3(x)sec^2(x)dx - int tan(x) sec^2(x) dx + int tan(x) dx`

For the first and second integral, apply u-substitution method. Let u be:

     `u = tan x`

Then, differentiate u.

     `du= sec^2(dx)`

Plugging them, the first and second integral becomes:

`= int u^3 du - int u du + int tan (x) dx`

Then, apply the integral formula `int x^n dx = x^(n+1)/(n+1) + C` and `int tan(theta)d theta = ln |sec (theta)| + C` .

`= u^4/4 - u^2/2 + ln |sec(x)| + C`

And, substitute back u = tan(x).

`= (tan^4(x))/4 - (tan^2(x))/2 + ln|sec(x)| + C`

Therefore, `int tan^5(x) dx= (tan^4(x))/4 - (tan^2(x))/2 + ln|sec(x)| + C` .