Question

How do you find the integral of `tan^3(2x) sec^100(2x) dx`?

Answer 1

`I=1/204sec^102(2x)-1/200sec^100(2x)+C`

Explanation:

We want to solve

`I=inttan^3(2x)sec^100(2x)dx`

Make a substitution `u=2x=>(du)/dx=2`

`I=1/2inttan^3(u)sec^100(u)du`

Use the identity `color(green)(tan^2(x)=sec^2(x)-1`

`I=1/2inttan(u)(sec^2(u)-1)sec^100(u)du`

Make a substitution `s=sec(u)=>(ds)/(du)=sec(u)tan(u)`

`I=1/2int(s^2-1)s^99du`

`color(white)(I)=1/2ints^101-s^99du`

`color(white)(I)=1/204s^102-1/200s^100`

Substitute back `s=sec(u)` and `u=2x`

`I=1/204sec^102(2x)-1/200sec^100(2x)+C`