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

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Answer 1 · 2018-03-23T14:17:25

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

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 $u=2x$

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

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