Question

What is the Integral of `tan^3 3x * sec3x dx`?

Answer 1

`inttan^3 3xsec3x"d"x=1/9sec^3 3x-1/3sec3x+"c"`

Explanation:

For `inttan^3 3xsec3x"d"x`, let `u=3x` and `"d"u=3"d"x`

Then `inttan^3 3xsec3x"d"x=1/3inttan^3usecu"d"u`

Now, use `tan^2x=sec^2x-1`

`1/3inttan^3usecu"d"u=1/3int(sec^2u-1)tanusecu"d"u=1/3intsec^3utanu"d"u-1/3intsecutanu"d"u`

For the first integral, we use the reverse chain rule, noting that `d/dx1/3sec^3u=sec^3utanu` and for the second, we use the fundamental theorem of calculus. So,

`1/3intsec^3utanu"d"u-1/3intsecutanu"d"u=1/9sec^3u-1/3secu`

Now we substitute `u=3x` to get a final answer of

`1/9sec^3 3x-1/3sec3x+"c"`