Question

What's the integral of `int tanx * sec^3x * dx`?

Answer 1

I found: `1/(3cos(x))+c`

Explanation:

I would write it as:
`intsin(x)/cos(x)*1/cos^3(x)dx=intsin(x)/cos^4(x)dx=`
we can set `cos(x)=t`
so that:

`d[cos(x)]=dt=sin(x)dx`

and our integral becomes:
`=-int1/t^4dt=`
`=-intt^-4dt=`
`=-t^(-4+1)/(-4+1)+c=`
`=1/3t^-3+c=1/(3t^3)+c=`
but `t=cos(x)` so:
`=1/(3cos^3(x))+c`