Question

What is the integral of `int sin^3x/cos^2x dx`?

Answer 1

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

Explanation:

Let us try writing it as:
`int(sin(x)*sin^2(x))/(cos^2(x))dx=`
`=int(sin(x)(1-cos^2(x)))/(cos^2(x))dx=`
`=int(sin(x))/(cos^2(x))dx-intsin(x)(cancel(cos^2(x)))/cancel((cos^2(x)))dx=`
let us use a little manipulation into the first integral:
`=-int(d[cos(x)])/cos^2(x)-intsin(x)dx=`
`=-int(cos(x))^-2d[cos(x)]-intsin(x)dx=`
`=(cos(x))^-1+cos(x)+c=1/cos(x)+cos(x)+c`