Question

Question #91a61

Answer 1

`-(cscx+sinx)`

Explanation:

`intcos^3x * csc^2x dx =intcos^2x*cosx *1/sin^2x dx`
`=int(1-sin^2x)*cosx /sin^2x dx`

Now let `z=sinx =>dz=cosxdx`

so changing the varable from x to z the integral becomes

`=int(1-z^2)/z^2 dz=intdz/z^2-intdz=z^(-2+1)/(-2+1)-z+c`
`=-1/z-z+c=-1/sinx-sinx=-(cscx+sinx)`