Question

How do you evaluate the integral of `int (1 + cos 4x)^(3/2) dx`?

Answer 1

`int(1+cos4x)^(3/2)dx=sqrt2/8(3sin2x+sin6x)+"c"`

Explanation:

`int(1+cos4x)^(3/2)dx=int(1+2cos^2 2x-1)^(3/2)dx=int2sqrt2cos^3 2xdx=sqrt2/2int(2cos^2x-1)^3dx`

After a very long expansion, we get it into the following form

`(sqrt2)/2int3cos2x+cos6xdx=sqrt2/2(1/2sin2x+1/6sin6x)+"c"`
`=sqrt2/8(3sin2x+sin6x)+"c"`