`int((cos^-1 2x)^4)/(sqrt[1-4x^2]) dx`
Put `u = cos^-1 2x`
`du=(-2)/(sqrt[1-4x^2]) dx`
So
` = - 1/2 intu^4 du` `=(-1/2) ((u^5)/(5))+K` `=(-(cos^-1 2x)^5)/(10)+K`
` = - 1/2 intu^4 du`
`=(-1/2) ((u^5)/(5))+K`
`=(-(cos^-1 2x)^5)/(10)+K`