Question

How do you find the integral of `cos^2(2x)dx`?

Answer 1

`\frac{1}{2}\(x+\frac{1}{4}\sin \(4x))+C`

Explanation:

`int cos ^2(2x)dx`

using the following identity,

`cos ^2(x)=frac{1+cos (2x)}{2}`

`=int frac{1+cos (2cdot 2x)}{2}dx`

taking the constant out,

`int acdot f(x)dx=acdot int f(x)dx`

so,`=frac{1}{2}int 1+cos (2cdot 2x)dx`

applying the sum rule,

`int f(x)pm g(x)dx=int f(x)dxpm int g(x)dx`

we have,`int 1dx` `=x`
and,
`int cos (2cdot 2x)dx=frac{1}{4}sin (4x)`

finally,
`=frac{1}{2}(x+frac{1}{4}sin (4x))`

adding constant,we get,
`frac{1}{2}(x+frac{1}{4}sin (4x))+C`