Question

Evaluate ∫ sin 4x cos2 4x dx.

Answer 1

We have the following integral:


`\int sin(4x)cos(24x)dx`

Changing variables:


`u=4x \rightarrow dx=(1)/(4)dx`

Therefore:


`(1)/(4)\int sin(u)cos(6u)du`

From trigonometrical identities:


`sin(\alpha)cos(\beta)= (sin(\beta+\alpha)-sin(\beta-\alpha))/(2)`

Thus:


`(1)/(4)\int (sin(7u)-sin(5u))/(2)du= (1)/(8) \int sin(7u)du-(1)/(8) \int sin(5u)du`

∴
`(-cos(7u))/(56)+ (cos(5u))/(40)+C`


`u \rightarrow 4x`

Finally:


`\boxed{ I= (cos(20x))/(40)-(cos(28x))/(56)+C}`