Question

What is the antiderivative from 0 to Pi/3 of (Cos(4x)+Sin(4x))^2​

Answer 1

The quickest method would probably be to just expand the integrand:

`(\cos4x+\sin4x)^2=\cos^24x+2\cos4x\sin4x+\sin^24x=1+\sin8x`

Then

`\displaystyle\int_0^(\pi/3)(1+\sin8x)\,\mathrm dx=\left(x-\frac18\cos8x\right)\bigg|_0^(\pi/3)`

`=\left(\frac\pi3-\frac18\cos\frac{8\pi}3\right)-\left(0-\frac18\right)=\boxed{\frac3{16}+\frac\pi3}`