hey guys im trying to integrate sin^4(x) can someone check over this and tell me if its correct:
= sin(4x)/32 - sin(2x)/4 + 3x/8
is that the correct answer for the integration of sin^4(x)
hey guys im trying to integrate sin^4(x) can someone check over this and tell me if its correct:
= sin(4x)/32 - sin(2x)/4 + 3x/8
is that the correct answer for the integration of sin^4(x)
You don't have to use de Moivre's formula to solve this.
$\displaystyle \sin^4 x = \frac{1-\cos 2x}{2}\cdot \frac{1-\cos 2x}{2}=\frac{1}{4}(1-2\cos 2x + \cos^2 2x)=$ $\displaystyle \frac{1}{4}\left(1-2\cos 2x + \frac{1}{2}(1+\cos 4x)\right)=\frac{1}{8}\cos 4x - \frac{1}{2} \cos 2x + \frac{3}{8}$