Question

Solve $\frac{1}{4}\int {\left(\frac{1+\cos 2x}{2}\right)}^{2}dx$.

Answer 1

Given: $\frac{1}{4}\cdot \int {\left(\frac{1+\cos \left(2x\right)}{2}\right)}^{2}dx$

Simplify the integral.

$\frac{1}{4}\cdot \int {\left(\frac{1+\cos \left(2x\right)}{2}\right)}^{2}dx$

$=\frac{1}{4}\cdot \int {\cos }^4\left(x\right)dx$

$=\frac{1}{4}\cdot \int {\cos }^3\left(x\right)\cos \left(x\right)dx$

Apply integration by parts,

$u={\cos }^3\left(x\right),v^{\prime }=\cos \left(x\right)$

$=\frac{1}{4}\cdot ({\cos }^3\left(x\right)\sin \left(x\right)-\int -3{\cos }^2\left(x\right){\sin }^2\left(x\right)dx)$

$=\frac{1}{4}\cdot ({\cos }^3\left(x\right)\sin \left(x\right)-(-\frac{3}{8}(x-\frac{1}{4}\sin \left(4x\right))))$

$=\frac{1}{4}\cdot ({\cos }^3\left(x\right)\sin \left(x\right)+\frac{3}{8}(x-\frac{1}{4}\sin \left(4x\right)))$

$=\frac{1}{4}\cdot ({\cos }^3\left(x\right)\sin \left(x\right)+\frac{3}{8}(x-\frac{1}{4}\sin \left(4x\right)))$

$=\frac{1}{4}({\cos }^3\left(x\right)\sin \left(x\right)+\frac{3}{8}(x-\frac{1}{4}\sin \left(4x\right)))+C$