Question

Solve $\int {\cos }^{4}2xdx$

Answer 1

We know that,

$\cos 4x=2{\cos }^{2}2x-1$

$\Rightarrow {\cos }^{2}2x=\frac{1+\cos 4x}{2}$

Squaring both side and we get,}

$\Rightarrow {\cos }^{4}2x={\left(\frac{1+\cos 4x}{2}\right)}^2$

$\Rightarrow {\cos }^{4}2x=\frac{1}{4}(1+{\cos }^{2}4x+2\cos 4x)$

$\Rightarrow {\cos }^{4}2x=\frac{1}{4}[1+\frac{1+\cos 8x}{2}+2\cos 4x]$

$\Rightarrow {\cos }^{4}2x=\frac{1}{8}(2+1+\cos 8x+4cos4x)$

$\Rightarrow {\cos }^{4}2x=\frac{1}{8}(3+\cos 8x+4cos4x)$

On integrating and we get,

$\int {\cos }^{4}2xdx=\int \frac{1}{8}(3+\cos 8x+4cos4x)dx$

$=\int \frac{3}{8}dx+\int \frac{\cos 8x}{8}dx+\int \frac{4\cos 4x}{8}dx$

$=\frac{3}{8}\int 1dx+\frac{1}{8}\int \cos 8xdx+\frac{1}{2}\int \cos 4xdx$

$=\frac{3}{8}x+\frac{1}{8}\frac{\sin 8x}{8}+\frac{1}{2}\frac{\sin 4x}{4}+C$

$=\frac{3}{8}x+\frac{\sin 8x}{16}+\frac{\sin 4x}{8}+C$

Hence, this is the answer.