Question

Find the integrals of the functions.
$\int co{s}^{4}2xdx.$

Answer 1

$\int co{s}^{4}2xdx=\int (co{s}^{2}2x{)}^{2}dx=\int {\left(\frac{1+cos4x}{2}\right)}^{2}dx=\frac{1}{4}\int (1+cos4x{)}^{2}dx=\frac{1}{4}\int (1+co{s}^{2}4x+2cos4x)dx=\frac{1}{4}[\int 1dx+\int co{s}^{2}4xdx+2\int cos4xdx]=\frac{1}{4}[\int 1dx+\int \frac{1+cos8x}{2}dx+2\int cos4xdx](\therefore co{s}^{2}x=\frac{1+cos2x}{2})=\frac{1}{4}[\int 1dx+\frac{1}{2}(\int 1dx+\int cos8xdx)+2\int cos4xdx]=\frac{1}{4}[x+\frac{1}{2}\{x+\frac{sin8x}{8}\}+\frac{2sin4x}{4}]+C=\frac{1}{4}x+\frac{x}{8}+\frac{sin8x}{64}+\frac{sin4x}{8}+C=[\frac{3x}{8}+\frac{sin8x}{64}+\frac{sin4x}{8}]+C$