Question

Evaluate $\int {\sin }^{4}xdx$

• A. $4si{n}^{3}x+c$
• B. $\frac{si{n}^{3}x(-cosx)}{4}+c$
• C. $\frac{sin\left(4x\right)-8sin\left(2x\right)+12x}{32}+c$
• D. $\frac{sin\left(4x\right)-8cos\left(2x\right)+4x}{16}+c$

Answer 1

Answer: (C)

$\frac{sin\left(4x\right)-8sin\left(2x\right)+12x}{32}+c$

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

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

$=\frac{1}{4}\int (1+\frac{1+\cos 4x}{2}-2\cos 2x)dx$

$=\frac{1}{8}\int (2+1+\cos 4x-4\cos 2x)dx$

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

$=\frac{1}{8}[3x+\frac{\sin 4x}{4}-\frac{4\sin 2x}{2}]+c$

$=\frac{3x}{8}+\frac{\sin 4x}{32}-\frac{\sin 2x}{4}+c$

$=\frac{\sin 4x-8\sin 2x+12x}{32}+c$

Option $C$ is correct.