Question

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

• A. $\frac{3x}{8}-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x$
• B. $\frac{3x}{8}+\frac{1}{4}\sin 2x-\frac{1}{32}\sin 4x$
• C. $\frac{3x}{8}+\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x$
• D. None of these

Answer 1

Answer: (A)

$\frac{3x}{8}-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x$

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

$={\int }^{}{\left({\sin }^{2}x\right)}^{2}dx$

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

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

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

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

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

$=\frac{1}{4}\left[x-\sin 2x\right]+\frac{1}{8}{\int }^{}\left(1+\cos 4x\right)dx$

$=\frac{1}{4}\left[x-\sin 2x\right]+\frac{1}{8}{\int }^{}\left(x+\frac{\sin 4x}{4}\right)dx+c$

$=\frac{x}{4}-\frac{1}{4}\sin 2x+\frac{x}{8}+\frac{1}{32}\sin 4x+c$

$=\frac{3x}{8}-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+c$