Question

The value of the integral $\underset{-\pi /4}{\overset{\pi /4}{\int }}\sin -4xdx$ is

• A. $-\frac{8}{3}$
• B. $\frac{3}{2}$
• C. $\frac{8}{3}$
• D. None of these

Answer 1

Answer: (A)

Let $I=\underset{-\pi /4}{\overset{\pi /4}{\int }}{\sin }^{-4}xdx$
$=\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{1}{{\sin }^{4}x}dx$
$=\underset{-\pi /4}{\overset{\pi /4}{\int }}{\text{cosec}}^{4}xdx={\int }_{-\pi /4}^{\pi /4}{\left({\text{cosec}}^{2}x\right)}^{2}dx$
$=\underset{-\pi /4}{\overset{\pi /4}{\int }}\left(1+{\cot }^{2}x\right){\text{cosec}}^{2}xdx$
Put $\cot x=t$
$\Rightarrow -{\text{cosec}}^{2}xdx=dt$
$\therefore I=-\underset{-1}{\overset{1}{\int }}\left(1+t^2\right)dt=-2\underset{0}{\overset{1}{\int }}\left(1+t^2\right)dt$
$=-2{\left[t+\frac{t^3}{3}\right]}_0^1$
$=-2\left[1+\frac{1}{3}\right]$
$=-\frac{8}{3}$