Question

The value of ${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}{\sin }^{-4}\left(x\right)dx$ is

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

Answer 1

The correct option is A

$-\frac{8}{3}$

Explanation for the correct option:

Compute the required value:

Given: ${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}{\sin }^{-4}\left(x\right)dx$

$={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}{\csc }^4\left(x\right)dx$

$={\int }_{\frac{-\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}{\csc }^2\left(x\right)(1+{\cot }^2\left(x\right))dx$

Let

$$\begin{gathered} \cot \left(x\right)=t \\ -{\csc }^2\left(x\right)·dx=dt \end{gathered}$$

When

$$\begin{gathered} x=\frac{\mathrm{\pi }}{4},t=1 \\ x=-\frac{\mathrm{\pi }}{4},t=-1 \end{gathered}$$

Then,

$$\begin{gathered} {\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}{\csc }^2\left(x\right)(1+{\cot }^2\left(x\right))dx \\ =-{\int }_{-1}^1(1+t^2)dt \\ =-{\int }_{-1}^1\left(1\right)·dt-{\int }_{-1}^1\left(t^2\right)·dt \\ =-{\left|t\right|}_{-1}^1-{\left|\frac{t^3}{3}\right|}_{-1}^1 \\ =-\left(1+1\right)-\frac{1}{3}\left(1+1\right) \\ =-2-\frac{2}{3} \\ =-\frac{8}{3} \end{gathered}$$

Hence, option $A$ is the correct answer.