Question

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\left[f\left(\cos x\right)\right]dx=$

• A. $2f\left(\mathrm{\pi }\right)$
• B. $2f\left(2\right)$
• C. $2f\left(1\right)$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Finding the value for the given integral:

Let $I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\left[f\left(\cos x\right)\right]dx$………$\left(1\right)$

$$\begin{gathered} ={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin \left(\mathrm{\pi }+\left(-\mathrm{\pi }\right)-\mathrm{x}\right)\left[f\left(\cos \left(\mathrm{\pi }+\left(-\mathrm{\pi }\right)-\mathrm{x}\right)\right)\right]dx\left[\sin ce{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right] \\ ={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}-\sin x\left[f\left(\cos x\right)\right]dx..........\left(2\right) \end{gathered}$$

By adding $\left(1\right)$ and $\left(2\right)$, we get

$$\begin{gathered} 2I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\left[f\left(\cos x\right)\right]dx+{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}-\sin x\left[f\left(\cos x\right)\right]dx \\ 2I=0 \\ I=0 \end{gathered}$$

Thus, option (D) is the correct answer