Question

Evaluate: ${\int }_{-\pi /4}^{\pi /4}{e}^{-x}\sin xdx$

• A. $-\frac{\sqrt{2}}{2}{e}^{-\pi /4}$
• B. $\frac{\sqrt{2}}{2}{e}^{-\pi /4}$
• C. $-\sqrt{2}(e^{-\pi /4}-e^{\pi /4})$
• D. $0$

Answer 1

The correct option is A $-\frac{\sqrt{2}}{2}{e}^{-\pi /4}$

$I={\int }_{-\pi /4}^{\pi /4}{e}^{-x}\sin xdx$

Using integration by parts:-

$I=e^{-x}\int \sin xdx-\int (\frac{d}{dx}{e}^{-x}.\int \sin xdx)dx$

$⟹I=-e^{-x}\cos x-\int e^{-x}\cos xdx$

$⟹I=-e^{-x}\cos x-\left[e^{-x}\int \cos xdx-\int (\frac{d}{dx}{e}^{-x}.\int \cos xdx)dx\right]$

$⟹I=-e^{-x}\cos x-e^{-x}\sin x-\int e^{-x}\sin xdx$

$⟹I=-e^{-x}\cos x-e^{-x}\sin x-I$

$⟹2I={[-e^{-x}\cos x-e^{-x}\sin x]}_{-\pi /4}^{\pi /4}$

$⟹2I=-2\frac{e^{-\pi /4}}{\sqrt{2}}$

$⟹I=\frac{-1}{\sqrt{2}}{e}^{-\pi /4}=\frac{-\sqrt{2}}{2}{e}^{-\pi /4}$

Hence, answer is option-(A).