MCQ
$\int_0^{\pi / 2} e ^x \sin x d x=$
  • A
    $\frac{1}{2}\left(e^{\frac{\pi}{2}}-1\right)$
  • $\frac{1}{2}\left( e ^{\frac{\pi}{2}}+1\right)$
  • C
    $\frac{1}{2}\left(1- e ^{\frac{\pi}{2}}\right)$
  • D
    $2\left( e ^{\frac{\pi}{2}}+1\right)$

Answer

Correct option: B.
$\frac{1}{2}\left( e ^{\frac{\pi}{2}}+1\right)$
(B)
Let $I =\int_0^{\frac{\pi}{2}} e ^x \sin x d x$
$=\left[\sin x \cdot e ^x\right]_0^{\pi / 2}-\int_0^{\frac{\pi}{2}} \cos x \cdot e ^x d x$
$\therefore \quad I =\left[ e ^x \sin x\right]_0^{\pi / 2}-\left[\cos x \cdot e ^x\right]_0^{\pi / 2}-\int_0^{\frac{\pi}{2}} \sin x \cdot e ^x d x$
$\begin{array}{l}\Rightarrow 2 I =\left[ e ^x(\sin x-\cos x)\right]_0^{\pi / 2} \\ \Rightarrow 2 I = e ^{\pi / 2}+1 \\ \Rightarrow I =\frac{ e ^{\pi / 2}+1}{2}\end{array}$

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