The correct option is D 1/2(e^π+1) ∫_6^π e^x nbsp; nbsp; sin nbsp; x nbsp; dx ∫ nbsp; e^x nbsp; sin nbsp; x nbsp; dx = sin n ...

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Factorization Method Form to Remove Indeterminate Form

∫0πexsin xdx=

Question

$\int_{0}^{π} e^{x} sin x d x =$

\frac{1}{2} e^{π}

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e^{π} + 1

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\frac{1}{2} (e^{π} - 1)

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\frac{1}{2} (e^{π} + 1)

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Solution

The correct option is D $\frac{1}{2} (e^{π} + 1)$
$\int_{6}^{π} e^{x} s i n x d x$
$\int e^{x} s i n x d x = s i n x e^{x} - \int c o s x e^{x}$
$= s i n x e^{x} - [c o s x e^{x} + \int s i x e^{x} d x]$
$= \int e^{x} s i n x d x = \frac{1}{2} (s i n x - c o s x) e^{x} + c$
$\int_{0}^{π} e^{x} s i n x d x = (\frac{1}{2} (x i n x - c o s x) \times e^{x} + c) \int_{0}^{π}$
$= \frac{1}{2} (0 + 2) e^{π} - (\frac{1}{2} (0 - 1))$
$= \frac{1}{2} (e^{π} + 1)$

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