-0-2 ex sin x dx- -Roorkee 1978-

Let ∫_0^π/2 e^x sin x dx= = [e^x cos x]_0^π/2 + ∫_0^π/2e^x cos x dx = [e^x cos x]_0^π/2 + [e^x sin x]_0^x/2 ∫^π/2_0 e^x sin x dx ∴ 2I = [e^x (sin x cos x)] ...

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Byju's Answer

Standard XI

Mathematics

Second Fundamental Theorem of Calculus

∫0π/2 ex sin ...

Question

$\int_{0}^{\frac{π}{2}} e^{x} s i n x d x =$ [Roorkee 1978]

\frac{1}{2} (e^{\frac{x}{2}} - 1)

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\frac{1}{2} (e^{\frac{x}{2}} + 1)

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\frac{1}{2} (1 - e^{\frac{x}{2}})

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2 (e^{\frac{x}{2}} + 1)

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Solution

The correct option is B $\frac{1}{2} (e^{\frac{x}{2}} + 1)$
$L e t \int_{0}^{\frac{π}{2}} e^{x} s i n x d x =$
$= - [e^{x} c o s x]_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} e^{x} c o s x d x$
$= - [e^{x} c o s x]_{0}^{\frac{π}{2}} + [e^{x} s i n x]_{0}^{\frac{x}{2}} - \int_{0}^{\frac{π}{2}} e^{x} s i n x d x$
$∴ 2 I = [e^{x} (s i n x - c o s x)]_{0}^{\frac{π}{2}} = (e^{\frac{π}{2}} + 1)$
$H e n c e \int_{0}^{\frac{π}{2}} e^{x} s i n x d x = \frac{1}{2} (e^{\frac{π}{2}} + 1)$

Suggest Corrections

∫π20 ex sin x dx= [Roorkee 1978]

The correct option is B 12(ex2+1)Let ∫π20 ex sin x dx= =−[ex cos x]π20+∫π20ex cos x dx =−[ex cos x]π20+[ex sin x]x20−∫π20 ex sin x dx ∴ 2I=[ex(sin x−cos x)]π20=(eπ2+1) Hence ∫π20 ex sin x dx=12(eπ2+1)

$\int_{0}^{\frac{π}{2}} e^{x} s i n x d x =$ [Roorkee 1978]