Definite Integration Question 53
Question: $ \int_0^{\pi /2}{e^{x}\sin xdx=} $
[Roorkee 1978]
Options:
A) $ \frac{1}{2}({e^{\pi /2}}-1) $
B) $ \frac{1}{2}({e^{\pi /2}}+1) $
C) $ \frac{1}{2}(1-{e^{\pi /2}}) $
D) $ 2({e^{\pi /2}}+1) $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ I=\int_0^{\pi /2}{e^{x}\sin xdx} $
= $ -[e^{x}\cos x]_0^{\pi /2}+\int_0^{\pi /2}{e^{x}\cos xdx} $
$ =-[e^{x}\cos x]_0^{\pi /2}+[e^{x}\sin x]_0^{\pi /2}-\int_0^{\pi /2}{e^{x}\sin xdx} $
$ \therefore $ $ 2I=[e^{x}(\sin x-\cos x)]_0^{\pi /2}=({e^{\pi /2}}+1) $
Hence $ \int_0^{\pi /2}{e^{x}\sin xdx=\frac{1}{2}({e^{\pi /2}}+1)} $ .
BETA
