Integral Calculus Question 328
Question: If $ \int_{{}}^{{}}{e^{x}\sin x\ dx=\frac{1}{2}e^{x}\ .\ a+c} $ , then $ a= $
[MP PET 1989]
Options:
A) $ \sin x-\cos x $
B) $ \cos x-\sin x $
C) $ -\cos x-\sin x $
D) $ \cos x+\sin x $
Show Answer
Answer:
Correct Answer: A
Solution:
Given that $ \int_{{}}^{{}}{e^{x}\sin x,dx}=\frac{1}{2}e^{x}a+c $ ..?(i) Let $ I=\int_{{}}^{{}}{e^{x}\sin x,dx}=-e^{x}\cos x+\int_{{}}^{{}}{e^{x}\cos x,dx+c} $ $ =-e^{x}\cos x+e^{x}\sin x-\int_{{}}^{{}}{e^{x}\sin x,dx+c} $
$ \Rightarrow 2I=e^{x}(-\cos x+\sin x)+c $ . Now from (i), we get $ \frac{1}{2}e^{x}a=\frac{1}{2}e^{x}(\sin x-\cos x)\Rightarrow a=\sin x-\cos x. $
BETA
