Integral Calculus Question 339
Question: $ \int_{{}}^{{}}{e^{x}\sin x\ dx=} $
[IIT 1978; AI CBSE 1980; MP PET 1999]
Options:
A) $ \frac{1}{2}e^{x}(\sin x+\cos x)+c $
B) $ \frac{1}{2}e^{x}(\sin x-\cos x)+c $
C) $ e^{x}(\sin x+\cos x)+c $
D) $ e^{x}(\sin x-\cos x)+c $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ I=\int_{{}}^{{}}{e^{x}\sin x,dx}=e^{x}\sin x-\int_{{}}^{{}}{e^{x}\cos x,dx+c} $ $ =e^{x}\sin x-e^{x}\cos x-\int_{{}}^{{}}{e^{x}\sin x,dx+c} $
Þ $ 2I=e^{x}(\sin x-\cos x)+c $
Þ $ I=\frac{1}{2}e^{x}(\sin x-\cos x)+c $ .
BETA
