Integral Calculus Question 444
Question: If $ I=\int_{{}}^{{}}{e^{x}\sin 2x\ dx} $ , then for what value of K, $ KI=e^{x}(\sin 2x-2\cos 2x)+ $ constant
[MP PET 1992]
Options:
A) 1
B) 3
C) 5
D) 7
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int_{{}}^{{}}{e^{x}\sin 2x,dx}=\sin 2x,.,e^{x}-2\int_{{}}^{{}}{\cos 2x,.,e^{x}dx} $
$ =\sin 2x,.,e^{x}-2\cos 2x,.,e^{x}-4\int_{{}}^{{}}{e^{x}\sin 2x,dx} $
$ \Rightarrow 5I=e^{x}(\sin 2x-2\cos 2x)+ $ Constant
Equating the given value, we get $ K=5. $
BETA
