Integral Calculus Question 341
Question: $ \int_{{}}^{{}}{e^{2x}(-\sin x+2\cos x)\ dx=} $
[DSSE 1987]
Options:
A) $ e^{2x}\sin x+c $
B) $ -e^{2x}\sin x+c $
C) $ -e^{2x}\cos x+c $
D) $ e^{2x}\cos x+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{e^{2x}(-\sin x+2\cos x),dx} $ $ =-\int_{{}}^{{}}{e^{2x}\sin x,dx}+2\int_{{}}^{{}}{e^{2x}\cos x,dx} $ $ =e^{2x}\cos x-2\int_{{}}^{{}}{e^{2x}\cos x,dx+2\int_{{}}^{{}}{e^{2x}\cos x,dx+c}} $ $ =e^{2x}\cos x+c. $ Aliter : $ \int_{{}}^{{}}{e^{2x}(2\cos x-\sin x),dx}=e^{2x}\cos x+c $ $ { \because ,\int_{{}}^{{}}{e^{kx}{ k,f(x)+{f}’(x) }dx=e^{kx}f(x)+c} } $
BETA
