Integral Calculus Question 64
Question: $ \int_{{}}^{{}}{\frac{3\sin x+2\cos x}{3\cos x+2\sin x}\ dx=} $
Options:
A) $ \frac{12}{13}x-\frac{5}{13}\log (3\cos x+2\sin x) $
B) $ \frac{12}{13}x+\frac{5}{13}\log (3\cos x+2\sin x) $
C) $ \frac{13}{12}x+\frac{5}{13}\log (3\cos x+2\sin x) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Write $ N^{r}=l(D^{r})+m $ (differential coefficient of $ D^{r}). $ Let 3sinx+2cosx = l(3cosx+2sinx)+m(?3sinx + 2cosx) Comparing coefficients of $ \sin x $ and $ \cos x $ on both sides $ 3=2l-3m $ and $ 2=3l+2m $ Solving, we get $ l=\frac{12}{13}, $ $ m=-\frac{5}{13}, $
$ \therefore I=l,\int_{{}}^{{}}{dx}+m\int_{{}}^{{}}{\frac{-3\sin x+2\cos x}{3\cos x+2\sin x},dx} $ =lx+mlog(3cosx+2sinx) $ =\frac{12}{13}x-\frac{5}{13} $ log(3cosx+2sinx).
BETA
