Integral Calculus Question 84
Question: If $ \int \frac{3\sin x+2\cos x}{3\cos x+2\sin x}dx=ax+b $ ln $ | 2\sin x+3\cos x |+C $ , then
Options:
A) $ a=-\frac{12}{13},b=\frac{15}{39} $
B) $ a=-\frac{7}{13},b=\frac{6}{13} $
C) $ a=\frac{12}{13},b=-\frac{15}{39} $
D) $ a=-\frac{7}{13},b=-\frac{6}{13} $
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Answer:
Correct Answer: C
Solution:
[c] Differentiating both sides, we get $ \frac{3\sin x+2\cos x}{3\cos x+2\sin x}= $ $ a+\frac{b(2\cos x-3\sin x)}{(2\sin x+3\cos x)} $ $ =\frac{\sin x(2a-3b)+\cos x,(3a+2b)}{(3\cos x+2\sin x)} $ Comparing like terms on both sides, we get $ 3=2a-3b,2=3a+2b $ Or $ a=\frac{12}{13},b=-\frac{15}{39} $
BETA
