Integral Calculus Question 455
Question: $ \int_{{}}^{{}}{\frac{\sin x\ dx}{{{(a+b\cos x)}^{2}}}=} $
Options:
A) $ \frac{1}{b}(a+b\cos x)+c $
B) $ \frac{1}{b(a+b\cos x)}+c $
C) $ \frac{1}{b}\log (a+b\cos x)+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ a+b\cos x=t\Rightarrow dx=-\frac{dt}{b\sin x}, $ then $ \int_{{}}^{{}}{\frac{\sin x}{{{(a+b\cos x)}^{2}}},dx=-\frac{1}{b}\int_{{}}^{{}}{\frac{1}{t^{2}},dt}=\frac{1}{b}\frac{1}{t}+c} $ $ =\frac{1}{b(a+b\cos x)}+c. $
BETA
