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
Correct Answer: B
Solution :
Put \[a+b\cos x=t\Rightarrow dx=-\frac{dt}{b\sin x},\] then \[\int_{{}}^{{}}{\frac{\sin x}{{{(a+b\cos x)}^{2}}}\,dx=-\frac{1}{b}\int_{{}}^{{}}{\frac{1}{{{t}^{2}}}\,dt}=\frac{1}{b}\frac{1}{t}+c}\] \[=\frac{1}{b(a+b\cos x)}+c.\]You need to login to perform this action.
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