A) \[\frac{\sin 2x}{2}+\sin x+C\]
B) \[\frac{-\sin 2x}{2}-\sin x+C\]
C) \[\frac{\sin 2x}{2}-\sin x+C\]
D) \[\frac{-\sin 2x}{2}+\sin x+C\]
Correct Answer: B
Solution :
[b] \[\int{\frac{\cos 5x+\cos 4x}{1-2\cos 3x}}dx\] \[=\int{\frac{\sin 3x\left( \cos 5x+\cos 4x \right)}{\sin 3x-\sin 6x}}dx\] \[=\int{\frac{\sin 3x\left( \cos 5x+\cos 4x \right)}{\sin 3x-\sin 6x}}dx\] \[=-\int{2\cos \frac{3x}{2}\cos \frac{x}{2}dx}\] \[=-\int{\left( \cos 2x+cos\,\,x \right)dx}\] \[=-\left( \frac{\sin 2x}{2}+\sin x \right)+C\]
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