A) \[\frac{\tan \,2x}{8}+c\]
B) \[\frac{\tan \,8x}{8}+c\]
C) \[\frac{\tan \,4x}{4}+c\]
D) \[\frac{\tan \,4x}{8}+c\]
Correct Answer: D
Solution :
Let \[I=\int{\frac{1}{1+\cos \,8x}}dx\] \[=\int{\frac{1}{2{{\cos }^{2}}\,4x}\,}dx\] \[=\frac{1}{2}\int{{{\sec }^{2}}4\,x\,dx}\] \[=\frac{1}{2}\frac{\tan \,4\,x}{4}+c\] \[=\frac{\tan \,4\,x}{8}+c\]
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