Integral Calculus Question 297
Question: $ \int_{{}}^{{}}{\frac{dx}{4{{\cos }^{3}}2x-3\cos 2x}}= $
Options:
A) $ \frac{1}{3}\log [\sec 6x+\tan 6x]+c $
B) $ \frac{1}{6}\log [\sec 6x+\tan 6x]+c $
C) $ \log [\sec 6x+\tan 6x]+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{dx}{4{{\cos }^{3}}2x-3\cos 2x}}=\int_{{}}^{{}}{\frac{dx}{\cos 6x}=\int_{{}}^{{}}{\sec 6xdx}} $ $ =\frac{1}{6}\log ,(\sec 6x+\tan 6x)+c. $
BETA
