Integral Calculus Question 22
Question: $ \int_{{}}^{{}}{\frac{1}{\cos x(1+\cos x)}}\ dx= $
Options:
A) $\log \left|(\sec x+\tan x)\right| +2 \tan \frac{x}{2}+c$
B) $\log \left|(\sec x+\tan x)\right| -2 \tan \frac{x}{2}+c$
C) $\log \left|(\sec x+\tan x)\right| +\tan \frac{x}{2}+c$
D) $\log \left|(\sec x+\tan x)\right| -\tan \frac{x}{2}+c$
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Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\frac{1}{\cos x(1+\cos x)}}dx=\int_{{}}^{{}}{\frac{dx}{\cos x}-\int_{{}}^{{}}{\frac{dx}{1+\cos x}}} $ $ =\int_{{}}^{{}}{\sec x\ dx-\frac{1}{2}\int_{{}}^{{}}{{{\sec }^{2}}\frac{x}{2}dx}} $ $ =\log |(\sec x+\tan x)-\tan \frac{x}{2}|+c. $
BETA
