Integral Calculus Question 385
Question: $ \int_{{}}^{{}}{\frac{x+\sin x}{1+\cos x}\ dx} $ is equal to
[Roorkee 1980; UPSEAT 1999]
Options:
A) ? $ x\tan \frac{x}{2}+c $
B) $ x\tan \ \frac{x}{2}+c $
C) $ x\tan x+c $
D) $ \frac{1}{2}x\tan x+c $
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Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{x+\sin x}{1+\cos x},dx=\frac{1}{2}\int_{{}}^{{}}{x{{\sec }^{2}}\frac{x}{2},dx+\int_{{}}^{{}}{\tan \frac{x}{2},dx}}} $ $ =\frac{1}{2}\frac{x\tan \frac{x}{2}}{\frac{1}{2}}-\int_{{}}^{{}}{\tan \frac{x}{2},dx}+\int_{{}}^{{}}{\tan \frac{x}{2},dx} $ $ =x\tan \frac{x}{2}+c $ . Trick : By inspection, $ \frac{d}{dx}{ x\tan \frac{x}{2}+c } $ $ =\frac{x}{2}{{\sec }^{2}}\frac{x}{2}+\tan \frac{x}{2}=\frac{1}{2}[ \frac{x}{{{\cos }^{2}}\frac{x}{2}}+\frac{2\sin \frac{x}{2}}{\cos \frac{x}{2}} ]=\frac{x+\sin x}{1+\cos x} $ .
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