A) ?\[x\tan \frac{x}{2}+c\]
B) \[x\tan \ \frac{x}{2}+c\]
C) \[x\tan x+c\]
D) \[\frac{1}{2}x\tan x+c\]
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, \[\frac{d}{dx}\left\{ x\tan \frac{x}{2}+c \right\}\] \[=\frac{x}{2}{{\sec }^{2}}\frac{x}{2}+\tan \frac{x}{2}=\frac{1}{2}\left[ \frac{x}{{{\cos }^{2}}\frac{x}{2}}+\frac{2\sin \frac{x}{2}}{\cos \frac{x}{2}} \right]=\frac{x+\sin x}{1+\cos x}\].
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