A) \[\int_{0}^{\pi /6}{\frac{xdx}{\tan x}}\]
B) \[\int_{0}^{\pi /6}{\frac{2x}{\tan x}}dx\]
C) \[\int_{0}^{\pi /2}{\frac{2xdx}{\tan x}}\]
D) \[\int_{0}^{\pi /6}{\frac{xdx}{\sin x}}\]
E) \[\int_{0}^{\pi /6}{\frac{2x}{\sin x}}dx\]
Correct Answer: B
Solution :
\[z=\int_{0}^{1}{\frac{2{{\sin }^{-1}}\frac{x}{2}}{x}}dx\] Put \[{{\sin }^{-1}}\frac{x}{2}=t\] \[\Rightarrow \] \[x=2\sin t\] \[\Rightarrow \] \[dx=2\cos tdt\] Also, \[x=0\Rightarrow t=0\] \[x=1\Rightarrow t=\frac{\pi }{6}\] \[\therefore \] \[I=\int_{0}^{\pi /6}{\frac{2t}{2\sin t}}.2\cos t\,dt\] \[=\int_{0}^{\pi /6}{\frac{2t}{\tan t}}dt=\int_{0}^{\pi /6}{\frac{2x}{\tan x}}dx\]
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