A) \[G(\pi /4)-G(\pi /16)\]
B) \[2[G(\pi /4)-G(\pi /16)]\]
C) \[\pi [G(1/2)-G(1/4)]\]
D) \[G(1/\sqrt{2})-G(1/2)\]
Correct Answer: A
Solution :
Let\[\frac{d}{dx}G\left( x \right)=\frac{{{e}^{\tan x}}}{x},x\in \left( 0,\frac{\pi }{2} \right)\] Now,\[I=\int\limits_{{}^{1}/{}_{4}}^{{}^{1}/{}_{2}}{\frac{2}{x}{{e}^{\tan \pi {{x}^{2}}}}}dx\] \[=\int\limits_{{}^{1}/{}_{4}}^{{}^{1}/{}_{2}}{\frac{2\pi }{\pi {{x}^{2}}}{{e}^{\tan \pi {{x}^{2}}}}}dx\] Let\[\pi {{x}^{2}}=t\Rightarrow 2\pi x\,dx=dt\] when\[x=\frac{1}{2},t=\frac{\pi }{4}\]and\[x=\frac{1}{4},t=\frac{\pi }{16}\] \[\therefore \]\[\left. I=\int\limits_{{}^{\pi }/{}_{16}}^{{}^{\pi }/{}_{4}}{\frac{{{e}^{\tan t}}}{t}dt}=g(t) \right|\begin{matrix} \frac{\pi }{4} \\ \frac{\pi }{16} \\ \end{matrix}\] \[=G\left( \frac{\pi }{4} \right)-G\left( \frac{\pi }{16} \right)\]
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