If frac{d}{{dx}},Gleft( x right) = frac{ | vedclass

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Question

$	ext { Let } \frac{d}{d x} G(x)=\frac{e^{	an x}}{x}, x \in\left(0, \frac{\pi}{2}ight)$

${m{Now}},I = \int\limits_{1/4}^{1/2} {\frac{2}{x}} {

Vedclass · Accepted Answer

$G\left( {\pi /4} \right) - G\left( {\pi /16} \right)$

Vedclass · Answer

$	ext { Let } \frac{d}{d x} G(x)=\frac{e^{	an x}}{x}, x \in\left(0, \frac{\pi}{2}ight)$

${m{Now}},I = \int\limits_{1/4}^{1/2} {\frac{2}{x}} {e^{	an \pi {x^2}}}dx$

$ = \int\limits_{1/4}^{1/2} {\frac{{2\pi x}}{{\pi {x^2}}}} {e^{	an \pi {x^2}}}dx$

$	ext { Let } \pi x^{2}=t \Rightarrow 2 \pi x d x=d t$

$	ext { When } x=\frac{1}{2}, t=\frac{\pi}{4} 	ext { and } x=\frac{1}{4}, t=\frac{\pi}{16}$

$	herefore I = \int\limits_{\pi /16}^{\pi /4} {\frac{{{e^{	an t}}}}{t}} dt = \left. {g(t)} ight|_{\frac{\pi }{{16}}}^{\frac{\pi }{4}}$

$=G\left(\frac{\pi}{4}ight)-G\left(\frac{\pi}{16}ight)$

If $\frac{d}{{dx}}\,G\left( x \right) = \frac{{{e^{\tan \,x}}}}{x},\,x \in \left( {0,\pi /2} \right)$, then $\int\limits_{1/4}^{1/2} {\frac{2}{x}} .{e^{\tan \,\left( {\pi \,{x^2}} \right)}}dx$ is equal to

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