If frac{d}{{dx}}G(x) = frac{{{e^{tan x}}}}{x} | vedclass

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(A) Given $\frac{d}{dx} G(x) = \frac{e^{	an x}}{x}$ for $x \in (0, \pi/2)$.Let $I = \int_{1/4}^{1/2} \frac{2}{x} e^{	an(\pi x^2)} dx$.Multiply and d

Vedclass · Accepted Answer

$G(\pi/4) - G(\pi/16)$

Vedclass · Answer

(A) Given $\frac{d}{dx} G(x) = \frac{e^{	an x}}{x}$ for $x \in (0, \pi/2)$.Let $I = \int_{1/4}^{1/2} \frac{2}{x} e^{	an(\pi x^2)} dx$.Multiply and divide by $\pi$ inside the integral: $I = \int_{1/4}^{1/2} \frac{2\pi x}{\pi x^2} e^{	an(\pi x^2)} dx$.Let $t = \pi x^2$. Then $dt = 2\pi x dx$.When $x = 1/4$,$t = \pi/16$. When $x = 1/2$,$t = \pi/4$.Substituting these into the integral,we get $I = \int_{\pi/16}^{\pi/4} \frac{e^{	an t}}{t} dt$.Since $\frac{d}{dt} G(t) = \frac{e^{	an t}}{t}$,the integral becomes $[G(t)]_{\pi/16}^{\pi/4} = G(\pi/4) - G(\pi/16)$.

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

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