If d d x G(x)= e^tan x x, x ∈(0, π / 2), then ∫_1 / 4^1 / 2 2 x · e^tan (π x^2 ) d x is equal to

Let $\frac{d}{d x} G(x)=\frac{e^{\tan x}}{x}, x \in\left(0, \frac{\pi}{2}\right)$dxd​G(x)=xetanx​,x∈(0,2π​) Now, $$\begin{aligned} I & =\int_{1 / 4}^{1 / 2} \frac{2}{x} e^{\tan \pi x^2} \cdot d x \\ & =\int_{1 / 4}^{1 / 2} \frac{2 \pi x}{\pi x^2} e^{\tan \pi x^2} \cdot d x \end{aligned}$$I​=∫1/41/2​x2​etanπx2⋅dx=∫1/41/2​πx22πx​etanπx2⋅dx​ Let $\pi x^2=t \Rightarrow 2 \pi x d x=d t$πx2=t⇒2πxdx=dt When $x=\frac{1}{2}, t=\frac{\pi}{4}$x=21​,t=4π​ and $x=\frac{1}{4}, t=\frac{\pi}{16}$x=41​,t=16π​ $$\begin{aligned} \therefore I & =\int_{\pi / 16}^{\pi / 4} \frac{e^{\tan t}}{t} d t=g(t) \mid \begin{array}{l} \frac{\pi}{4} \\ \frac{\pi}{16} \end{array} \\ & =G\left(\frac{\pi}{4}\right)-G\left(\frac{\pi}{16}\right) \end{aligned}$$∴I​=∫π/16π/4​tetant​dt=g(t)∣4π​16π​​=G(4π​)−G(16π​)​