Mathematics - Definite Integration Question with Solution | TestHub

Let $\frac{d}{dx}G(x)=\frac{e^{\tan x}}{x},x\in \left(0,\frac{\pi }{2}\right)$ Now, $$\begin{array}{rl}I& ={\int }_{1/4}^{1/2}\frac{2}{x}{e}^{\tan \pi x^2}\cdot dx\\ & ={\int }_{1/4}^{1/2}\frac{2\pi x}{\pi x^2}{e}^{\tan \pi x^2}\cdot dx\end{array}$$ Let $\pi x^2=t\Rightarrow 2\pi xdx=dt$ When $x=\frac{1}{2},t=\frac{\pi }{4}$ and $x=\frac{1}{4},t=\frac{\pi }{16}$ $$\begin{array}{rl}\therefore I& ={\int }_{\pi /16}^{\pi /4}\frac{e^{\tan t}}{t}dt=g(t)\mid \begin{array}{l}\frac{\pi }{4}\\ \frac{\pi }{16}\end{array}\\ & =G\left(\frac{\pi }{4}\right)-G\left(\frac{\pi }{16}\right)\end{array}$$