Question

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

• A. $ G\left(\pi/4\right)-G\left(\pi/16\right)$
• B. $2\left[G\left(\pi/4\right)-G\left(\pi/16\right)\right]$
• C. $\pi\left[G\left(1/2\right)-G\left(1/4\right)\right]$
• D. $G\left(1/\sqrt{2}\right)-G\left(1/2\right)$

Answer 1

The Correct Option is A

Let $\frac{d}{dx}G\left(x\right) = \frac{e^{tan\,x}}{x}, x\in\left(0, \frac{\pi}{2}\right)$ Now, $I = \int\limits^{1/2}_{1/4} \frac{2}{x} e^{tan\left(\pi\,x^2\right)}.dx \int\limits^{1/2}_{1/4} \frac{2\pi}{\pi x^{2}} e^{tan\left(\pi \,x^2\right)}.dx$ Let $\pi x^{2} = t \Rightarrow 2\pi x \,dx = dt$ When $x = \frac{1}{2}, t = \frac{\pi}{4}$ and $x = \frac{1}{4}, t = \frac{\pi }{16}$ $I = \int\limits^{\pi/4}_{\pi/16} \frac{e^{tan\,t}}{t}dt = g\left(t\right) |\begin{matrix}\frac{\pi}{4}\\ \frac{\pi}{16}\end{matrix}$ $= G \left(\frac{\pi }{4}\right)-G \left(\frac{\pi }{16}\right)$