Question

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

Answer: (A)

Let $\frac{d}{dx}G\left(x\right)=\frac{e^{tanx}}{x},x\in \left(0,\frac{\pi }{2}\right)$
Now, $I=\underset{1/4}{\overset{1/2}{\int }}\frac{2}{x}{e}^{tan\left(\pi x^2\right)}.dx<br/>\underset{1/4}{\overset{1/2}{\int }}\frac{2\pi }{\pi x^2}{e}^{tan\left(\pi x^2\right)}.dx$
Let $\pi x^2=t\Rightarrow 2\pi xdx=dt$
When $x=\frac{1}{2},t=\frac{\pi }{4}$ and $x=\frac{1}{4},t=\frac{\pi }{16}$
$I=\underset{\pi /16}{\overset{\pi /4}{\int }}\frac{e^{tant}}{t}dt=g\left(t\right)|\begin{array}{c}\frac{\pi }{4}\\ \frac{\pi }{16}\end{array}$
$=G\left(\frac{\pi }{4}\right)-G\left(\frac{\pi }{16}\right)$