Question

If there differentiation is given as $\dfrac{d}{dx}G(x)=\dfrac{{{e}^{\tan x}}}{x},x\in \left( 0,\dfrac{\pi }{2} \right)$, then $\int\limits_{1/4}^{1/2}{\dfrac{2}{x}{{e}^{\tan \left( \pi {{x}^{2}} \right)}}}dx$ is equal to:
a) $G\left( \dfrac{\pi }{4} \right)-G\left( \dfrac{\pi }{16} \right)$
b) \[G\left( \dfrac{1}{\sqrt{2}} \right)-G\left( \dfrac{1}{2} \right)\]
c) \[\pi \left[ G\left( \dfrac{1}{2} \right)-G\left( \dfrac{1}{4} \right) \right]\]
d) $2\left[ G\left( \dfrac{\pi }{4} \right)-G\left( \dfrac{\pi }{16} \right) \right]$

Answer 1

Hint: We are given a function of derivative as: $\dfrac{d}{dx}G(x)=\dfrac{{{e}^{\tan x}}}{x},x\in \left( 0,\dfrac{\pi }{2} \right)$ We need to solve the given integral as: $\int\limits_{1/4}^{1/2}{\dfrac{2}{x}{{e}^{\tan \left( \pi {{x}^{2}} \right)}}}dx$. Let us assume that $I=\int\limits_{1/4}^{1/2}{\dfrac{2}{x}{{e}^{\tan \left( \pi {{x}^{2}} \right)}}}dx$. Now, multiply and divide I by \[\pi x\]. Then, let $\pi {{x}^{2}}=t$ and solve the integral. As we know that: \[\int\limits_{a}^{b}{F(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)\]. Now, use this identity to find the solution for $\int\limits_{1/4}^{1/2}{\dfrac{2}{x}{{e}^{\tan \left( \pi {{x}^{2}} \right)}}}dx$

Complete step-by-step solution:
As we have: $\dfrac{d}{dx}G(x)=\dfrac{{{e}^{\tan x}}}{x},x\in \left( 0,\dfrac{\pi }{2} \right).................(1)$
Since we have assumed that: $I=\int\limits_{1/4}^{1/2}{\dfrac{2}{x}{{e}^{\tan \left( \pi {{x}^{2}} \right)}}}dx...................(2)$
Now, multiply and divide the equation (2) by \[\pi x\], we get:
$I=\int\limits_{1/4}^{1/2}{\dfrac{2\pi x}{\pi {{x}^{2}}}{{e}^{\tan \left( \pi {{x}^{2}} \right)}}}dx……….........(3)$
Now, let: $\pi {{x}^{2}}=t..................(4)$
Now, differentiate the equation (4), we get:
$2\pi xdx=dt................(5)$
Therefore,
$\begin{align}
  & x=\dfrac{1}{4}\to t=\dfrac{\pi }{16} \\
 & x=\dfrac{1}{2}\to t=\dfrac{\pi }{4} \\
\end{align}$
Now, put the value of the equation (4) and the equation (5) in the equation (3), we get:
$I=\int\limits_{\pi /16}^{\pi /4}{\dfrac{{{e}^{\tan \left( t \right)}}}{t}}dt.................(6)$
As we know that, from the equation (1), $\dfrac{d}{dx}G(x)=\dfrac{{{e}^{\tan x}}}{x}$
Now, integrate the equation (1), we get:
$\int{\dfrac{d}{dx}G(x)}dx=\int{\dfrac{{{e}^{\tan x}}}{x}}dx................(7)$
Since we know that:
$\int{\dfrac{d}{dx}}F(x)dx=F(x)$
So, we can write the equation (7) as:
$G(x)=\int{\dfrac{{{e}^{\tan x}}}{x}}dx...................(8)$
Now, compare equation (6) with the equation (8), we get:
$\int\limits_{\pi /16}^{\pi /4}{\dfrac{{{e}^{\tan \left( t \right)}}}{t}}dt=G(t)..............(9)$
Now, by using the formula: \[\int\limits_{a}^{b}{F(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)\]
We get:
$\begin{align}
  & \int\limits_{\pi /16}^{\pi /4}{\dfrac{{{e}^{\tan \left( t \right)}}}{t}}dt=\left\{ G(t) \right\}_{\pi /16}^{\pi /4} \\
 & =G\left( \dfrac{\pi }{4} \right)-G\left( \dfrac{\pi }{16} \right)
\end{align}$
Hence, option (a) is the correct answer.

Note: While applying the substitution method to solve the integral, be careful to change the upper limit and lower limit of the integral. Students might forget to change the limits which give a wrong answer. Also, we are given a derivative of the function whose integral is required. So, try to make an equation so that we can use the formula: $\int{\dfrac{d}{dx}}F(x)dx=F(x)$. It makes the solution easier.