Q58 / 90 · JEE Main 2012 (12 May Online) · Mathematics
If $\frac{d}{d x} G(x)=\frac{e^{\tan x}}{x}, x \in(0, \pi / 2)$, then $\int_{1 / 4}^{1 / 2} \frac{2}{x} \cdot e^{\tan \left(\pi x^2\right)} d x$ is equal to
- Correct answerA$G(\pi / 4)-G(\pi / 16)$
- B$2[G(\pi / 4)-G(\pi / 16)]$
- C$\pi[G(1 / 2)-G(1 / 4)]$
- D$G(1 / \sqrt{2})-G(1 / 2)$
Solution
Let $\frac{d}{d x} G(x)=\frac{e^{\tan x}}{x}, x \in\left(0, \frac{\pi}{2}\right)$
Now,
$$
\begin{aligned}
I & =\int_{1 / 4}^{1 / 2} \frac{2}{x} e^{\tan \pi x^2} \cdot d x \\
& =\int_{1 / 4}^{1 / 2} \frac{2 \pi x}{\pi x^2} e^{\tan \pi x^2} \cdot d x
\end{aligned}
$$
Let $\pi x^2=t \Rightarrow 2 \pi x d x=d t$
When $x=\frac{1}{2}, t=\frac{\pi}{4}$ and $x=\frac{1}{4}, t=\frac{\pi}{16}$
$$
\begin{aligned}
\therefore I & =\int_{\pi / 16}^{\pi / 4} \frac{e^{\tan t}}{t} d t=g(t) \mid \begin{array}{l}
\frac{\pi}{4} \\
\frac{\pi}{16}
\end{array} \\
& =G\left(\frac{\pi}{4}\right)-G\left(\frac{\pi}{16}\right)
\end{aligned}
$$
Hence, the correct answer is option: 'A'.