Question

Let $I=\underset{0}{\overset{\pi /4}{\int }}\sin 4x\cdot e^{{\tan }^{2}x}dx$. Then which of the following is/are CORRECT? (where [.] denotes greatest integer function)

• A. $[$ I $]=1$
• B. $[I]=0$
• C. $I=2-\frac{e}{2}$
• D. $I=1-\frac{e}{4}$

Answer 1

Answer: (B), (C)

Let $\underset{0}{\overset{\frac{\pi }{4}}{\int }}2\cdot \sin 2x\cdot \cos 2x\cdot e^{{\tan }^{2}x}dx=I$, then
$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}2\cdot \frac{2\tan x}{1+{\tan }^{2}x}\cdot \frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}\cdot e^{{\tan }^{2}x}dx$
$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{2\left(1-{\tan }^{2}x\right)}{\left(1+{\tan }^{2}x\right)}\cdot \frac{2\tan x\cdot {\sec }^{2}x}{\left(1+{\tan }^{2}x\right)}\cdot e^{{\tan }^{2}x}dx$
${\tan }^{2}x=t$
$2\tan x\cdot {\sec }^{2}xdx=dt$
$\text{ At }x=0,t=0$
$x=\frac{\pi }{4},t=1$
$I=\underset{0}{\overset{1}{\int }}\frac{2(1-t)}{(1+t{)}^3}\cdot e^{t}dt=-2\underset{0}{\overset{1}{\int }}\frac{t-1}{(1+t{)}^3}\cdot e^{t}dt=-2\underset{0}{\overset{1}{\int }}\frac{(t+1)-2}{(1+t{)}^3}\cdot e^{t}dt=-2\underset{0}{\overset{1}{\int }}{e}^t\left[\frac{1}{(t+1{)}^2}+\frac{-2}{(1+t{)}^3}\right]dt$
$=-2{\left[\frac{e^t}{(1+t{)}^2}\right]}_0^1=-2\left(\frac{e}{4}-1\right)=2-\frac{e}{2}$