Question

Let $I_1=\underset{0}{\overset{\pi /3}{\int }}(2\sec x{)}^{100}dx,$ and $I_2=\underset{0}{\overset{\ln (2+\sqrt{3})}{\int }}(e^x+e^{-x}{)}^{99}dx$. Then the value of $\frac{I_1}{I_2}$ is

• A. $2$
• B. $\frac{1}{2}$
• C. $1$
• D. $2^{99}$

Answer 1

The correct option is A $2$

$I_1={\int }_0^{\pi /3}(2\sec x{)}^{100}dx{I}_2={\int }_0^{\ln (2+\sqrt{3})}(e^x+e^{-x}{)}^{99}dx$
Let $e^x=\sec t+\tan t$
$\Rightarrow e^{x}dx=\sec t(\sec t+\tan t)dt$
$\Rightarrow dx=\sec tdt$
$I_2={\int }_0^{\pi /3}\sec t\{\sec t+\tan t+\sec t-\tan t{\}}^{99}dt$
$={\int }_0^{\pi /3}\sec t(2\sec t{)}^{99}dt$
$=\frac{1}{2}{\int }_0^{\pi /3}(2\sec t{)}^{100}dt$
$I_2=\frac{1}{2}{I}_1$
$\Rightarrow \frac{I_1}{I_2}=2$