Question

The value of $\underset{0}{\overset{\ln 2}{\int }}\sqrt{e^x-1}dx$ is

• A. $1-\frac{\pi }{4}$
• B. $2-\frac{\pi }{2}$
• C. $2-\frac{\pi }{4}$
• D. $\ln 2-\frac{\pi }{2}$

Answer 1

The correct option is B $2-\frac{\pi }{2}$

$I=\underset{0}{\overset{\ln 2}{\int }}\sqrt{e^x-1}dx$
Put $u=\sqrt{e^x-1}$
$\Rightarrow du=\frac{e^x}{2\sqrt{e^x-1}}dx$
$\Rightarrow \frac{2u}{u^2+1}du=dx$
So we can write,
$I=\underset{0}{\overset{1}{\int }}\frac{2u^2}{u^2+1}du$
$\Rightarrow \frac{I}{2}=\underset{0}{\overset{1}{\int }}\frac{u^2}{u^2+1}du$
$\Rightarrow \frac{I}{2}=\underset{0}{\overset{1}{\int }}\frac{u^2+1-1}{u^2+1}du$
$\Rightarrow \frac{I}{2}=\underset{0}{\overset{1}{\int }}1-\frac{1}{u^2+1}du$
$\Rightarrow \frac{I}{2}={[u-{\tan }^{-1}u]}_0^1$
$\Rightarrow I=2-\frac{\pi }{2}$