Question

$\stackrel{\ln \pi }{\underset{\ln \pi -\ln 2}{\int }}\frac{e^x}{1-\cos \left(\frac{2}{3}{e}^x\right)}dx$ is equal to

• A. $\sqrt{3}$
• B. $-\sqrt{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. $-\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A $\sqrt{3}$

Now,

$\stackrel{\ln \pi }{\underset{\ln \pi -\ln 2}{\int }}\frac{e^x}{1-\cos \left(\frac{2}{3}{e}^x\right)}dx$

$=\stackrel{\ln \pi }{\underset{\ln \frac{\pi }{2}}{\int }}\frac{e^x}{1-\cos \left(\frac{2}{3}{e}^x\right)}dx$

Put $\frac{2e^x}{3}=z$.

or, $e^{x}dx=\frac{3}{2}dz$, again when $x=\ln \pi$ then $z=\frac{2}{3}\pi$ and $x=\ln \frac{\pi }{2}$ then $z=\frac{\pi }{3}$.

So the given integration becomes,

$=\stackrel{\frac{2\pi }{3}}{\underset{\frac{\pi }{3}}{\int }}\frac{\frac{3}{2}}{1-\cos \left(z\right)}dz$

$=\frac{3}{4}\stackrel{\frac{2\pi }{3}}{\underset{\frac{\pi }{3}}{\int }}\cos e{c}^2\frac{z}{2}dz$

$=\frac{3}{4}\times 2{[-\cot \frac{z}{2}]}_{\frac{\pi }{3}}^{\frac{2\pi }{3}}$

$=\frac{3}{2}\times [\sqrt{3}-\frac{1}{\sqrt{3}}]$

$=\sqrt{3}$.