Mathematics

$$\displaystyle \overset{\ln \pi}{\underset{\ln\pi - \ln2}{\int}} \dfrac{e^x}{1 - \cos \left(\tfrac{2}{3}e^x\right)}dx$$ is equal to


ANSWER

$$\sqrt{3}$$


SOLUTION
Now,
$$\displaystyle \overset{\ln \pi}{\underset{\ln\pi - \ln2}{\int}} \dfrac{e^x}{1 - \cos \left(\dfrac{2}{3}e^x\right)}dx$$

$$=\displaystyle \overset{\ln \pi}{\underset{\ln\tfrac{\pi}{2}}{\int}} \dfrac{e^x}{1 - \cos \left(\dfrac{2}{3}e^x\right)}dx$$

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

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

So the given integration becomes,

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

$$=\dfrac{3}{4}\displaystyle \overset{ \tfrac{2\pi}{3}}{\underset{\tfrac{\pi}{3}}{\int}}\cos ec^2 \dfrac{z}{2}dz$$

$$=\dfrac{3}{4}\times 2\left[-\cot \dfrac{z}{2}\right]_{\tfrac{\pi}{3}}^{\tfrac{2\pi}{3}}$$

$$=\dfrac{3}{2}\times\left[\sqrt3-\dfrac{1}{\sqrt 3}\right] $$

$$=\sqrt 3$$.
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Single Correct Medium Published on 17th 09, 2020
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