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

$\sqrt{3}$

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int^{2n\pi}_{0}\max {(\sin x,\sin^{-1}(\sin x))}dx$ equals to (wher, $n\ \in\ I$)
• A. $\dfrac {n(\pi^{2}-4)}{4}$
• B. $\dfrac {n(\pi^{2}-8)}{4}$
• C. $\dfrac {n(\pi^{2}-4)}{4}$
• D. $\dfrac {n(\pi^{2}-4)}{2}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } \right) } dx }$
• A. $\dfrac{\pi}{2}$
• B. $0$
• C. None of these
• D. $\dfrac{\pi}{2}-1$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
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1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\int _{ }^{ }{ \cfrac { \log { x } }{ { \left( x+1 \right) }^{ 2 } } } dx$ is
• A. $\cfrac { \log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C$
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