Mathematics

# Evaluate the given integral.$\displaystyle\int{\dfrac{\sqrt{1-\cos{2x}}}{2}}dx$

##### SOLUTION
$I=\displaystyle\int{\dfrac{\sqrt{1-\cos{2x}}}{2}}dx$

$=\displaystyle\int{\dfrac{\sqrt{1-1+2{\sin}^{2}{x}}}{2}}dx$          ....$(cos2x=1-2sin^2x)$

$=\displaystyle\int{\dfrac{\sqrt{2{\sin}^{2}{x}}}{2}}dx$

$=\dfrac{\sqrt{2}}{2}\displaystyle\int{\sin{x}dx}$

$=\dfrac{\sqrt{2}\times \sqrt{2}}{2\sqrt{2}}\times-\cos{x}+c$

$=\dfrac{-1}{\sqrt{2}}\cos{x}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int { \cfrac { dx }{ \sqrt { { e }^{ 2x } } -1 } }$ equals to
• A. $\cos ^{ -1 }{ { e }^{ x } }$
• B. $\ln \left|e^x\right|-x+C$
• C. $\sec ^{ -1 }{ { e }^{ x } }$
• D. $\ln \left|e^x-1\right|-x+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \lim _{ n\rightarrow \infty }{ \left[ \frac { { 1 }^{ 2 } }{ { 1 }^{ 3 }+{ n }^{ 3 } } +\frac { { 2 }^{ 2 } }{ { 2 }^{ 3 }+{ n }^{ 3 } } +...+\frac { 1 }{ 2n } \right] }$ is equal to
• A. $\displaystyle \frac { 1 }{ 2 } \log { 2 }$
• B. $\log { 2 }$
• C. none of these
• D. $\displaystyle \frac { 1 }{ 3 } \log { 2 }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Solve: $\displaystyle \overset{1}{\underset{0}{\int}} \dfrac{dx}{\sqrt{x + 1} + \sqrt{x}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Integrate the function    $x\cos^{-1}x$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\displaystyle \int {\frac{{\left( {{e^{2x}} - 1} \right)}}{{{e^{2x}} + 1}}} dx$
• A. $\log (e^x-e^{-x})+C$
• B. $\log (e^{2x}+e^{-2x})+C$
• C. $\log (e^{2x}-e^{-2x})+C$
• D. $\log (e^x+e^{-x})+C$