Mathematics

# The value of $\displaystyle \int_{0}^{3\pi /4}$ $\sqrt{1-sin2x} \ dx$

$2\sqrt{2}+1$

##### SOLUTION
$\displaystyle \int _{ 0 }^{ \frac { 3\pi }{ 4 } }{ \sqrt { 1-\sin { 2x } } } dx=\int _{ 0 }^{ \frac { 3\pi }{ 4 } }{ \sqrt { { \left( \sin { x } -\cos { x } \right) }^{ 2 } } } dx$
$\displaystyle =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \cos { x } -\sin { x } \right) dx } +\int _{ \frac { \pi }{ 4 } }^{ \frac { 3\pi }{ 4 } }{ \left( \sin { x } -\cos { x } \right) } dx$
$\displaystyle =\left[ \sin { x } +\cos { x } \right] _{ 0 }^{ \frac { \pi }{ 4 } }{ + }\left[ -\cos { x } -\sin { x } \right] _{ \frac { \pi }{ 4 } }^{ \frac { 3\pi }{ 4 } }$
$\displaystyle =\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } +1+\frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \\ =2\sqrt { 2 } +1$

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

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$\displaystyle\int\dfrac{x}{\sqrt{3x^2+4}}dx$.