Mathematics

# Solve:$\int _{ 0 }^{ \log { 2 } }{ \cos { 2x } dx }$=

##### SOLUTION
$\displaystyle\int_{0}^{\log{2}}{\cos{2x}dx}$

Let $t=\sin{2x}\Rightarrow dt=2\cos{2x}dx\Rightarrow \cos{2x}dx=\dfrac{dt}{2}$

When $x=0\Rightarrow t=0$

When $x=\log{2}\Rightarrow t=\sin{2\log{2}}$

$=\displaystyle\int_{0}^{\sin{2\log{2}}}{\dfrac{dt}{2}}$

$=\dfrac{1}{2}\left[t\right]_{0}^{\sin{2\log{2}}}$

$=\dfrac{1}{2}\left[\sin{2\log{2}}-0\right]$

$=\dfrac{1}{2}\sin{2\log{2}}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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