Mathematics

# $\displaystyle\int_{0}^{\pi/2} \sin x \cos x\ dx$

##### SOLUTION
$\displaystyle\int_{0}^{\pi/2} \sin x \cos x\ dx$

Let $t=\sin x \implies dt=\cos x dx$

$t_1 =sin {\dfrac {\pi}{2}}=1=upper \ limit$

$t_2=sin {0}=0=lower \ limit$

Replacing we get,
$\displaystyle \int _0^1 t \ dt \\ \left.\dfrac {t^2}2 \right] _0^1 =\dfrac 12$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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