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
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