Mathematics

Integrate the following:
$$\displaystyle \int\limits_0^{\pi /2} {{{\sin }^3}x} \,\,\cos x\,\,dx$$


SOLUTION
$$I = \displaystyle \int^{\pi/2}_0\sin^3c\cos x\ dx$$

$$= \displaystyle \int_0^{\pi/2} \sin^3\left(\dfrac{\pi}{2}-x\right) \cos \left(\dfrac{\pi}{2} - x\right)dx$$

$$= \displaystyle \int_{0}^{\pi/2} \cos^3x\sin x\, dx$$

$$\therefore I + I = \displaystyle \int_0^{\pi/2} (\sin^3x\cos x + \cos^3x\sin x)dx$$

$$2I = \int_0^{\pi/2}\sin x \cos x(\sin^2x + \cos^2x)dx$$

$$= \int_0^{\pi/2} \sin x \cos x dx$$

$$= \left[ \dfrac{\sin^2x}{2}\right]_0^{\pi/2} = \dfrac{1}{2}$$

$$I = \dfrac{1}{4}$$

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Subjective Medium Published on 17th 09, 2020
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