Find the value of \(\rm \int_{0}^{\pi\over2}\sin^2x\cos x\) dx

Concept:

• ∫ sin x dx = -cos x + C
• ∫ cos x dx = sin x + C
• ∫ xⁿ dx = \(\rm {x^{n+1}\over n+1}\) + C

Calculation:

I = \(\rm ∫\sin^2x\cos x\)dx

Let sin x = t

dt = cos x dx

⇒ I = ∫ t² (dt)

⇒ I =  \(\rm t^3\over3\) + c

⇒ I = \(\rm {\sin^3x\over3}\) + c

Putting the limits 

⇒ I = \(\rm \left[{\sin^3x\over3}+c\right]^{\pi\over2}_0\)

⇒ I = \(\rm \left[{\sin^3{\pi\over2}\over3}+c\right]-\left[{\sin^3(0)\over3}+c\right]\)

⇒ I = \(\boldsymbol {\rm {1\over3}}\)