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Question

Mathematics Question on integral

Prove: ∫0π2sin3xdx=23∫^{\frac{π}{2}}_0 sin^3xdx=\frac{2}{3}

Answer

Let I=∫0π2sin3xdx=23I=∫^{\frac{π}{2}}_0 sin^3xdx=\frac{2}{3}
I=∫0π2sin2x.sinxdxI=∫^{\frac{π}{2}}_0 sin^2x.sinxdx
=∫0π2(1−cos2x)sinxdx=∫^{\frac{π}{2}}_0(1-cos^2x)sinxdx
=∫0π2sinxdx−∫0π2cos2x.sinxdx=∫^{\frac{π}{2}}_0 sinxdx-∫^{\frac{π}{2}}_0 cos^2x.sinxdx
=[−cosx]0π2+[cos3x3]0π2=[-cosx]^{\frac{π}{2}}_0+[\frac{cos^3x}{3}]^{\frac{π}{2}}_0
=1+13[−1]=1−13=23=1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}
Hence,the given result is proved.