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  • Prove the following (Exercises 34 to 39) Integral o to pi/2 sin^3 x dx equals 2/3

Prove the following (Exercises 34 to 39)

    Q37.    \int_0^\frac{\pi}{2}\sin^3 x dx =\frac{2}{3}

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\\Let\ I= \int_{0}^{\frac{\pi }{2}}sin^{3}xdx\\ I=\int_{0}^{\frac{\pi }{2}}sinx(1-sin^{2}x)dx\\ I=\int_{0}^{\frac{\pi }{2}}sinxdx-\int_{0}^{\frac{\pi }{2}}cos^{2}xsinxdx\\ I=I_{1}-I_{2}

\\I_{1}=[-cosx]_{0}^{\frac{\pi }{2}}\\ I_{1}=-0-(-1)=1

For I2 let cosx=t, -sinxdx=dt

The limits change to 0 and 1

\\I_{2}=-\int_{1}^{0}t^{2}dt\\ I_{2}=-[\frac{t^{3}}{3}]{_{1}}^{0}\\ I_{2}=0-(-\frac{1}{3})\\ I_{2}=\frac{1}{3}

I1-I2=2/3

Hence proved.

Posted by

Sayak

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