Why-is-it-equal-1-2-0-pi-sin-2p-udu-0-pi-2-sin-2p-udu-

Question Number 170871 by sciencestudent last updated on 02/Jun/22

W h y i s i t e q u a l ?

\frac{1}{2} \underset{0}{\int^{π}} {s i n}^{2 p} u d u = \underset{0}{\int^{\frac{π}{2}}} {s i n}^{2 p} u d u

Answered by thfchristopher last updated on 02/Jun/22

\int_{0}^{π} \sin^{2 p} u d u

Let u = \frac{π}{2} - x, d u = - d x

When u = π, x = - \frac{π}{2}

u = 0, x = \frac{π}{2}

∴ \int_{0}^{π} \sin^{2 p} u d u

= - \int_{\frac{π}{2}}^{- \frac{π}{2}} \sin^{2 p} (\frac{π}{2} - x) d x

= \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2 p} x d x

Since \cos^{2 p} (\frac{π}{2}) = \cos^{2 p} (- \frac{π}{2}),

∴ it is an even function

\Rightarrow \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2 p} x d x = 2 \int_{0}^{\frac{π}{2}} \cos^{2 p} x d x

= - 2 \int_{\frac{π}{2}}^{0} \cos^{2 p} (\frac{π}{2} - u) d u

= 2 \int_{0}^{\frac{π}{2}} \sin^{2 p} u d u

\Rightarrow \int_{0}^{π} \sin^{2 p} u d u = 2 \int_{0}^{\frac{π}{2}} \sin^{2 p} u d u

\Rightarrow \frac{1}{2} \int_{0}^{π} \sin^{2 p} u d u = \int_{0}^{\frac{π}{2}} \sin^{2 p} u d u