Mathematics, 23.08.2019 11:20, lolgirl9757
(a) use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integral from 0 to pi/2 of sin(x)^(n-2)dx where n> =2 is an integer.
(b)use part (a) to evaluate integral from 0 to pi/2 of sin(x)^3dx and integral from 0 to pi/2 of sin(x)^5dx.
(c) use part (a) to show that, for odd powers of sine, integral from 0 to pi/2 sin(x)^(2n 1)dx is (2*4*6**2n)/[3*5**(2n 1)].
Answers: 1
![I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\ = [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\ =0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\ ](/tpl/images/0190/7701/29c04.png)
![\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\ = \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\ = \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\ ](/tpl/images/0190/7701/94d05.png)
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(a) use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integr...
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