Question

(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*7...*(2n 1)].

Answer 1

Hello,

a)
$I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\ = [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\ =0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx$
$I(1+n-1)= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\ I= \dfrac{n-1}{n} *\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx$

b)
$\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\ = \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\ = \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3}$

$\int\limits^{ \frac{\pi}{2} }_0 {sin^{5}(x)} \, dx \\ = \dfrac{4}{5}*\dfrac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx = \dfrac{8}{15}$

c)

$I_n= \dfrac{n-1}{n} * I_{n-2} \\ I_{2n+1}= \dfrac{2n+1-1}{2n+1} * I_{2n+1-2} \\ = \dfrac{2n}{2n+1} * I_{2n-1} \\ = \dfrac{(2n)*(2n-2)}{(2n+1)(2n-1)} * I_{2n-3} \\ = \dfrac{(2n)*(2n-2)*...*2}{(2n+1)(2n-1)*...*3} * I_{1} \\\\ I_1=1$