Question

Using the principle mathematical induction , prove that $\forall nϵN$
$I_n={\int }_0^{\pi /2}co{s}^{n}xsinnxdx$
$=\frac{1}{2^{n+1}}[2+\frac{2^2}{2}+\frac{2^3}{3}....+\frac{2^n}{n}]$

Answer 1

p (1) ${\int }_0^{\pi /2}$ cos x sin x dx
$=\frac{1}{2}[si{n}^{2}x{]}_0^{\pi /2}=\frac{1}{2}=\frac{1}{2^{1+1}}\left[\frac{2}{1}\right]$
Assume p (m) to hold good.
p(m + 1) = ${\int }_0^{\pi /2}$ $co{s}^{m+1}$ x sin (m + 1) x dx
= ${\int }_0^{\pi /2}$ $co{s}^m$ x[sin (m + 1) x cos x] dx
= ${\int }_0^{\pi /2}$ $co{s}^m$ x[cos(m + 1) x sin x + sin mx]
Using the formula of sin (A - B)
i.e., sin mx = sin [(m + 1)x - x]
= sin (m + 1) x cos x - cos (m + 1) x sin x....
$therfore$ p (m + 1) = ${\int }_0^{\pi /2}$ $(co{s}^m$ x sin x )cos (m + 1) x + p (m), by
Integrating by parts.
$=\left[\frac{co{s}^{m+1}x}{m+1}cos\right(m+1\left)x\right]+\frac{1}{m+1}\int co{s}^{m+1}x\times$ -(m + 1) sin (m + 1) x + p (m)
p(m + 1) = $\frac{1}{m+1}$ - p(m + 1) + p(m)
2p (m + 1) = $\frac{1}{m+1}+,\frac{1}{2^{m+1}}[2++\frac{2^2}{2}+...+\frac{2^m}{m}]$
$=\frac{1}{2^{m+1}}[2+\frac{2^2}{2}+...+\frac{2^m}{m}+\frac{2^{m+1}}{m+1}]$
$\therefore$ p(m + 1) = $\frac{1}{2^{m+2}}[2+\frac{2^2}{2}+...+\frac{2^{m+1}}{m+}]$