Question

Let $I_m={\int }_0^{\pi }\frac{1-cosmx}{1-cosx}dx$
Use mathematical induction to prove that
$I_m=m\pi ,m0$ , , 1, 2.....

Answer 1

$p\left(1\right)={\int }_0^{\pi }\frac{1-cosx}{1-cosx}dx=[x{]}_0^{\pi }=1.\pi =\pi$
$p\left(2\right)={\int }_0^{\pi }\frac{1-cos2x}{1-cosx}dx={\int }_0^{\pi }\frac{2si{n}^{2}x}{2si{n}^2\left(x/2\right)}$
${\int }_0^{\pi }4.co{s}^2\frac{x}{2}dx$
$={\int }_0^{\pi }2(1+cosx)dx=2.\pi +0=2\pi$
Let us assume that p (m) = mn and we will establish that p(m + 1) = (m + 1)$\pi$
$\therefore p(m+1)={\int }_0^{\pi }\frac{1-cos(m+1)}{1-cosx}dx=(m+1)\pi$
Now cos (m + 1) x + cos (m - 1) x = 2 cos mx cos x
$\therefore$ -cos(m + 1)x
= cos(m - 1) x - 2 cos mx cos x
or 1 - cos (m + 1)x
= 1 + cos(m - 1) x - 2 cos mx cos x
= -[1 - cos(m - 1) x] + 2 cos mx (1 - cos x) + 2 - 2 cos mx
Now divide throughout by 1 - cos x and integrate
$\therefore$ p(m + 1) (by 1)
= -p(m - 1) + 2 p (m) + ${\int }_0^{\pi }$ 2 cos mx dx
= -(m - 1) $\pi$ + 2 $m\pi$ + 2 $[sinmx{]}_0^{\pi }$
= (m + 1) $\pi$ + 0 = (m + 1) $\pi$