Let `I_(m)=int_(0)^(pi)((1-cos mx)/(1-cos x))dx` use mathematical induction to prove that `l_(m)= m pi, m=0,1,2`......

Explanation

`becauseI_(m)=int_(0)^(pi)((1-cosmx)/(1-cosx))dx`
Step I For `m=1,l_(1) int_(0)^(pi)((1-cosx)/(1-cosx))dx`
`therefore I_(1) = pi and "for" m=2`,
`I_(2)int_(0)^(pi)((1-cos2x)/(1-cosx))dx`
`=int_(0)^(pi)(2sin^2x(1+cosx))/((1-cosx)(1+cosx))dx`
`=int_(0)^(pi)(2sin^2x(1+cosx))/(sin^2x)dx=2int_(0)^(pi)(1+cosx)dx`
`=2[x + sin x]_(0)^(pi)=2(pi+0)-(0+0)=2pi` which are true , therefore , `I_1 and I_2` are true.
Step II Assume `I_(k+1)=int_(0)^(pi)(1-cos(k+1)x)/(1-cosx)dx`
`therefore I_(k+1)-I_(k)=int_(0)^(pi)(coskx-cos (k+1)x)/(1-cosx)dx`
`=int_(0)^(pi)(2sin((2k+1)/(2))x.sin((x)/(2)))/(2sin^2((x)/(2)))dx`
`=int_(0)^(pi)(sin((2k+1)/2)x)/(sin((x)/(2)))dx`.....(iii)
Similarly ,`I_(k)-I_(k-1)=int_(0)^(pi)(sin((2k-1)/(2))x)/(sin((x)/(2)))dx`....(iv)
On subtracting Eq.(iv) from Eq.(iii) , we get
`I_(k+1) -2I_(k)+I_(k-1)=int_(0)^(pi)(sin((2k+1)/(2))x-sin((2k-1)/(2))x)/(sin((x)/(2)))dx`
`=int_(0)^(pi)(2cos(kx)sin((x)/(2)))/(sin((x)/(2)))dx2 int_(0)^(pi)cos kxdx =2[(sinkx)/(k)]_(0)^(pi)=0`
`rArr l_(k+1)=2I_(k)-I_(k-1)=2kpi-(k-1)pi` [by assumption step]
`=kpi+pi=(k=1)pi`
This show that the result is true for `m=k+1`. Hence , by the principle of mathematical induction the result is true for all `m in N`.