Using the principle of mathematical induction to prove that `int_(0)^(pi//2)(sin^2nx)/(sinx)dx=1+(1)/(3)+(1)/(5)+.....+(1)/(2n-1)`

Explanation

Let `P(n): int _(0)^(pi//2)(sin^2nx)/(sinx)dx=1+(1)/(3)+(1)/(5)+......+(1)/(2n-1)` ..........(i)
Step I For n =1
LHS of Eq. 9i) `= int_(0)^(pi//2)(sin^2x)/(sinx)dx=int_(0)^(pi//2)sin xdx=-[cosx]_(0)^(pi//2)=-(0-1)=1` and RHS of Eq. (i) =1
Therefore , P(1) is true .
Step II Assume it is true for n=k, then
`P(k):int_(0)^(pi//2)(sin^2kx)/(sinx)dx=1+(1)/(3)+(1)/(5)+.......+(1)/(2k-1)`
Step III For `n=k+1`,
`P(k+1):int_(0)^(pi//2)(sin^2(k+1)x)/(sinx)dx=1+(1)/(3)+(1)/(5)+.....+(1)/(2k-1)+(1)/(2k+1)`
LHS `=int_(0)^(pi//2)(sin(k+1)x)/(sinx)kdx`
`=int_(0)^(pi//2)(sin^2(k+1)x-sin^2kx+sin^2kx)/(sinx)dx`
`=int_(0)^(pi//2)(sin^2(k+1)x-sin^2kx)/(sinx)dx+int_(0)^(pi//2)(sin^2kx)/(sinx)dx`
`=int_(0)^(pi//2)(sin(2k+1)xsinx)/(sinx)dx+P(k)` [by assumption step]
`=int_(0)^(pi//2)sin(2k+1)xdx+P(k)`
`=-[(cos (2k+1)x)/(2k+1)]_(0)^(pi//2)+P(k)`
`=-(1)/((2k+1))[cos(pik+(pi)/(2))-1]+P(k)`
`=-(1)/((2k+1))[-sinpik-1]+P(k)`
`=-(1)/(2k+1)[-0-1]+P(k)`
`=(1)/((2k+1))+1+(1)/(3)+(1)/(5)+......+(1)/((2k-1))` [by assumption step]
`=1+(1)/(3)+(1)/(5)+.....+(1)/((2k-1))+(1)/((2k+1))=RHS`
This shows that the result is true for `n=k+1`. Hence , by the principle of mathematical induction , the result is true for all `n in N`,