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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)` |
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Answer» 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`, |
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