If `I_(n)=int_(0)^(pi)x^(n)sinxdx`, then find the value of `I_(5)+20I_

1.	If `I_(n)=int_(0)^(pi)x^(n)sinxdx`, then find the value of `I_(5)+20I_(3)`.
Answer» Correct Answer - `pi^(5)` `I_(m)=int_(0)^(pi)x^(m) sin x dx` `=[-x^(m)cosx]_(0)^(pi)+n int_(0)^(pi)x^(n-1)cos x dx` `=pi^(n)+n[x^(n-1)sinx ]_(0)^(pi)-n(n-1)int_(0)^(pi)x^(n-2)sin x dx` `implies I_(m)=pi^(n)+n.0-n(n-1)I_(n-2)` Put `n=5` `I_(5)=pi^(5)-20I_(3)` `I_(5)+20I_(3)=pi^(5)`

Discussion

Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)` Making the substitution `"tan"1/2x=t`, we have `I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.
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