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Question Number 25682 by abdo imad last updated on 13/Dec/17
we give ∫_0 ^∞  t^(a−1) (1 + t)^(−1) dt =π (sin(πa))^(−1)  with 0<a<1 find the value of ∫_0 ^∞  (1 +x^(16) )^(−1) dx
wegive0ta1(1+t)1dt=π(sin(πa))1with0<a<1findthevalueof0(1+x16)1dx
Answered by ajfour last updated on 13/Dec/17
If  x^(16) =t  and   let  ∫(dx/(1+x^(16) )) = I  Then  I=∫(dt/(16x^(15) (1+t)))     =(1/(16))∫_0 ^(  ∞)  ((t^(−15/16)  dt)/(1+t))     =(1/(16))∫_0 ^(  ∞)  ((t^((1/(16))−1) dt)/(1+t)) =(1/(16))×(π/(sin ((π/(16)))))  sin (π/(16)) =(√((1−cos (π/8))/2))                =(√((1−(√((1+1/(√2))/2)))/2))  .
Ifx16=tandletdx1+x16=IThenI=dt16x15(1+t)=1160t15/16dt1+t=1160t1161dt1+t=116×πsin(π16)sinπ16=1cos(π/8)2=11+1/222.

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