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0-pi-2-dx-1-tan-5-x-




Question Number 151423 by peter frank last updated on 21/Aug/21
∫_0 ^(π/2) (dx/(1+tan^5 x))
0π2dx1+tan5x
Answered by Olaf_Thorendsen last updated on 21/Aug/21
I = ∫_0 ^(π/2) (dx/(1+tan^5 x))  I = ∫_0 ^(π/2) ((cos^5 x)/(cos^5 x+sin^5 x)) dx    (1)  Let u = (π/2)−x :  I = ∫_0 ^(π/2) ((sin^5 u)/(sin^5 u+cos^5 u)) du    (2)  (((1)+(2))/2) : I = (π/4)
I=0π2dx1+tan5xI=0π2cos5xcos5x+sin5xdx(1)Letu=π2x:I=0π2sin5usin5u+cos5udu(2)(1)+(2)2:I=π4

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