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0-pi-2-cos-3-x-1-cos-2-x-dx-




Question Number 149903 by bramlexs22 last updated on 08/Aug/21
 Ω = ∫_0 ^(π/2)  ((cos^3 x)/( (√(1−cos^2 x)))) dx
Ω=π20cos3x1cos2xdx
Answered by Ar Brandon last updated on 08/Aug/21
Ω=∫_0 ^(π/2) ((cos^3 x)/( (√(1−cos^2 x))))dx=∫_0 ^(π/2) ((1−sin^2 x)/( sinx))cosxdx      =∫_0 ^1 ((1/u)−u)du=[lnu−(u^2 /2)]_0 ^1 =−(1/2)+∞  divergent
Ω=0π2cos3x1cos2xdx=0π21sin2xsinxcosxdx=01(1uu)du=[lnuu22]01=12+divergent

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