Just use the "Owl Transform" on it!

[@cablethelarryguy]

I wish you used the first method. I find all these formulae overwhelming. I still love it.

[@owlsmath]

Good point. Makes sense. The first method is more straightforward and no formulas to remember

[@holyshit922]

My solution t^3 = tan(x) substitution then by parts with u = t , dv = 3t^2/(t^3 + 1)^2dt After integration by parts 1/(1+t^3) = 1/2 *1/(1+t^3)+1/2*1/(1+t^3) then in one of the integrals 1/2*1/(1+t^3) use t = 1/u substitution then add both integrals

[@holyshit922]

In my opinion solution above is the easiest No Beta and Gamma functions and stuff like this

[@owlsmath]

nice!

[@slavinojunepri7648]

Excellent

[@owlsmath]

Thanks!

[@reaperskyfall6691]

This can be mcqs material

[@owlsmath]

Sorry what’s mcqs? I’ll try to google it 😆

[@owlsmath]

Multiple choice question?

[@suryamgangwal8315]

if we use 2/3 as x wont the answer be negative?

[@owlsmath]

No, because sin(2pi/3) = sin(pi/3) = sqrt(3)/2

[@suryamgangwal8315]

@@owlsmath yeahhhh, I was things of sin(-x)=-sin(x).

[@maxvangulik1988]

I=int[0,pi/2](cbrt(tan(x))/(1+sin(2x)))dx t=tan(x) dx=dt/(1+t^2) sin(2x)=2t/(1+t^2) I=int[0,♾️](cbrt(t)/(1+2t+t^2))dt I=int[0,♾️](t^(1/3)/(1+t)^2)dt ẞ(x,y)=int[0,♾️](p^(x-1)/(1+p)^(x+y))dp I=ẞ(4/3,2/3)=Ř(4/3)Ř(2/3)/Ř(2) Ř(2)=1!=1 Ř(4/3)=1/3•Ř(1/3) I=Ř(1/3)Ř(2/3)/3 Ř(x)Ř(1-x)=pi•csc(pi•x) I=pi/3•csc(pi/3) I=2pi/3sqrt(3)