okay "smart guys"

Probably zero, due to symmetry.

(2(tanx)^(3/2))/3

>>9083723
Do your own homework anon.

>>9083723
Trick question. The letters are all next to each other so it's already integrated.

>>9083738
The function only has positive real and imaginary parts. The definite integral from negative to positive infinite is definately not 0.

>>9083723
tan(x) = u^2
sec^2(x)dx = 2udu
dx = 2udu/(sec^2(x)) = 2udu/(1 + tan^2(x)) = 2udu/(1+u^4)
int sqrt(tan(x)) dx = int 2u^2 du/(1 + u^4)
its time to factorize 1 + u^4. (-1)^(1/4) = sqrt(2)/2 + i*sqrt(2)/2, -sqrt(2)/2 + i*sqrt(2)/2, -sqrt(2)/2 - i*sqrt(2)/2, sqrt(2)/2 - i*sqrt(2)/2. This gives us 1 + u^4 = (u - (sqrt(2)/2 + i*sqrt(2)/2))*(u - (-sqrt(2)/2 + i*sqrt(2)/2))*(u - (-sqrt(2)/2 - i*sqrt(2)/2))*(u - (sqrt(2)/2 - i*sqrt(2)/2)). Multiplying conjugate terms, we get (u^2 - sqrt(2)*u + 1)*(u^2 + sqrt(2)*u + 1) = 1 + u^4. So, now we need to decompose 2*u^2/(u^2 - sqrt(2)*u + 1)*(u^2 + sqrt(2)*u + 1) = (A*u + B)/(u^2 - sqrt(2)*u + 1) + (C*u + D)/(u^2 + sqrt(2)*u + 1). This gives us (A*u + B)*(u^2 + sqrt(2)*u + 1) + (C*u + D)*(u^2 - sqrt(2)*u + 1) = 2*u^2. This gives us (A + C)*u^3 = 0, (sqrt(2)*A + B - sqrt(2)*C + D)*u^2 = 2*u^2, (A + sqrt(2)*B + C - sqrt(2)*D)*u = 0, B + D = 0. So (sqrt(2)*A + B + sqrt(2)*A - B)*u^2 = 2*u^2, so A = sqrt(2)/2, and C = -sqrt(2)/2. Also, B = D = 0. So 2*u^2/(1 + u^4) = sqrt(2)/2[u/(u^2 - sqrt(2)*u + 1) - u/(u^2 + sqrt(2)*u + 1)]. Now, we can just integrate these rational functions separately.

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168KB, 500x313px

>>9083845
ugliest post award

>>9083864
>can't compile LaTeX in his brain instantly to visualize gorgeous equations

yeah me neither

>>9084208
>this is [math]\rm \LaTeX[/math]