if we let
u = 1+7tanx
du = 7sec^2 x dx
and you have ∫ u^(-1/3) 1/7 du
if we let
u = sin√x
du = cos√x * 1/(2√x)
and you have ∫ 1/√u 2 du
integrate from 0 to pi/4 (sec^2x)/((1+7tanx)^2)^1/3
integrate form pi^2/36 to pi^2/4 (cos(x^1/2))/(xsin(x^1/2))^1/2
integrate from 0 to pi/3 (tanx)/(2secx)^1/2
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