Intergral of cube root tanx = Integral [(tan x)^1/3 . dx] Using chain Rule it becomes =>{ (tan x)^(1/3 + 1)/ (1/3 + 1)} * Integral (tanx.dx) To keep it simple let (tanx)^(4/3) / (4/3) = A => A * Integral(tanx.dx) - Equation 1 => A*Integral[(sinx/cosx)dx] Let cosx = u - Differentiating both sides. =du/dx = -sinx SO, du = -sinx.dx Put -sinx dx = du and cosx = u in equation 1 => A*Integral(-du/u) => A* -( ln(|u|) + c) Substituting values of A and u back in the equation => (tanx)^(4/3) / (4/3) * (-ln(cosx)+ c)

Dread it, run from it, the end of summer arrives all the same. And now it's here. It’s finally August.…

APPROVED BY CLIENTS

Every assingment I've needed helped with has been dealt with an efficient and wonderful manner. I will continue to use AssignmentExpert in the near future as needed. Great services every time.

## Comments

vivek chhabra11.08.11, 18:44Intergral of cube root tanx = Integral [(tan x)^1/3 . dx] Using chain Rule it becomes =>{ (tan x)^(1/3 + 1)/ (1/3 + 1)} * Integral (tanx.dx) To keep it simple let (tanx)^(4/3) / (4/3) = A => A * Integral(tanx.dx) - Equation 1 => A*Integral[(sinx/cosx)dx] Let cosx = u - Differentiating both sides. =du/dx = -sinx SO, du = -sinx.dx Put -sinx dx = du and cosx = u in equation 1 => A*Integral(-du/u) => A* -( ln(|u|) + c) Substituting values of A and u back in the equation => (tanx)^(4/3) / (4/3) * (-ln(cosx)+ c)

## Leave a comment