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)

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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)

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