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Calculate integral of y=x*tan^2x?
Solution
Since we know that (tan x)^2+1 = (tan x)', we'll add and subtract 1 inside the expression of the
given function;
Int f(x)dx = Int x[(tan x)^2+1-1]dx
We'll remove the brackets:
Int x[(tan x)^2+1] - Int xdx
We'll note Int x[(tan x)^2+1] = I1 and Int xdx = I2.
Since I2 is an elementary formula, we'll solve it first:
I2 = x^2/2 + C
For solving I1 we'll apply integration by parts:
Int udv = uv - Int vdu
u = x => du = dx
dv = [(tan x)^2+1] = (tan x)'dx
v = tan x
I1 = x*tanx - Int tan xdx
I1 = x*tanx + ln (cos x)
Int f(x)dx = I1 - I2
Int f(x)dx = x*tanx + ln (cos x) - x^2/2 + C

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Calculate integral of y=xtan^2xSolutionSince we know that (t.pdf

  • 1. Calculate integral of y=x*tan^2x? Solution Since we know that (tan x)^2+1 = (tan x)', we'll add and subtract 1 inside the expression of the given function; Int f(x)dx = Int x[(tan x)^2+1-1]dx We'll remove the brackets: Int x[(tan x)^2+1] - Int xdx We'll note Int x[(tan x)^2+1] = I1 and Int xdx = I2. Since I2 is an elementary formula, we'll solve it first: I2 = x^2/2 + C For solving I1 we'll apply integration by parts: Int udv = uv - Int vdu u = x => du = dx dv = [(tan x)^2+1] = (tan x)'dx v = tan x I1 = x*tanx - Int tan xdx I1 = x*tanx + ln (cos x) Int f(x)dx = I1 - I2 Int f(x)dx = x*tanx + ln (cos x) - x^2/2 + C