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Integration of square root tan x dx?
for solving this ..the first thing to do is substitute tanx=t^2 then x=tan inverse t^2 then solve the integral..
Integration by parts of x tanx?
XtanX dx formula uv - int v du u = x du = dx dv = tanX dx v = ln(secX) x ln(secX) - int ln(secx) dx = X ln(secx) - x ln(secx) - x + C -----------------------------------------
What is the limit of x- sin x cos x over tan x -x as x tends to zero?
It is minus 1 I did this: sinx/cos x = tan x sinx x = cosx tanx you have (x - sinxcosx) / (tanx -x) (x- cos^2 x tan x)/(tanx -x) let x =0 -cos^2 x (tanx) /tanx = -cos^x -cos^2 (0) = -1
How do you simplify cosx plus sinx tanx?
to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities.. Cosx + Sinx Tanx Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx Cosx + Sin2x/ Cos x <------------- do the LCD Cosx (Cosx/Cosx) + Sin2x/Cosx (Cos2x+Sin2x)/Cosx 1/Cosx <--------- From Sin2x + Cos2x =1 or Secx <-------- answer Comment if you have questions...:))
What is integration of secx tanx?
Will try integration by parts. uv - int[v du] u = sec(x)----------------du = sec(x) tan(x) dv = tan(x)---------------v = ln[sec(x)] sec(x) ln[sex(x)] - int[lnsec(x) dx] = sec(x) ln[sec(x)] - xlnsec(x) - x + C ===========================
