The integral of tan(x) dx = ln | sec(x) | + c
to solve... tan(x) = sin(x)/cos(x)
the integral of (sin(x)/cos(x) dx) ... let u = cos(x) then du = -sin(x) dx
= the integral of (1/u -du)
= -ln | u | + c
= -ln | cos(x) | + c
= ln | (cos(x))^-1 | + c ... or ... ln | 1/cos(x) | + c
= ln | sec(x) | + c
Chat with our AI personalities
The integral of sqrt(tan(x)) is rather complex and is hard to show with the formatting allowed on Answers.com. See the related links for a representation of the answer.
for solving this ..the first thing to do is substitute tanx=t^2 then x=tan inverse t^2 then solve the integral..
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 -----------------------------------------
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
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...:))