calculus
posted by Alex .
using tangent half substitution show that integral of
sec(x)=ln(secx+tanx)

calculus 
Anonymous
If tan(x/2)=u then
dx=2du/(1+u^2), sec(x)=(1+u^2)/(1u^2)
Integral of sec(x)dx=Integral of 2du/(1u^2)=ln((1+u)/(1u))
(1+tan(x/2))/(1tan(x/2))=
(cos(x/2)+sin(x/2))/(cos(x/2)sin(x/2))=
Multiply numerator and denominator by
cos(x/2)+sin(x/2)
=(1+sin(x))/cos(x)=sec(x)+tan(x)
Respond to this Question
Similar Questions

Calculus  Integration
Hello! I really don't think I am understanding my calc hw. Please help me fix my errors. Thank you! 1. integral from 0 to pi/4 of (tanx^2)(secx^4)dx It says u = tan x to substitute So if I use u = tan x, then my du = secx^2 then I … 
calculus
find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it. Use the chain rule. Let y(u) = ln u u(x) = sec x + tan x … 
Calculus
find the derivative of f(x)=tanx4/secx I used the quotient rule and got (sec^2x4)(secx)(tanx4)(cosx)/sec^2x but I'm pretty sure that's wrong. Help. Thanks. 
Calculus
I think I have the correct answers for the following problems. For anyone who has the time I would appreciate it if you could tell me if I am correct/incorrect. Thank you. 1). Differentiate ã(x)sin x = (2xcosx+sinx)/2ã(x) 2). … 
Calc
Hello im trying to integrate tan^3 dx i have solved out the whole thing but it doesnt match up with the solution.. this is what i did: first i broke it up into: integral tan^2x (tanx) dx integral (sec^2x1)(tanx) dx then i did a u … 
Calc
Hello im trying to integrate tan^3 dx i have solved out the whole thing but it doesnt match up with the solution.. this is what i did: first i broke it up into: integral tan^2x (tanx) dx integral (sec^2x1)(tanx) dx then i did a u … 
Calc
Evaluate the integral using any method: (Integral)sec^3x/tanx dx I started it out and got secx(1tan^2x)/tanx. I know I just have to continue simplifying and finding the integral, but I'm stuck on the next couple of steps. Also, I have … 
Math
I can't find the integral for (tanx)^(6)*(secx)^(2) I tried splitting up tanx into (tanx)^2*(tanx)^4 and let the latter equal (secx)^2  1. Please help, thanks! 
Calculus
Int tanx sec^2x dx can be taken as (by putting it in form of Int xdx) Int secx.secx tanx dx=(sec^2x)/2 Int tanx.sec^2x dx=(tan^2)/2. Which one is correct and why is the difference? 
Math
Evaluate the integral of (secx)^2 * (tanx)^3 *dx I started out with letting u=secx and du=secx*tanx*dx , but then I am kind of stuck because I don't know how to factor out just secx*tanx?