Integral of Trig. Function, Power

Date: 7/19/96 at 11:25:22
From: Suryadie Gemilang
Subject: Integral of Trig. Function, Power

Can you solve this one?
Integral of dx/cos^4 x

Thanks
Sg

Date: 7/19/96 at 17:12:26
From: Doctor Anthony
Subject: Re: Integral of Trig. Function, Power

 I = INT[dx/cos^4(x)] = INT[sec^4(x)*dx]

                      = INT[(1+tan^2(x))*sec^2(x) dx]   

Integrate by parts

  I = (1+tan^2(x))*tan(x) - INT[tan(x)*2tan(x)*sec^2(x)*dx]

    =  (1+tan^2(x))*tan(x) - 2*INT[tan^2(x)*sec^2(x)*dx]

    =  (1+tan^2(x))*tan(x) - 2*Int[(sec^2(x)-1)*sec^2(x)*dx]

    = tan(x)+tan^3(x) - 2*INT[{sec^4(x) - sec^2(x)}*dx]

  I = tan(x)+tan^3(x) - 2*I + 2*INT[sec^2(x)*dx]

  3I = tan(x)+tan^3(x) + 2*tan(x) + Const

   I = (1/3)(tan(x)+tan^3(x)) + (2/3)tan(x) + C

     =  tan(x) + (1/3)tan^3(x) + C


-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/