Integration by PartsDate: 05/27/97 at 03:07:55 From: ajit Subject: indefinite integrals INT[sec^3 x]dx Date: 05/27/97 at 10:33:44 From: Doctor Anthony Subject: Re: indefinite integrals If we factor sec^3 into sec^2(x)*sec(x), we can solve the integral using parts. Our integral becomes: INT[sec(x)*sec^2(x)dx]. INT[sec^3(x)dx] = I = sec(x)tan(x) - INT[tan(x)sec(x)tan(x)dx] I = sec(x)tan(x) - INT[sec(x)(sec^2(x)-1)dx] I = sec(x)tan(x) - INT[sec^3(x)dx] + INT[sec(x)dx] 2I = sec(x)tan(x) + INT[sec(x)dx] 2I = sec(x)tan(x) + INT[sec(x)(sec(x)+tan(x))/(sec(x)+tan(x)) dx] Let u = sec(x)+tan(x), then du = (sec(x)tan(x) + sec^2(x))dx: 2I = sec(x)tan(x) + INT[du/u] 2I = sec(x)tan(x) + ln(sec(x)+tan(x)) + C I = (1/2)sec(x)tan(x) + (1/2)ln[sec(x)+tan(x)] + C -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/