Results 1 to 3 of 3

Math Help - Trig Integration -- secant and tangent problem

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    23

    Trig Integration -- secant and tangent problem

    Indefinite integral of (sec(x))^1/2(tan^3(x)).... I know that tan^3(x) becomes tan^2(x) and tanx.... so it will be the Indefinite integral of (sec(x))^-1/2 (tan^2(x)) secx tanx dx ... I dont understand why my book says that there will be a secx in there, why isn't it just (sec(x))^1/2 (tan^2(x)) tanx dx?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Sep 2009
    Posts
    23
    Quote Originally Posted by bgonzal8 View Post
    Indefinite integral of (sec(x))^1/2(tan^3(x)).... I know that tan^3(x) becomes tan^2(x) and tanx.... so it will be the Indefinite integral of (sec(x))^-1/2 (tan^2(x)) secx tanx dx ... I dont understand why my book says that there will be a secx in there, why isn't it just (sec(x))^1/2 (tan^2(x)) tanx dx?

    Nevermind I figured it out... sec-1/2 (x) +sec x = -1/2 +1 = 1/2 .. sec^1/2 (x) is the original factor...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member DeMath's Avatar
    Joined
    Nov 2008
    From
    Moscow
    Posts
    473
    Thanks
    5
    Quote Originally Posted by bgonzal8 View Post
    Indefinite integral of (sec(x))^1/2(tan^3(x)).... I know that tan^3(x) becomes tan^2(x) and tanx.... so it will be the Indefinite integral of (sec(x))^-1/2 (tan^2(x)) secx tanx dx ... I dont understand why my book says that there will be a secx in there, why isn't it just (sec(x))^1/2 (tan^2(x)) tanx dx?
    Hint:

    \int {{{\sec }^{1/2}}x\,{{\tan }^3}x\,dx}  = \int {\frac{{{{\tan }^3}x}}{{{{\cos }^{1/2}}x}}\,dx}  = \int {\frac{{{{\sin }^3}x}}<br />
{{{{\cos }^{1/2}}x\,{{\cos }^3}x}}\,dx}  = \int {\frac{{{{\sin }^3}x}}<br />
{{{{\cos }^{7/2}}x}}\,dx}  =

    =  - \int {\frac{{\sin x\left( {{{\cos }^2}x - 1} \right)}}<br />
{{{{\cos }^{7/2}}x}}\,dx}  = \left\{ \begin{gathered}<br />
  \cos x = u, \hfill \\<br />
   - \sin x\,dx = du \hfill \\ <br />
\end{gathered}  \right\} = \int {\frac{{{u^2} - 1}}<br />
{{{u^{7/2}}}}\,du}  =  \ldots
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tangent and Secant line
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 2nd 2012, 11:53 AM
  2. Replies: 8
    Last Post: February 20th 2010, 06:52 PM
  3. tangent/secant lines.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 7th 2009, 05:44 PM
  4. tangent and secant
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 29th 2008, 04:24 AM
  5. Replies: 0
    Last Post: January 16th 2008, 04:49 PM

Search Tags


/mathhelpforum @mathhelpforum