Results 1 to 5 of 5

Math Help - Integral of sec^3 (theta) d(theta)

  1. #1
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597

    Integral of sec^3 (theta) d(theta)

    How do I integrate sec^3 (theta) d(theta) ?

    I am doing another problem and this is the step I am stuck on. I know all my previous steps are correct because math software told me but I'd like to be able to manually do this.

    Any help would be greatly appreciated!
    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2009
    Posts
    202
    Quote Originally Posted by s3a View Post
    How do I integrate sec^3 (theta) d(theta) ?

    I am doing another problem and this is the step I am stuck on. I know all my previous steps are correct because math software told me but I'd like to be able to manually do this.

    Any help would be greatly appreciated!
    Thanks in advance!
    1)express sec as cos
    2)multiply numerator and denominator by cos(theta)
    3)denominator becomes an even power of cos. so express it as sin
    4)substitute sin(theta)=t
    5)use partial fractions
    6)integrate
    Follow Math Help Forum on Facebook and Google+

  3. #3
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597
    I subsitute for u instead of t because I am more familiar with it and I am at Integral of 1/(1-u^2)^2 du but am a bit lost now. Could you show me what to do with the partial fractions?

    I don't know if the following is right:

    1 = A(u+1)(u-1)^2 + B(u-1)^2 + C(u+1)^2 * (u-1) + D(u+1)^2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,059
    Thanks
    894
    Quote Originally Posted by s3a View Post
    How do I integrate sec^3 (theta) d(theta) ?
    I always thought the method of parts was a neat way to do it ...

    \int \sec{t} \cdot \sec^2{t} \, dt

    u = \sec{t} ... du = \sec{t}\tan{t} \, dt

    dv = \sec^2{t} \, dt ... v = \tan{t}

    \int \sec^3{t} \, dt = \sec{t}\tan{t} - \int \tan^2{t}\sec{t} \, dt

    \int \sec^3{t} \, dt = \sec{t}\tan{t} - \int (sec^2{t}-1)\sec{t} \, dt

    \int \sec^3{t} \, dt = \sec{t}\tan{t} - \int sec^3{t}-\sec{t} \, dt

    \int \sec^3{t} \, dt = \sec{t}\tan{t} - \int sec^3{t} \, dt + \int \sec{t} \, dt

    2\int \sec^3{t} \, dt = \sec{t}\tan{t} + \int  \sec{t} \, dt

    2\int \sec^3{t} \, dt = \sec{t}\tan{t} + \ln|\sec{t}+\tan{t}| + C<br />

    \int \sec^3{t} \, dt = \frac{1}{2}\left[\sec{t}\tan{t} + \ln|\sec{t}+\tan{t}|\right] + C
    Follow Math Help Forum on Facebook and Google+

  5. #5
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597
    Just a question, what if I had a higher power though. Like what if I had Integral (sec^5(x))dx is there any easy method for that?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 8th 2010, 03:13 PM
  2. Replies: 0
    Last Post: April 29th 2010, 10:24 AM
  3. Replies: 2
    Last Post: March 29th 2010, 07:38 AM
  4. Replies: 3
    Last Post: February 6th 2009, 04:19 PM
  5. Replies: 1
    Last Post: January 23rd 2009, 10:53 AM

Search Tags


/mathhelpforum @mathhelpforum