Results 1 to 5 of 5

Math Help - Tough integral

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    14

    Tough integral

    Well, probably not so tough. I haven't done much integrating in a while and I ran into an integral of (sint*tant) in my homework. I couldn't rearrange it in any helpful way, so I'm hoping there is an identity for it. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by griffsterb View Post
    Well, probably not so tough. I haven't done much integrating in a while and I ran into an integral of (sint*tant) in my homework. I couldn't rearrange it in any helpful way, so I'm hoping there is an identity for it. Thanks.
    note that \sin t \tan t = \sin t \cdot \frac {\sin t}{\cos t} = \frac {\sin^2 t}{\cos t} = \frac {1 - \cos^2 t}{\cos t}=  \sec t - \cos t. you should know the integrals of each of those
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2008
    Posts
    14
    Bah, should have seen that sin squared identity in the numerator! Shakin the rust off :) Thanks so much!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by griffsterb View Post
    Bah, should have seen that sin squared identity in the numerator! Shakin the rust off Thanks so much!
    you're welcome

    keep at it!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member polymerase's Avatar
    Joined
    May 2007
    From
    Sydney
    Posts
    267
    Or you can apply by parts:

    u=\tan x \to du=\sec^2x dv=\sin x \to v=-\cos x

    \int \sin x\tan x=-\cos x\tan x+\int \cos x\sec^2 x\;dx
    \sec^2x\cos x = \frac{\cos x}{\cos^2x}=\sec x

    Therefore, \int \cos x\sec^2 x\;dx=\int \sec x \;dx=\ln|\sec x+\tan x|+C

    Thus, \int \sin x\tan x\;dx=\ln|\sec x+ \tan x|-\sin x +C
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tough integral
    Posted in the Calculus Forum
    Replies: 10
    Last Post: April 29th 2010, 11:23 AM
  2. Tough Integral.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 19th 2009, 11:23 PM
  3. a tough integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 10th 2009, 02:39 AM
  4. tough integral?.
    Posted in the Calculus Forum
    Replies: 0
    Last Post: December 7th 2008, 05:07 PM
  5. tough integral?.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 19th 2008, 06:17 PM

Search Tags


/mathhelpforum @mathhelpforum