Results 1 to 6 of 6

Math Help - integrals.help please

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    3

    integrals.help please

    i've been doing this for hours, please help

    integrals.help please-tan33x-sec43x.png

    thx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    \begin{aligned}<br />
{\displaystyle{\int}}\tan^3\,3x\,\sec^4\,3x\,dx &=  {\displaystyle{\int}}\tan^3\,3x\,\sec^2\,3x\,\sec^  2\,3x\,dx \\<br />
&= {\displaystyle{\int}}\tan^3\,3x\,(1 + \tan^2\,3x)\sec^2\,3x\,dx \\<br />
&= {\displaystyle{\int}}(\tan^3\,3x\, + \tan^5\,3x\,)\sec^2\,3x\,dx \\<br />
&= {\displaystyle{\int}}\tan^3\,3x\,\sec^2\,3x\,dx + {\displaystyle{\int}}\tan^5\,3x\,\sec^2\,3x\,dx \\<br />
\end{aligned}

    \begin{aligned}<br />
u &= \tan\,3x \\<br />
du &= 3\sec^2\,3x\,dx \\<br />
\frac{1}{3}du &= \sec^2\,3x\,dx<br />
\end{aligned}

    Can you take it from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,905
    Thanks
    765
    Hello, kenziefc!

    We have to guess what the integral is supposed to be.

    Here's my approach . . .


    \int\tan^3\!3x\!\cdot\!\sec^4\!3x\!\cdot\!dx . . . . . my guess

    We have: . \int \sec^3\!3x\!\cdot\!\tan^2\!3x\!\cdot\!(\sec3x\!\cd  ot\!\tan3x\!\cdot\!dx)

    . . . . . =\;\int\sec^3\!3x\!\cdot\!(\sec^2\!3x - 1)\!\cdot\!(\sec3x\!\cdot\!\tan3x\!\cdot\!dx)

    . . . . . =\;\int(\sec^5\!3x - \sec^3\!3x)\!\cdot\!(\sec3x \tan3x\,dx)


    Let: u = \sec3x

    . . du \,=\,3\sec3x\tan3x\,dx \quad\Rightarrow\quad \sec3x\tan3x\,dx \,=\,\frac{1}{3}\,du


    Substitute: . \int (u^5-u^3)\cdot\frac{1}{3}\,du \;=\;\frac{1}{3}\int(u^5 - u^3)\,du


    I assume you can finish it . . .

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2010
    Posts
    3
    thx!!
    helped a lot!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,400
    Thanks
    1848
    Another way to do this (without so much guessing): convert to sine and cosine:
    \int tan^3(3x)sec^4(3x)dx= \int \frac{sin^3(3x)}{cos^3(3x)}\frac{1}{cos^4(3x)}dx= \int \frac{sin^3(3x)}{cos^7(3x)}dx

    Now, since that has sin(3x) to an odd power, the standard method is to factor one out for the differential and convert to cos(3x): \int\frac{sin^3(3x)}{cos^7(3x)}dx= \int\frac{1- cos^2(3x)}{cos^7(3x)}(sin(x)dx).

    Let u= cos(x) so that du= -sin(x)dx and the integral becomes -\int\frac{1- u^2}{u^7}du= \int (u^{-5}- u^{-7})du.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    We could also do with 'straightforward' substitution by letting \displaystyle u = \sec^4{3x},

    Then \displaystyle \frac{du}{dx} = 12\tan{3x}\sec^4{3x} \Rightarrow dx = \frac{du}{12\tan{3x}\sec^4{3x}}

    Thus \displaystyle \int\tan^3{3x}\sec^4{3x}\;{dx} = \int\dfrac{\tan^3{3x}\sec^4{3x}}{12\tan{3x}\sec^4{  3x}}\;{du} = \frac{1}{12}\int\tan^2{3x}\;{du}.

    Now, since \displaystyle u = \sec^4{3x} = (\tan^2{3x}+1)^2, we have \displaystyle \tan^2{3x} = \sqrt{u}-1.

    Thus \displaystyle \frac{1}{12}\int\tan^2{3x}\;{du} = \frac{1}{12}\int \sqrt{u}-1 \;{du} = ...

    Everyone so generously left something for the OP to do, and so would I.
    Last edited by TheCoffeeMachine; August 6th 2010 at 05:50 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Contour Integrals (to Evaluate Real Integrals)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 17th 2011, 10:23 PM
  2. integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2010, 02:54 PM
  3. Replies: 1
    Last Post: December 6th 2009, 08:43 PM
  4. Integrals and Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 9th 2009, 05:52 PM
  5. Some integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 20th 2008, 02:41 AM

Search Tags


/mathhelpforum @mathhelpforum