Results 1 to 4 of 4

Math Help - algebraic/trigonometric substitution

  1. #1
    Junior Member
    Joined
    Mar 2007
    Posts
    27

    algebraic/trigonometric substitution

    \int\frac{e^-x}{(9e^-2x+1)^3/2}\, dx

    \int\frac{dz}{(z^2-6z+18)^3/2}

    \int\frac{ln^3w}{w\sqrt{ln^2w-4}}\,dw
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by cazimi View Post
    \int\frac{e^{-x}}{(9e^{-2x}+1)^{3/2}}\, dx
    Substitute u=e^{-x},

    \int {\frac{{e^{ - x} }}<br />
{{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = - \int {\frac{{du}}<br />
{{\left( {9u^2 + 1} \right)\sqrt {9u^2 + 1} }}} .

    Make another substitution defined by u=\frac1\tau,

    \int {\frac{{e^{ - x} }}<br />
{{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = \int {\frac{\tau }<br />
{{\left( {9 + \tau ^2 } \right)\sqrt {9 + \tau ^2 } }}\,d\tau } .

    One more substitution according to \varphi = \sqrt {9 + \tau ^2 } ,

    \int {\frac{{e^{ - x} }}<br />
{{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = \int {\varphi ^{ - 2} \,d\varphi } = - \frac{1}<br />
{\varphi } + k.

    Back substitute:

    \begin{aligned}<br />
\varphi &= \sqrt {9 + \frac{1}<br />
{{u^2 }}}\\<br />
&= \sqrt {9 + \frac{1}<br />
{{e^{ - 2x} }}}\\<br />
&= \frac{{\sqrt {9e^{ - 2x}  + 1} }}<br />
{{e^{ - x} }}.<br />
\end{aligned}

    And we happily get \int {\frac{{e^{ - x} }}<br />
{{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = - \frac{{e^{ - x} }}<br />
{{\sqrt {9e^{ - 2x} + 1} }} + k.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Second one

    On the second integral,

    First, I think you meant
     \int\frac{dz}{(z^2-6z+18)^{3/2}} , not \int\frac{dz}{(z^2-6z+18)^3/2} (you forgot to bracket the exponent).

    First, we want to complete the square in the denominator: z^2-6z+18=z^2-6z+9+9=(z-3)^2+9

    So using u=z-3, we get  \int\frac{dz}{(z^2-6z+18)^{3/2}} = \int\frac{du}{(u^2+3^2)^{3/2}}

    From here, you can use the trigonometric substitution u=3 \tan \theta, which gives u^2+3^2 = 3^2 (\tan^2 \theta +1) = (3 \sec \theta)^2 and du = 3 \sec^2 \theta d\theta
    And thus
     \int\frac{dz}{(z^2-6z+18)^{3/2}} = \int\frac{3 \sec^2 \theta d\theta}{(3 \sec \theta)^3} =  \int\frac{d\theta}{9 \sec \theta} = \frac{1}{9} \int \cos \theta d\theta

    You can take it from there.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Well, the third

    Let u=\ln(w) thus: \frac{du}{dw}=\frac{1}{w}

    Then \int\frac{ln^3w}{w\sqrt{ln^2w-4}}dw=\int\frac{u^3}{\sqrt{u^2-4}}du

    Now z=\sqrt{u^2-4} so: \frac{dz}{du}=\frac{u}{\sqrt{u^2-4}}

    Therefore: \int\frac{u^3}{\sqrt{u^2-4}}du=\int(z^2+4)dz
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 1st 2011, 01:30 PM
  2. Algebraic Substitution Calculator
    Posted in the Math Software Forum
    Replies: 1
    Last Post: March 22nd 2011, 11:26 AM
  3. Trigonometric Expression As Algebraic
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 28th 2008, 02:59 PM
  4. Integration - Algebraic substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 31st 2007, 05:58 AM
  5. Integration - Algebraic substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 31st 2007, 05:44 AM

Search Tags


/mathhelpforum @mathhelpforum