Results 1 to 9 of 9

Math Help - Find the Indicated Integrals

  1. #1
    Junior Member
    Joined
    May 2008
    Posts
    50

    Find the Indicated Integrals

    I'm supposed to find several indefinite integrals, but I am not sure exactly how to do each, especially as they involve natural logs and trigonmetric functions.

    The three are:

    \int ln(x^4)/x dx

    \int ((e^t^x) cos (e^t))/(3+5 sin (e^t))

    and lastly

    \int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi
    Quote Originally Posted by Pikeman85 View Post
    \int ln(x^4)/x dx
    \frac{\ln(x^4)}{x}=\frac{4\ln x}{x}=4\times \frac{1}{x}\times \ln x and as the derivative of x\mapsto \ln x is x\mapsto\frac{1}{x}...
    \int ((e^t^x) cos (e^t))/(3+5 sin (e^t))
    Is it \int \frac{\mathrm{e}^{tx} \cos (\mathrm{e}^t)}{3+5 \sin (\mathrm{e}^t)}\,\mathrm{d}{\color{red}x} or \int \frac{\mathrm{e}^{tx} \cos (\mathrm{e}^t)}{3+5 \sin (\mathrm{e}^t)}\,\mathrm{d}{\color{red}t} ?

    \int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))
    \int_0^{\frac{4}{5}} \frac{\arcsin \left(\frac{5}{4}x\right)}{\sqrt{16-25x^2}}\,\mathrm{d}x
    Remember that \arcsin'x=\frac{1}{\sqrt{1-x^2}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Arrow

    Quote Originally Posted by Pikeman85 View Post
    I'm supposed to find several indefinite integrals, but I am not sure exactly how to do each, especially as they involve natural logs and trigonmetric functions.

    The three are:

    \int ln(x^4)/x dx

    \int ((e^t^x) cos (e^t))/(3+5 sin (e^t))

    and lastly

    \int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))

    for the frist one

    \int \frac{\ln(x^4)}{x}dx=4\int\frac{\ln(x)}{x}dx

    Let u=\ln(x) \implies du=\frac{1}{x}dx

    4\int udu=2u^2+C=2\left( \ln(x)\right)^2+C

    For number 2 you have both x's and t's but no dx or dt what variable are integrating with respect to?

    For the last one

    \int_{0}^{4/5}\frac{\sin^{-1\left( \frac{5}{4}x\right)}}{\sqrt{16-25x^2}}dx

    let x=\frac{4}{5}\sin(t) \implies dx=\frac{4}{5}\cos(t)dt

    \int_{0}^{4/5}\frac{\sin^{-1\left( \frac{5}{4}x\right)}}{\sqrt{16-25x^2}}dx=\int_{0}^{\frac{\pi}{2}}\frac{t}{\sqrt{1  6-16\sin^2(t)}}\left( \frac{4}{5}\cos(t)dt \right)=

    \frac{1}{5}\int_{0}^{\pi/2}tdt=\frac{1}{5} \left[ \frac{1}{2} \left( \frac{\pi}{2}\right)^2-0\right]=\frac{\pi^2}{40}

    Here is a web page with some La tex code for you
    Helpisplaying a formula - Wikipedia, the free encyclopedia
    Good luck.
    Last edited by TheEmptySet; June 19th 2008 at 11:59 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,746
    Thanks
    648
    Hello, Pikeman85!

    These all require "simple" substitutions.
    . . The trick is to recognize them.


    \int \frac{\ln(x^4)}{x}\,dx
    We have: . \int\frac{4\ln(x)}{x}\,dx \;=\;4\int \ln x\,\frac{dx}{x}

    Let u \:=\:\ln(x) \quad\Rightarrow\quad du \:=\:\frac{dx}{x}

    Substitute: . 4\int u\,du . . . etc.



    \int \frac{e^{t}\cos(e^t)\,dt}{3+5\sin(e^t)}
    Let u \:=\:3+5\sin(e^t)\quad\Rightarrow\quad du \:=\:5e^t\cos(e^t)\,dt \quad\Rightarrow\quad e^t\cos(e^t)\,dt \:=\:\frac{1}{5}\,du

    Substitute: . \int\frac{\frac{1}{5}\,du}{u} \;=\;\frac{1}{5}\int \frac{du}{u} . . . etc.



    \int^{\frac{4}{5}}_0 \frac{\sin^{-1}\!\left(\frac{5}{4}x\right)}{\sqrt{16-25x^2}}\,dx
    The denominator is: . \sqrt{16\left(1 - \frac{25}{16}x^2\right)} \;=\;4\sqrt{1 - \left(\frac{5}{4}x\right)^2}

    The integral becomes: . \int^{\frac{4}{5}}_0 \frac{\sin^{-1}\!\left(\frac{5}{4}x\right)\,dx} {4\sqrt{1 - \left(\frac{5}{4}x\right)^2}} . = \;\;\frac{1}{4}\int^{\frac{4}{5}}_0\sin^{-1}\!\left(\frac{5}{4}x\right)\cdot\frac{dx}{\sqrt{  1 - \left(\frac{5}{4}x\right)^2}}

    \text{Let }u \:=\:\sin^{-1}\!\left(\frac{5}{4}x\right) \quad\Rightarrow\quad<br />
du \:=\:\frac{\frac{5}{4}\,dx}{\sqrt{1-\left(\frac{5}{4}x\right)^2}} <br />
\quad\Rightarrow\quad \frac{dx}{\sqrt{1-\left(\frac{5}{4}x\right)^2}} \:=\:\frac{4}{5}\,du

    Substitute: . \frac{1}{4}\int^{\frac{4}{5}}_0 u\cdot\frac{4}{5}\,du \;=\;\frac{1}{5}\int^{\frac{4}{5}}_0 u\,du . . . etc.



    Edit: I'm way too slow this time . . . *sigh*
    .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2008
    Posts
    50
    What about this integral?

    \int x^2 \sqrt{(7 + x^3)} dx

    The answer I got for it is  (x^3/3) \sqrt{(7x+(x^4/4))}

    Which is not correct. How am I doing this incorrectly? I imagine my substitution is wrong.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by Pikeman85 View Post
    What about this integral?

    \int x^2 sqrt(7 + x^3) dx

    The answer I got for it is (x^3/3) sqrt(7x+(x^4/4))

    Which is not correct. How am I doing this incorrectly? I imagine my substitution is wrong.
    One can notice that the derivative of 7+x^3 is 3x^2 so x^2 \sqrt{7 + x^3} looks like u'(x)\sqrt{u(x)} which can easily be integrated : you may try to substitute u(x)=x^3+7.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Well, a faster substitution is also z^2=x^3+7.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    May 2008
    Posts
    50
    x^3 \sqrt{7+x^3}

    I got this, but it did not work. I'm missing a step here I think

    I don't get substitution very well
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Pikeman85 View Post
    What about this integral?

    \int x^2 \sqrt{(7 + x^3)} dx

    The answer I got for it is  (x^3/3) \sqrt{(7x+(x^4/4))}

    Which is not correct. How am I doing this incorrectly? I imagine my substitution is wrong.
    Let u=7+x^3 \implies du=3x^2dx \iff \frac{du}{3}=x^2dx

    Now subbing these into your integral gives

    \int x^2\sqrt{7+x^3}dx=\int \sqrt{u}\left( \frac{du}{3}\right)=\frac{1}{3}\int u^\frac{1}{2}du=\frac{2}{9}u^\frac{3}{2}+C=\frac{2  }{9}(7+x^3)^\frac{3}{2}+C
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using Integrals to Find Work
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 7th 2011, 06:57 PM
  2. using integrals to find volume
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 23rd 2010, 03:43 AM
  3. using integrals to find volume
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 20th 2010, 06:20 PM
  4. Replies: 4
    Last Post: May 10th 2010, 09:29 AM
  5. how to find these contour integrals?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 20th 2009, 10:19 PM

Search Tags


/mathhelpforum @mathhelpforum