Page 1 of 2 12 LastLast
Results 1 to 15 of 22

Math Help - Problems with some integration

  1. #1
    Senior Member
    Joined
    Jul 2006
    Posts
    364

    Problems with some integration

    Hi, im new here, I have some problems with the following integrations:
    I've tried various things with all of them, but cant figure them out. I'd love some pointers, hopefully with explanations

    1. \int \frac {\sin^3(x)} {\cos^2(x)}dx
    2. \int^{\frac{\pi}{2}}_{0} \frac {\sin(3x)} {\cos(3x)} dx with a suitable substitution is? Apparently the answer is: \int^{1}_{0} \frac {1}{3u} du , but im not sure what i should pull put as 'u'. Any ideas?
    3. The area of the region bounded by the axis, the line x=1 and the curve: y = \frac {1}{\sqrt{4-3x^2}}, so i presume we need to get the definite integral from 0-1 of that curve, but once again i cant figure out the answer.

    I'll add to this list as more problems arise.

    Thanks for the help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by scorpion007

    1. \int \frac {\sin^3(x)} {\cos^2(x)}dx
    \int \frac {\sin^3(x)} {\cos^2(x)}dx =\int (\sin(x))^2\ \frac{sin}{(cos(x))^2}dx.

    Then integrate by parts with:

    u=(\sin(x))^2, and dv=\frac{sin}{(cos(x))^2},

    ( dv is the derivative of 1/\cos(x))

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by scorpion007
    2. \int^{\frac{\pi}{2}}_{0} \frac {\sin(3x)} {\cos(3x)} dx with a suitable substitution is? Apparently the answer is: \int^{1}_{0} \frac {1}{3u} du , but im not sure what i should pull put as 'u'. Any ideas?)
    u=\cos(3x)

    look plausible.

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by scorpion007
    3. The area of the region bounded by the axis, the line x=1 and the curve: y = \frac {1}{\sqrt{4-3x^2}}, so i presume we need to get the definite integral from 0-1 of that curve, but once again i cant figure out the answer.
    By the look of it you want:

    \int_0^1\ \frac {1}{\sqrt{4-3x^2}}\ dx=\int_0^1\ \frac {1}{2\ \sqrt{1-(3/4)x^2}}\ dx.

    Try the change of variable \sqrt{3/4}\ x=\sin(u)

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    Hey, thanks for all the help

    Could i get a little more help as to how to differentiate \sin^2(x)? I've forgotten how to do it, and i need to be able to do it for the first question to solve by parts.

    EDIT: nevermind, sorry, i remembered how to differentiate sin^2x. and i got the correct answer
    Time for the other Q's.
    Last edited by scorpion007; July 22nd 2006 at 01:10 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    ok, got the first 2 questions sorted.

    But for number 3, im getting the incorrect answer, this is what i did briefly:
    \frac {1}{2} \int^{1}_{0} \frac{1}{\sqrt{1- \frac{3}{4}x^2}} dx = \frac {1}{2} \Bigl[ \arcsin (\frac {\sqrt{3}}{2}x) \Bigr]^{1}_{0}
    Hence:
    <br />
\frac{1}{2} \Bigl[ \frac{\pi}{3} - 0 \Bigr] = \frac{\pi}{6}<br />
    right?
    This is apparently incorrect, as the answer is suppose to be \frac {\sqrt{3}\pi}{9}

    Did i screw up somewhere?

    EDIT: Nevermind, i missed a division by sqrt(3)/2, which caused it to be wrong..
    Last edited by scorpion007; July 22nd 2006 at 03:06 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    ok, now i have another problem, different question:

    The area of the region bounded by the axes, the line x = a, where 0 < a < 1, and the curve with equation y = \arccos (x) is determined by:

    Apparently this is the answer, but I don't understand how they derived it:
    a\arccos(a) + \int^{\frac{\pi}{2}}_{\arccos(a)} \cos(x) dx

    Any ideas?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by scorpion007
    ok, now i have another problem, different question:

    The area of the region bounded by the axes, the line x = a, where 0 < a < 1, and the curve with equation y = \arccos (x) is determined by:

    Apparently this is the answer, but I don't understand how they derived it:
    a\arccos(a) + \int^{\frac{\pi}{2}}_{\arccos(a)} \cos(x) dx

    Any ideas?
    Looks like making the substitution \cos(u)=x, and then
    integrating by parts.

    RonL
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645
    Hello, scorpion007!

    The area of the region bounded by the axes, the line x = a,\;0 < a < 1,
    and the curve with equation y = \arccos (x) is determined by:

    Apparently this is the answer, but I don't understand how they derived it:
    a\arccos(a) + \int^{\frac{\pi}{2}}_{\arccos(a)} \cos(x) dx

    Any ideas? . . . yes

    They were very inconsiderate . . .

    If they had written: . a\arccos(a) + \int^{\frac{\pi}{2}}_{\arccos(a)} \cos (y)\,dy

    . . we might have had a chance of figuring it out.


    The integration limits made me suspicious . . .
    Code:
            |
         π/2* *
            |::::::*  (a, arccos(a))
            + - - - - o
            |:::::::::: *
            |::::::::::   *
          - + - - - - + - -*- -
            0         a

    They integrated with respect to y ... and didn't tell us!

    We have: . x = \cos(y)

    The area of the rectangle is: . a\cdot\arccos(a)

    The area of the upper segment is: . \int^{\frac{\pi}{2}}_{arccos(a)} \cos y\,dy

    . . Diabolical!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    thanks for the help guys.

    Soroban, your reasoning seems very logical to me, and looks like that's what it *should* be. Unfortunately they say the answer is cos(x) dx, instead of your cos(y) dy. Which is obviously not correct according to your diagram. So im very confused.

    CaptainBlack, do you have any further advice on this perhaps? I'm not sure where to go from your substitution...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645
    Hello, scorpion007!

    I hope you agree that the area is: . a\cdot\arccos(a) + \int^{\frac{\pi}{2}}_{arccos(a)} \cos y\,dy


    Recall that, with a definite integral, the variable itself doesn't matter.

    For example: . \int^b_b f(\theta)\,d\theta \;= \;\int^b_a f(t)\,dt \;=\;\int^b_af(x)\,dx \;= \int^b_af(y)\,dy


    So my answer is equivalent to theirs . . . and mine is clearer.


    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    Here's also another question i couldnt solve. Its multiple choice. See if you can get it.

    f is a continuous, differentiable function with f(x) = f(-x) such that f(x) \ge 0 for x \in [0, a] and f(x) < 0 for x \in (a, b]. It is also known that \int^{b}_{0} f(x) dx = 0 and \int^{a}_{0} f(x) dx = A. Which of the following is false?

    A: \int^{b}_{-b} f(x) dx = 0
    B: -\int^{-a}_{0} f(x) dx = A
    C: -\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx - \int^{-a}_{0} f(x) dx = 3A
    D: \Bigl\vert \int^{b}_{a} f(x) dx \Bigr\vert + \Bigl\vert \int^{a}_{0} f(x) dx \Bigr\vert = 2A
    E: \int^{-a}_{-b} f(x) dx = 0

    I got confused just by the sheer amount of info in this question.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    Soroban, ah ok, i see what you mean. EDIT: nevermind, im an idiot :P, figured it out
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by scorpion007
    Here's also another question i couldnt solve. Its multiple choice. See if you can get it.

    f is a continuous, differentiable function with f(x) = f(-x) such that f(x) \ge 0 for x \in [0, a] and f(x) < 0 for x \in (a, b]. It is also known that \int^{b}_{0} f(x) dx = 0 and \int^{a}_{0} f(x) dx = A. Which of the following is false?

    A: \int^{b}_{-b} f(x) dx = 0(
    Is true, as:

    <br />
\int^{b}_{-b} f(x) dx=\int^{b}_{0} f(x) dx+\int^{0}_{-b} f(x) dx<br />

    <br />
=\int^{b}_{0} f(x) dx-\int^{0}_{b} f(-u) du<br />
    by change of variable u=-x in the second integral

    <br />
=\int^{b}_{0} f(x) dx-\int^{0}_{b} f(u) du<br />
    by symetry of f(u) about u=0, ie f(u)=f(-u)

    <br />
=\int^{b}_{0} f(x) dx+\int^{b}_{a} f(u) du=2\int^{b}_{0} f(x) dx=0<br />
    by interchanging the upper and lower limits of integration.

    RonL
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by scorpion007
    Here's also another question i couldnt solve. Its multiple choice. See if you can get it.

    f is a continuous, differentiable function with f(x) = f(-x) such that f(x) \ge 0 for x \in [0, a] and f(x) < 0 for x \in (a, b]. It is also known that \int^{b}_{0} f(x) dx = 0 and \int^{a}_{0} f(x) dx = A. Which of the following is false?

    B: -\int^{-a}_{0} f(x) dx = A
    Is true:

    -\int^{-a}_{0} f(x) dx=\int^{a}_{0} f(-u) du

    by change of variable u=-x

    <br />
=\int^{a}_{0} f(u) du<br />
    by symetry of f(x)
    <br />
=A<br />

    as this is given in the question.

    RonL
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  2. Problems with integration word problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 25th 2010, 05:39 PM
  3. integration problems
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2009, 04:01 AM
  4. integration problems
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 27th 2008, 04:31 AM
  5. Integration problems
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 26th 2008, 12:02 AM

Search Tags


/mathhelpforum @mathhelpforum