Results 1 to 8 of 8

Math Help - integral

  1. #1
    Newbie AgentSmith's Avatar
    Joined
    Dec 2011
    Posts
    12

    integral

    \int \sin(101x)\sin^{99}(x) dx

    Wolfram Alpha says that the answer is \frac{\sin(100x)\sin^{100}(x)}{100}+C, but it does not show steps (unfortunately).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,409
    Thanks
    1294

    Re: integral

    Quote Originally Posted by AgentSmith View Post
    \int \sin(101x)\sin^{99}(x) dx

    Wolfram Alpha says that the answer is \frac{\sin(100x)\sin^{100}(x)}{100}+C, but it does not show steps (unfortunately).
    Thinking out loud... Maybe integration by parts?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461

    Re: integral

    Quote Originally Posted by AgentSmith View Post
    \int \sin(101x)\sin^{99}(x) dx

    Wolfram Alpha says that the answer is \frac{\sin(100x)\sin^{100}(x)}{100}+C, but it does not show steps (unfortunately).
    Would you not try a simple one like \int sin (2x) sin^2 (x) dx to see the steps and then use for your problem?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: integral

    Euler said " e^{ix}=\cos(x)+i\sin(x)"

    We have

    e^{ix}=\cos(x)+i\sin(x) \quad [1]
    e^{-ix}=\cos(x)-i\sin(x) \quad[2]

    Subtract [2] from [1]

    e^{ix}-e^{-ix}=2i\sin(x)
    \implies \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}
    \implies \sin^{99}(x)= \Big( \frac{e^{ix}-e^{-ix}}{2i}\Big)^{99}

    \begin{align*} I= \int \sin(101x)\sin^{99}(x)  \ dx &=\text{Im}\Big( \int e^{101ix} \Big( \frac{e^{ix}-e^{-ix}}{2i}\Big)^{99} dx\Big)\\ &= \text{Im}\Big( \int \frac{e^{101ix}}{(2i)^{99}} \Big( e^{ix}-e^{-ix}\Big)^{99} dx\Big) \\ &= \text{Im}\Big( \int \frac{e^{101ix}}{(-i)2^{99}} \Big( \frac{e^{2ix}-1}{e^{ix}}\Big)^{99} dx\Big) \\ &= \text{Im}\Big( \int \frac{e^{101ix}}{(-i)2^{99}e^{99ix}} \Big( {e^{2ix}-1}\Big)^{99} dx\Big) \\ &=\text{Im}\Big( \int \frac{e^{2ix}}{(-i)2^{99}} \Big( {e^{2ix}-1}\Big)^{99} dx\Big) \end{align*}

    Substitute u=e^{2ix} and du=2i e^{2ix} \ dx.

    \begin{align*} I &= \text{Im}\Big( \int \frac{1}{2^{100}} ( {u-1})^{99} dx\Big) \\ &= \text{Im}\Big(  \frac{1}{2^{100}} \frac{( {u-1})^{100}}{100} \Big) \\ &= \text{Im}\Big(  \frac{1}{2^{100}} \frac{e^{100ix}}{(2i)^{100}}\frac{2^{100}}{e^{100i  x}}\frac{( {e^{2ix}-1})^{100}}{100} \Big) \\ &= \text{Im}\Big(  \frac{e^{100ix}}{100} \Big(\frac{ {e^{2ix}-1}}{2i e^{ix}}\Big)^{100} \Big) \\ &= \text{Im}\Big(  \frac{e^{100ix}}{100} \sin^{100}(x) \Big) \\ &= \text{Im }\Big(  \frac{ \cos(100x) \sin^{100}(x)}{100} +i\frac{\sin(100x)\sin^{100}(x)}{100} \Big) \\ &= \frac{\sin(100x)\sin^{100}(x)}{100}+C\end{align*}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie AgentSmith's Avatar
    Joined
    Dec 2011
    Posts
    12

    Re: integral

    Quote Originally Posted by sbhatnagar View Post
    Euler said " e^{ix}=\cos(x)+i\sin(x)"

    We have

    e^{ix}=\cos(x)+i\sin(x) \quad [1]
    e^{-ix}=\cos(x)-i\sin(x) \quad[2]

    Subtract [2] from [1]

    e^{ix}-e^{-ix}=2i\sin(x)
    \implies \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}
    \implies \sin^{99}(x)= \Big( \frac{e^{ix}-e^{-ix}}{2i}\Big)^{99}

    \begin{align*} I= \int \sin(101x)\sin^{99}(x)  \ dx &=\text{Im}\Big( \int e^{101ix} \Big( \frac{e^{ix}-e^{-ix}}{2i}\Big)^{99} dx\Big)\\ &= \text{Im}\Big( \int \frac{e^{101ix}}{(2i)^{99}} \Big( e^{ix}-e^{-ix}\Big)^{99} dx\Big) \\ &= \text{Im}\Big( \int \frac{e^{101ix}}{(-i)2^{99}} \Big( \frac{e^{2ix}-1}{e^{ix}}\Big)^{99} dx\Big) \\ &= \text{Im}\Big( \int \frac{e^{101ix}}{(-i)2^{99}e^{99ix}} \Big( {e^{2ix}-1}\Big)^{99} dx\Big) \\ &=\text{Im}\Big( \int \frac{e^{2ix}}{(-i)2^{99}} \Big( {e^{2ix}-1}\Big)^{99} dx\Big) \end{align*}

    Substitute u=e^{2ix} and du=2i e^{2ix} \ dx.

    \begin{align*} I &= \text{Im}\Big( \int \frac{1}{2^{100}} ( {u-1})^{99} dx\Big) \\ &= \text{Im}\Big(  \frac{1}{2^{100}} \frac{( {u-1})^{100}}{100} \Big) \\ &= \text{Im}\Big(  \frac{1}{2^{100}} \frac{e^{100ix}}{(2i)^{100}}\frac{2^{100}}{e^{100i  x}}\frac{( {e^{2ix}-1})^{100}}{100} \Big) \\ &= \text{Im}\Big(  \frac{e^{100ix}}{100} \Big(\frac{ {e^{2ix}-1}}{2i e^{ix}}\Big)^{100} \Big) \\ &= \text{Im}\Big(  \frac{e^{100ix}}{100} \sin^{100}(x) \Big) \\ &= \text{Im }\Big(  \frac{ \cos(100x) \sin^{100}(x)}{100} +i\frac{\sin(100x)\sin^{100}(x)}{100} \Big) \\ &= \frac{\sin(100x)\sin^{100}(x)}{100}+C\end{align*}
    thank you! I never thought of using euler's formula to express sin(x).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: integral

    Once you have seen the Wolfram Alpha solution, you can reconstruct it like this. Start with the addition formula \sin(101x) = \sin(100x)\cos x + \cos(100x)\sin x. then

    \begin{aligned}\sin(101x)\sin^{99}x &= \sin(100x)\cos x\sin^{99}x + \cos(100x)\sin^{100} x \\ &= \sin(100x)\frac d{dx}\Bigl(\frac{\sin^{100}x}{100}\Bigr) + \frac d{dx}\Bigl(\frac{\sin(100x)}{100}\Bigr)\sin^{100}x \\ &= \frac d{dx}\Bigl(\frac{\sin(100x)\sin^{100}x}{100}\Bigr)  \quad\text{by the product rule.} \end{aligned}

    Therefore \int\sin(101x)\sin^{99}x \,dx = \frac{\sin(100x)\sin^{100}x}{100} + C.

    (But I think you would be unlikely to find that method if you had not already seen the answer.)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2011
    Posts
    83
    Thanks
    7

    Re: integral

    How does Wolfram Alpha (Mathematica) even solve integrals like this one?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: integral

    Quote Originally Posted by Ridley View Post
    How does Wolfram Alpha (Mathematica) even solve integrals like this one?
    Mathematica's Integrate function represents the fruits of a huge amount of mathematical and computational research. It doesn't do integrals the way people do. Instead, it uses powerful, general algorithms that often involve very sophisticated math. There are a couple of approaches that it most commonly takes. One involves working out the general form for an integral, then differentiating this form and solving equations to match up undetermined symbolic parameters. Even for quite simple integrands, the equations generated in this way can be highly complex and require Mathematica's strong algebraic computation capabilities to solve. Another approach that Mathematica uses in working out integrals is to convert them to generalized hypergeometric functions, then use collections of relations about these highly general mathematical functions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Volume integral to a surface integral ... divergence theorem
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: October 7th 2010, 06:11 PM
  2. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  3. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  4. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  5. Replies: 0
    Last Post: September 10th 2008, 07:53 PM

Search Tags


/mathhelpforum @mathhelpforum