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Math Help - 2 integrals...

  1. #1
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    2 integrals...

    problem 1:   \int \frac{\sin 2x}{\sin ^{4}x+\cos^{4}x } dx

    problem 2 :  \int\limits_{0}^{\pi} \frac{\cos (3x)}{\cos x} dx

    please write the full solutions because i'm a integration newbie
    and a very happy christmas and new year,from India
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  2. #2
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    Hello, earthboy!

    The first one requires Olympic-level gymnastics.
    I hope you're up for it . . .


    \displaystyle (1)\;\int \frac{\sin 2x}{\sin^4\!x+\cos^4\!x}\,dx

    To the denominator, add and subtract 2\sin^2\!x\cos^2\!x

    \sin^4\!x {\bf + 2\sin^2\!x\cos^2\!x} + \cos^4\!x {\bf \:-\: 2\sin^2\!x\cos^2\!x}

    . . =\;(\sin^4\!x + 2\sin^2\!x\cos^2\!x + \cos^4\!x) - \frac{1}{2}\cdot4\sin^2\!x\cos^2\!x}

    . . =\;\underbrace{(\sin^2\!x + \cos^2\!x)^2}_{\text{This is }1^2}  - \frac{1}{2}\underbrace{(2\sin x\cos x)^2}_{\text{This is }\sin^22x} \;\;=\;\;1 - \frac{1}{2}\sin^22x

    . . =\;\; \frac{1}{2}\left[2 - \sin^22x\right]  \;\;=\;\;\frac{1}{2}\left[1 + 1 - \sin^22x\right] \;\;=\;\; \frac{1}{2}\left[1 + \cos^2\!2x\right]


    The integral becomes: . \displaystyle \int\frac{\sin 2x\,dx}{\frac{1}{2}(1 + \cos^2\!2x)} \;\;=\;\;2\int\frac{\sin2x\,dx}{1 + \cos^2\!2x}


    Let: u \:=\:\cos2x \quad\Rightarrow\quad du \:=\:\text{-}2\sin2x\,dx \quad\Rightarrow\quad \sin2x\,dx \:=\:\text{-}\frac{1}{2}du

    Substitute: . \displaystyle 2\int \frac{-\frac{1}{2}du}{1 + u^2} \;=\;-\int\frac{du}{1+u^2} \;\;=\;\;-\arctan u + C


    Back-substitute: . -\arctan(\cos 2x) + C

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  3. #3
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    Quote Originally Posted by earthboy View Post
    problem 1:   \int \frac{\sin 2x}{\sin ^{4}x+\cos^{4}x } dx

    problem 2 :  \int\limits_{0}^{\pi} \frac{\cos (3x)}{\cos x} dx

    please write the full solutions because i'm a integration newbie
    and a very happy christmas and new year,from India
    Here's an alternative for (1)

    \left[sin^2x+cos^2x\right]\left[sin^2x+cos^2x\right]=1=sin^4x+cos^4x+2(sinxcosx)^2

    \displaystyle\Rightarrow\ sin^4x+cos^4x=1-2(sinxcosx)^2=1-\frac{1}{2}(2sinxcosx)^2=1-\frac{1}{2}sin^22x

    \displaystyle\int{\frac{sin2x}{sin^4x+cos^4x}}dx=\  int{\frac{sin2x}{1-\frac{1}{2}sin^22x}}dx=\int{\frac{sin2x}{1-\frac{1}{2}\left[1-cos^22x\right]}}dx

    u=cos2x\Rightarrow\ du=-2sin2xdx

    \displaystyle\ -\frac{1}{2}\int{\frac{du}{1-\frac{1}{2}\left[1-u^2\right]}=-\int{\frac{du}{1+u^2}}


    For (2)

    cos3x=cos(2x+x)

    Use the identity cos(A+B)=cosAcosB-sinAsinB

    cos3x=cos2xcosx-sin2xsinx=\left[2cos^2x-1\right]cosx-2sin^2xcosx

    =2cos^3x-cosx-2\left[1-cos^2x\right]cosx

    =2cos^3x+2cos^3x-cosx-2cosx=4cos^3x-3cosx

    The integral is

    \displaystyle\int_{0}^{\pi}\left[{4cos^2x-3}\right]dx
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    Quote Originally Posted by earthboy View Post
    problem 1:  I =  \int \frac{\sin 2x}{\sin ^{4}x+\cos^{4}x } dx
    By the fact that a^4+b^4 = (a^2+b^2)^2-2(ab)^2 we have:

    \sin^4{x}+\cos^4{x} = \left(\sin^2{x}+\cos^2{x}\right)^2-2\left(\sin{x}\cos{x}\right)^2 = 1-2\left(\frac{1}{2}\sin{2x}\right)^2 = 1-\frac{1}{2}\sin^2{2x}.


    \begin{aligned}  \\&   \int\frac{\sin{2x}}{1-\frac{1}{2}\sin^2{2x}}\;{dx}  = \int\frac{2\sin{2x}}{2-\sin^2{2x}}\;{dx}  =  \int\frac{2\sin{2x}}{\left(\sqrt{2}-\sin{2x}}\right)\left(\sqrt{2}+\sin{2x}\right)}\;{  dx}   \\& =  \int\frac{\left(\sqrt{2}+\sin{2x}\right)-\left(\sqrt{2}-\sin{2x}\right)}{\left(\sqrt{2}-\sin{2x}}\right)\left(\sqrt{2}+\sin{2x}\right)}\;{  dx} = \int\frac{1}{\sqrt{2}-\sin{2x}}}\:{dx}-\int\frac{1}{\sqrt{2}+\sin{2x}}}\;{dx}\end{aligned  }

    Put x = \arctan{t}, then:

     \displaystyle   \int\frac{1}{\sqrt{2}-\sin{2x}}}\;{dx}  =   \int\frac{1}{\sqrt{2}t^2-2t+\sqrt{2}}\;{dt}  \displaystyle = \int\frac{\sqrt{2}}{\left(\sqrt{2}t-1\right)^2+1}\;{dt} = \frac{\sqrt{2}}{\sqrt{2}}\arctan\left(\sqrt{2}t-1\right)+k_{1}

     \displaystyle   \int\frac{1}{\sqrt{2}+\sin{2x}}}\;{dx}  =   \int\frac{1}{\sqrt{2}t^2+2t+\sqrt{2}}\;{dt}  \displaystyle  = \int\frac{\sqrt{2}}{\left(\sqrt{2}t+1\right)^2+1}\  ;{dt} = \frac{\sqrt{2}}{\sqrt{2}}\arctan\left(\sqrt{2}t+1\  right)+k_{2}

    Thus I = \arctan\left(\sqrt{2}t-1\right)-\arctan\left(\sqrt{2}t+1\right) = \arctan\left(\sqrt{2}\tan{x}-1\right)-\arctan\left(\sqrt{2}\tan{x}+1\right)+k.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by earthboy View Post
    problem 1:   \int \frac{\sin 2x}{\sin ^{4}x+\cos^{4}x } dx

    problem 2 :  \int\limits_{0}^{\pi} \frac{\cos (3x)}{\cos x} dx
    Note that:


    \displaystyle \begin{aligned}\int\frac{\sin(2x)}{\sin^4(x)+\cos^  4(x)}\text{ }dx &= \int\frac{2\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}\te  xt{ }dx\\ &= \int\frac{\frac{2\sin(x)\cos(x)}{\cos(x)^4}}{\tan^  4(x)+1}\text{ }dx\\ &=\int\frac{2\tan(x)\sec^2(x)}{\tan^4(x)+1}\tex  t{ }dx\end{aligned}

    Let \tan^2(x)=u\implies du=2\tan(x)\sec^2(x) dx so that


    \displaystyle \begin{aligned}\int\frac{\sin(2x)}{\sin^4(x)+\cos^  4(x)}\text{ }dx &= \int\frac{2\tan(x)\sec^2(x)}{\tan^4(x)+1}\text{ }dx\\ &= \int\frac{du}{u^2+1}\\ &= \arctan(u)+C\\ &= \arctan\left(\tan^2(x)\right)+C\end{aligned}
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    Quote Originally Posted by earthboy View Post
    problem 1:   \int \frac{\sin 2x}{\sin ^{4}x+\cos^{4}x } dx

    problem 2 :  \int\limits_{0}^{\pi} \frac{\cos (3x)}{\cos x} dx

    wow! so many responses! what about the second problem??? please tell if its right or wrong

     \int\limits_{0}^{\pi} \frac{\cos (3x)}{\cos x} dx <br />
= \int\limits_{0}^{2\pi}\cos 2x dx - \int\limits_{0}^{\pi} \frac{\sin 2x \sin x}{\cos x} dx <br />
= -2 \int\limits_{0}^{\pi} \sin^{2} x dx <br />
= -2 \int\limits_{0}^{\pi} \frac {1 - \cos 2x}{2} dx <br />
= - \pi
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  7. #7
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    Quote Originally Posted by earthboy View Post
    wow! so many responses! what about the second problem??? please tell if its right or wrong

     \int\limits_{0}^{\pi} \frac{\cos (3x)}{\cos x} dx <br />
= \int\limits_{0}^{2\pi}\cos 2x dx - \int\limits_{0}^{\pi} \frac{\sin 2x \sin x}{\cos x} dx <br />
= -2 \int\limits_{0}^{\pi} \sin^{2} x dx <br />
= -2 \int\limits_{0}^{\pi} \frac {1 - \cos 2x}{2} dx <br />
= - \pi
    Using the cos(A+B) identity..

    \displaystyle\int_{0}^{\pi}\frac{cos3x}{cosx}dx=\i  nt_{0}^{\pi}\frac{cos2xcosx-sin2xsinx}{cosx}dx

    =\displaystyle\int_{0}^{\pi}cos2xdx-\int_{0}^{\pi}\frac{2sinxcosxsinx}{cosx}dx

    =\displaystyle\int_{0}^{\pi}cos2xdx-\int_{0}^{\pi}2sin^2xdx=\int_{0}^{\pi}cos2xdx-\int_{0}^{\pi}(1-cos2x)dx

    =\displaystyle\int_{0}^{\pi}(2cos2x-1)dx

    Using

    \displaystyle\frac{d}{dx}sin2x=2cos2x

    \displaystyle\int_{0}^{\pi}(2cos2x-1)dx=sin(2{\pi})-{\pi}=-{\pi}

    gives an incorrect answer for the area between the curve and the x-axis,
    since the graph of the function crosses the x-axis twice between 0 and {\pi}.

    Hence, the x-axis crossing points are at

    \displaystyle\ 2cos2x-1=0\Rightarrow\ cos2x=\frac{1}{2}

    \displaystyle\Rightarrow\ 2x=\frac{\pi}{3},\;\;x=\frac{\pi}{6}

    and at

    \displaystyle\ 2x=\frac{5{\pi}}{3},\;\;x=\frac{5{\pi}}{6}

    Inverting the negative sign for the integral in the middle part, the area is

    \displaystyle\int_{0}^{\frac{\pi}{6}}(2cos2x-1)dx-\int_{\frac{\pi}{6}}^{\frac{5{\pi}}{6}}(2cos2x-1)dx+\int_{\frac{5{\pi}}{6}}^{\pi}(2cos2x-1)dx

    which gives

    \displaystyle\ 2\frac{3\sqrt{3}-{\pi}}{6}+\frac{2{\pi}+3\sqrt{3}}{3}=2\sqrt{3}+\fr  ac{\pi}{3}
    Attached Thumbnails Attached Thumbnails 2 integrals...-2cos2x-1.jpg  
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Archie Meade View Post

    gives an incorrect answer for the area between the curve and the x-axis,
    since the graph of the function crosses the x-axis twice between 0 and {\pi}.
    I concur with the OP's answer. I get -\pi also.
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  9. #9
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    @ Archie Meade, but we are not calculating the area, right? Just the definite integral!
    Quote Originally Posted by earthboy View Post
    please tell if its right or wrong
     \int\limits_{0}^{\pi} \frac{\cos (3x)}{\cos x} dx <br />
= \int\limits_{0}^{2\pi}\cos 2x dx - \int\limits_{0}^{\pi} \frac{\sin 2x \sin x}{\cos x} dx <br />
= -2 \int\limits_{0}^{\pi} \sin^{2} x dx <br />
= -2 \int\limits_{0}^{\pi} \frac {1 - \cos 2x}{2} dx <br />
= - \pi
    Looks good to me!
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  10. #10
    MHF Contributor Drexel28's Avatar
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    I'd like to remark that one can do \displaystyle I_n=\int_0^{\pi}\frac{\cos(nx)}{\cos(x)}\text{ }dx in general and what is interesting (not really 'unbelievable' though if you see why) is that I_n=0,\text{ }n\text{ even}.
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  11. #11
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    Yes, definite integral answer is fine for average value over the interval.

    Alternative for area (application of integration).
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    Quote Originally Posted by Drexel28 View Post
    I'd like to remark that one can do \displaystyle I_n=\int_0^{\pi}\frac{\cos(nx)}{\cos(x)}\text{ }dx in general and what is interesting (not really 'unbelievable' though if you see why) is that I_n=0,\text{ }n\text{ even}.
    Let me try that.

    Spoiler:
    Putting x\mapsto \pi-x, we have:

    \displaystyle I_{n} = \int_{0}^{\pi}\frac{\cos{nx}}{\cos{x}}\;{dx} = -\int_{\pi}^{0}\frac{\cos\left(n\pi-nx\right)}{\cos\left(\pi-x\right)}\;{dx} = \int_{0}^{\pi}\frac{\cos\left(n\pi-nx\right)}{\cos\left(\pi-x\right)}\;{dx}.

    Noting that \cos\left(n\pi-nx\right) = (-1)^n\cos{nx} and that \cos\left(\pi-x\right) = -\cos{x}, we have:

    \displaystyle 2I_{n} = \int_{0}^{\pi}\frac{\cos{nx}}{\cos{x}}\;{dx}+\int_  {0}^{\pi}\frac{\cos\left(n\pi-nx\right)}{\cos\left(\pi-x\right)}\;{dx} = \int_{0}^{\pi}\frac{\cos{nx}-(-1)^n\cos{nx}}{\cos{x}}\;{dx}.

    If n is even, then (-1)^n\cos{nx} = \cos{nx}, and so:

    \displaystyle I_{n} =  \frac{1}{2} \int_{0}^{\pi}\frac{\cos{nx}-\cos{nx}}{\cos{x}}\;{dx} = 0.
    Last edited by TheCoffeeMachine; December 26th 2010 at 04:22 PM.
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  13. #13
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    Let me try that.

    Spoiler:
    Putting x\mapsto \pi-x, we have:

    \displaystyle I_{n} = \int_{0}^{\pi}\frac{\cos{nx}}{\cos{x}}\;{dx} = -\int_{\pi}^{0}\frac{\cos\left(n\pi-nx\right)}{\cos\left(\pi-x\right)}\;{dx} = \int_{0}^{\pi}\frac{\cos\left(n\pi-nx\right)}{\cos\left(\pi-x\right)}\;{dx}.

    Noting that \cos\left(n\pi-nx\right) = (-1)^n\cos{nx} and that \cos\left(\pi-x\right) = -\cos{x}, we have:

    \displaystyle 2I_{n} = \int_{0}^{\pi}\frac{\cos{nx}}{\cos{x}}\;{dx}+\int_  {0}^{\pi}\frac{\cos\left(n\pi-nx\right)}{\cos\left(\pi-x\right)}\;{dx} = \int_{0}^{\pi}\frac{\cos{nx}-(-1)^n\cos{nx}}{\cos{x}}\;{dx}.

    If n is even, then (-1)^n\cos{nx} = \cos{nx}, and so:

    \displaystyle I_{n} =  \frac{1}{2} \int_{0}^{\pi}\frac{\cos{nx}-\cos{nx}}{\cos{x}}\;{dx} = 0.
    Good, good. But you can shorten it to

    Spoiler:


    \displaystyle \begin{aligned}I_{2k} &= \int_0^{\pi}\frac{\cos(2k x)}{\cos(x)}\text{ }dx\\ &=\int_{0}^{\pi}\frac{\cos(2k(\pi-x))}{\cos(\pi-x)}\text{ }dx\\ &= \int_0^{\pi}\frac{\cos(2 k x)}{-\cos(x)}\text{ }dx\\ &= -I_{2k}\end{aligned}
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