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Thread: Integrals!

  1. #1
    Senior Member polymerase's Avatar
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    Integrals!

    $\displaystyle \int \frac{sin\;3x}{1+cos^2\;x}\;dx$

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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by polymerase View Post
    $\displaystyle \int \frac{sin\;3x}{1+cos^2\;x}\;dx$
    You can easily show that $\displaystyle \sin 3x = 3\sin x - 4\sin ^3 x.$

    Now try to find an arctangent:

    $\displaystyle \int {\frac{{\sin 3x}}
    {{1 + \cos ^2 x}}\,dx} = \int {\frac{{3\sin x - 4\sin ^3 x}}
    {{1 + \cos ^2 x}}\,dx} = \int {\frac{{\left( {8\sin x - 4\sin ^3 x} \right) - 5\sin x}}
    {{1 + \cos ^2 x}}\,dx} .$

    Note that $\displaystyle 8\sin x - 4\sin ^3 x = 4\sin x\left( {2 - \sin ^2 x} \right) = 4\sin x\left( {1 + \cos ^2 x} \right).$

    The rest follows.
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  3. #3
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    Hello, polymerase!

    I found an approach . . . but there must be a better way!


    $\displaystyle \int \frac{\sin3x}{1+\cos^2\!x}\,dx$
    Multiple-angle identity: .$\displaystyle \sin3\theta \:=\:3\sin\theta - 4\sin^3\!\theta$

    We have: .$\displaystyle \int\frac{3\sin x-4\sin^3\!x}{1 + (1-\sin^2\!x)}\,dx \;=\;\int\frac{3\sin x-4\sin^3\!x}{2-\sin^2\!x}\,dx \;=\;\int\frac{4\sin^3\!x-3\sin x}{\sin^2\!x-2}\,dx$

    Long division: .$\displaystyle \int\left(4\sin x + \frac{5\sin x}{\sin^2\!x-2}\right)dx \;=\;\int\left(4\sin x + \frac{5\sin x}{(1-\cos^2\!x) - 2}\right)dx$

    . . and we have: .$\displaystyle 4\int\sin x\,dx - 5\int\frac{\sin x}{1 + \cos^2\!x}\,dx \;=\;-4\cos x - 5\int\frac{\sin x}{1+\cos^2\!x}\,dx$


    Let: .$\displaystyle u \:= \:\cos x\quad\Rightarrow\quad du \:=\: -\sin x\,dx$

    Substitute: .$\displaystyle -4\cos x + 5\int\frac{du}{1+u^2} \;=\;-4\cos x + 5\arctan u + C$


    Back-substitute: .$\displaystyle \boxed{-4\cos x + 5\arctan(\cos x) + C}$

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