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Math Help - Integration of 2/(3sin(2x)+4cos(2x))

  1. #1
    Member ssadi's Avatar
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    Integration of 2/(3sin(2x)+4cos(2x))

    Hello:
    I am stuck at this sum:
    \int\frac{2}{3 sin 2x+4 cos 2x}
    Help please
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  2. #2
    Air
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    Junior Member Air's Avatar
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    Quote Originally Posted by ssadi View Post
    Hello:
    I am stuck at this sum:
    \int\frac{2}{3 sin 2x+4 cos 2x}
    Help please
    \int\frac{2}{3 \sin 2x+4 \cos 2x}\ \mathrm{d}x

    Let u=2x \implies \mathrm{d}x= \frac{\mathrm{d}u}{2}

    \therefore \int \frac{1}{3 \sin u+4 \cos u}\ \mathrm{d}u

    You have to use T-Substitution and solve: t=\tan \left(\frac{u}{2}\right)

    Note: \cos u = \frac{1-t^2}{1+t^2} and \sin u = \frac{2t}{1+t^2}.
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  3. #3
    Member ssadi's Avatar
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    Quote Originally Posted by Air View Post
    \int\frac{2}{3 \sin 2x+4 \cos 2x}\ \mathrm{d}x

    Let u=2x \implies \mathrm{d}x= \frac{\mathrm{d}u}{2}

    \therefore \int \frac{1}{3 \sin u+4 \cos u}\ \mathrm{d}u

    You have to use T-Substitution and solve: t=\tan \left(\frac{u}{2}\right)

    Note: \cos u = \frac{1-t^2}{1+t^2} and \sin u = \frac{2t}{1+t^2}.
    The t substitution, it props up everywhere and yet no mention of its uses are in the book. Thanks a lot.
    t=tan x, can you give me any site address where it and the places where it can apply is discussed, I have yet to get used to this substitution.
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  4. #4
    Member ssadi's Avatar
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    Quote Originally Posted by Air View Post
    \int\frac{2}{3 \sin 2x+4 \cos 2x}\ \mathrm{d}x

    Let u=2x \implies \mathrm{d}x= \frac{\mathrm{d}u}{2}

    \therefore \int \frac{1}{3 \sin u+4 \cos u}\ \mathrm{d}u

    You have to use T-Substitution and solve: t=\tan \left(\frac{u}{2}\right)

    Note: \cos u = \frac{1-t^2}{1+t^2} and \sin u = \frac{2t}{1+t^2}.
    I noticed I gave you your first thanks
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Actually, those 2x work well, since

    \begin{aligned}<br />
   4\cos 2x+3\sin 2x&=2\left( 2{{\cos }^{2}}x-2{{\sin }^{2}}x+3\sin x\cos x \right) \\ <br />
 & =2\Big(\left( 2{{\cos }^{2}}x-\sin x\cos x \right)+\left( 4\sin x\cos x-2{{\sin }^{2}}x \right)\Big) \\ <br />
 & =2(2\cos x-\sin x)(2\sin x+\cos x).<br />
\end{aligned}

    Then the integral becomes,

    \int{\frac{dx}{(2\cos x-\sin x)(2\sin x+\cos x)}}=\int{\frac{{{(2\cos x-\sin x)}^{2}}+{{(2\sin x+\cos x)}^{2}}}{5(2\cos x-\sin x)(2\sin x+\cos x)}\,dx}, which equals \frac{1}{5}\left( \int{\frac{2\cos x-\sin x}{2\sin x+\cos x}\,dx}+\int{\frac{2\sin x+\cos x}{2\cos x-\sin x}\,dx} \right),

    and these are natural logs.
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