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Math Help - Integral. #6

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    Integral. #6

    Evaluate : \displaystyle \int_0^1 sin(\alpha x) sin(\beta x) \, dx

    where \displaystyle \alpha and \displaystyle \beta are roots for the equation \displaystyle 2x=tan(x).

    Using the forumla \displaystyle sin(A) sin(B)=\frac{1}{2} ( cos(A-B) - cos(A+B) ) is not allowed.

    Final Answer = 0
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    Well using that identity is the most obvious and easy way to solve the problem... Why can't you use it?
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    because the problem say that
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    Use \sin(x)=\frac{e^{ix}-e^{-ix}}{2i} then.
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    Behold, the power of SARDINES!
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    Quote Originally Posted by TWiX View Post
    Evaluate : \displaystyle \int_0^1 sin(\alpha x) sin(\beta x) \, dx

    where \displaystyle \alpha and \displaystyle \beta are roots for the equation \displaystyle 2x=tan(x).

    Using the forumla \displaystyle sin(A) sin(B)=\frac{1}{2} ( cos(A-B) - cos(A+B) ) is not allowed.

    Final Answer = 0

    Use integration by parts twice.

    u = \sin(\alpha x) \implies du=\alpha \cos(\alpha x)dx \qaud dv=\sin(\beta x) \implies v=-\frac{1}{\beta}\cos(\beta x)

    I= \int\sin(\alpha x) \sin(\beta x)dx

    I=-\frac{1}{\beta}\sin(\alpha x)\cos(\beta x)+\frac{\alpha}{\beta }\int \cos(\alpha x)\cos(\beta x)dx

    u = \cos(\alpha x) \implies du=-\alpha \sin(\alpha x)dx \qaud dv=\cos(\beta x) \implies v=\frac{1}{\beta}\sin(\beta x)

    I=-\frac{1}{\beta}\sin(\alpha x)\cos(\beta x)+\frac{\alpha}{\beta }\left( \frac{1}{\beta}\sin(\beta x)\cos(\alpha x) + \frac{\alpha}{\beta}\int \sin(\alpha x) \sin(\beta x)dx\right)

    This gives

    \left(1 -\frac{\alpha^2}{\beta^2} \right)I= -\frac{1}{\beta}\sin(\alpha x)\cos(\beta x) +\frac{\alpha}{\beta^2}\sin(\beta x)\cos(\alpha x)

    -\beta \left(1 -\frac{\alpha^2}{\beta^2} \right)I=\sin(\alpha x)\cos(\beta x)\left[1-\frac{\alpha}{\beta}\frac{\tan(\beta x)}{\tan(\alpha x)} \right]

    Now if you evaluate for 0 to 1 the last factor gives

    1-\frac{\alpha}{\beta}\frac{\tan(\beta x)}{\tan(\alpha x)}]

    Since alpha and beta satify the equation

    2x=\tan(x) we get

    1-\frac{\alpha}{\beta}\frac{2\beta}{2\alpha}=0
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