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Math Help - Integration

  1. #1
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    Integration

    Hello everyone,

    Im having difficulties with this integral.

    Integral x csc x cot x dx

    It cant be subsitution because dy/dx of csc x is -cscx and cotx.
    It got to be integration by parts but i have x csc x and cot x,
    anyone can give me a little bit so I can start it off. Thanks.
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by ff4930 View Post
    Integral x csc x cot x dx
    \int x\csc x\cot x\,dx=\int\frac{x\cos x}{\sin^2x}\,dx

    Define \alpha=\sin x

    Next step is integration by parts.
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  3. #3
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    ugh, Im still stumped. I know how you got to where you are tho.

    Can I move the sin^2x up and become (sin^2x)^-1?
    let u be xcosx and dv = (sin^2x)^-1
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  4. #4
    Math Engineering Student
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    \int\frac{x\cos x}{\sin^2x}\,dx

    \alpha=\sin x\implies d\alpha=\cos x\,dx, the integral becomes to

    \int\frac{x\cos x}{\sin^2x}\,dx=\int\frac{\arcsin\alpha}{\alpha^2}  \,d\alpha

    Now let's apply integration by parts, so set u=\arcsin\alpha\implies du=\frac1{\sqrt{1-\alpha^2}}\,d\alpha and dv=\alpha^{-2}\,d\alpha\implies v=-\frac1\alpha, the new integral becomes to

    \int\frac{\arcsin\alpha}{\alpha^2}\,d\alpha=-\frac{\arcsin\alpha}\alpha+\int\frac1{\alpha\sqrt{  1-\alpha^2}}\,d\alpha

    The remaining integral is easy to solve, so its answer is

    \ln|\alpha|-\ln\left(\sqrt{1-\alpha^2}+1\right)

    Finally

    \int\frac{\arcsin\alpha}{\alpha^2}\,d\alpha=-\frac{\arcsin\alpha}\alpha+\ln|\alpha|-\ln\left(\sqrt{1-\alpha^2}+1\right)

    Back substitute and you are done.
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