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Math Help - Interesting Integrals. #1

  1. #1
    Member Miss's Avatar
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    Interesting Integrals. #1

    Evaluate:

    \int_0^{\pi} \frac{x \, sin(x) }{1+cos^2(x)} \, dx.
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  2. #2
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    Put x\mapsto \pi - x to get
    \int_0^{\pi} \frac{x \sin x }{1+\cos^2 x} dx = \int_0^{\pi} \frac{(\pi-x) \sin x }{1+\cos^2 x} dx = -\int_0^{\pi} \frac{x \sin x }{1+\cos^2 x} dx + \pi \int_0^{\pi} \frac{\sin x }{1+\cos^2 x} dx
    And thus
    \int_0^{\pi} \frac{x \sin x }{1+\cos^2 x} dx = \frac{\pi}{2} \int_0^{\pi} \frac{\sin x }{1+\cos^2 x} dx
    The latter integral can easily be evaluated as \frac{\pi}{2} by letting x \mapsto \cos x. Thus your integral is:
    \int_0^{\pi} \frac{x \sin x }{1+\cos^2 x} dx = \frac{\pi^2}{4}
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  3. #3
    MHF Contributor chisigma's Avatar
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    Integrating by parts You obtain first...

    \int_{0}^{\pi} \frac{x\cdot \sin x}{1+\cos^{2} x}\cdot dx = |-x\cdot \tan^{-1} \cos x|_{0}^{\pi} + \int_{0}^{\pi} \tan^{-1} \cos x\cdot dx (1)

    Now You perform in the second integral the substitution \xi= x - \frac{\pi}{2} so that it becomes...

    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^{-1} \sin \xi\cdot d\xi (2)

    Now the integrand function in (2) is an odd function, so that the integral (2) vanishes and is...

    \int_{0}^{\pi} \frac{x\cdot \sin x}{1+\cos^{2} x}\cdot dx = |-x\cdot \tan^{-1} \cos x|_{0}^{\pi} = \frac{\pi^{2}}{4} (3)

    Kind regards

    \chi \sigma
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  4. #4
    Math Engineering Student
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    write \int_{0}^{\pi }{\arctan (\cos x)\,dx}=\int_{0}^{\pi }{\int_{0}^{1}{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dt}\,dx}=\int_{0}^{1}{\int_{0}^{\pi }{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dx}\,dt}, and in the last integral substitute x\mapsto \pi-x, you'll see it vanishes as well.
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  5. #5
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    i donot know double integrals

    Quote Originally Posted by Krizalid View Post
    write \int_{0}^{\pi }{\arctan (\cos x)\,dx}=\int_{0}^{\pi }{\int_{0}^{1}{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dt}\,dx}=\int_{0}^{1}{\int_{0}^{\pi }{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dx}\,dt}, and in the last integral substitute x\mapsto \pi-x, you'll see it vanishes as well.
    your solutions go over my head!
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