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Thread: Interesting Integrals. #1

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

    Evaluate:

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

    $\displaystyle \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 $\displaystyle \xi= x - \frac{\pi}{2}$ so that it becomes...

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
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    write $\displaystyle \int_{0}^{\pi }{\arctan (\cos x)\,dx}=\int_{0}^{\pi }{\int_{0}^{1}{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dt}\,dx}=\int_{0}^{1}{\int_{0}^{\pi }{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dx}\,dt},$ and in the last integral substitute $\displaystyle x\mapsto \pi-x,$ 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 $\displaystyle \int_{0}^{\pi }{\arctan (\cos x)\,dx}=\int_{0}^{\pi }{\int_{0}^{1}{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dt}\,dx}=\int_{0}^{1}{\int_{0}^{\pi }{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dx}\,dt},$ and in the last integral substitute $\displaystyle x\mapsto \pi-x,$ you'll see it vanishes as well.
    your solutions go over my head!
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