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Evaluating $int_0^l (lx-x^2)sinleft(fracnpi xlright),mathrmdx$



The Next CEO of Stack OverflowEvaluating the integral, $int_0^infty lnleft(1 - e^-xright) ,mathrm dx $Evaluating the definite integral $int_0^infty fracmathrme^xleft(mathrme^x-1right)^2,x^n ,mathrmdx$Evaluating $int_0^largefracpi4 logleft( cos xright) , mathrmdx $Can we simplify $int_0^pileft(fracsin nxsin xright)^mdx$?Assistance evaluating $int_0^2pifrac12left(2 sin theta + cos thetaright)dtheta$Evaluating $int_0^infty int_0^infty fracx^2 + y^21 + (x^2 - y^2)^2 e^-2xy :mathrmdx :mathrmdy$Closed form of $int_0^1int_0^1int_0^1fracleft(1-x^yright)left(1-x^zright)ln x(1-x)^3,mathrm dx;mathrm dy;mathrm dz$Evaluate :$int_0^1 frac1left(1+sqrtfrac1x-1right)(x^2-x-1)mathrm dx$Evaluating the integral: $intlimits_0^inftyleft(fracsin(ax)xright)^2 dx , a neq 0 $How to integrate $int_0^pi/2frac1sin^2(x)-sin(x)cos(x)+cos^2(x),mathrmdx$










-2












$begingroup$


I am stuck on evaluating the following definite integral:



$$int_0^l (lx-x^2)sinleft(fracnpi xlright) , mathrmdx$$



Aprreciate any help!










share|cite|improve this question











$endgroup$











  • $begingroup$
    Use Integration by parts en.wikipedia.org/wiki/Integration_by_parts
    $endgroup$
    – Mostafa Ayaz
    Mar 18 at 13:12















-2












$begingroup$


I am stuck on evaluating the following definite integral:



$$int_0^l (lx-x^2)sinleft(fracnpi xlright) , mathrmdx$$



Aprreciate any help!










share|cite|improve this question











$endgroup$











  • $begingroup$
    Use Integration by parts en.wikipedia.org/wiki/Integration_by_parts
    $endgroup$
    – Mostafa Ayaz
    Mar 18 at 13:12













-2












-2








-2





$begingroup$


I am stuck on evaluating the following definite integral:



$$int_0^l (lx-x^2)sinleft(fracnpi xlright) , mathrmdx$$



Aprreciate any help!










share|cite|improve this question











$endgroup$




I am stuck on evaluating the following definite integral:



$$int_0^l (lx-x^2)sinleft(fracnpi xlright) , mathrmdx$$



Aprreciate any help!







integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 18 at 13:31









StubbornAtom

6,30831440




6,30831440










asked Mar 18 at 13:10









mallorie000383mallorie000383

1




1











  • $begingroup$
    Use Integration by parts en.wikipedia.org/wiki/Integration_by_parts
    $endgroup$
    – Mostafa Ayaz
    Mar 18 at 13:12
















  • $begingroup$
    Use Integration by parts en.wikipedia.org/wiki/Integration_by_parts
    $endgroup$
    – Mostafa Ayaz
    Mar 18 at 13:12















$begingroup$
Use Integration by parts en.wikipedia.org/wiki/Integration_by_parts
$endgroup$
– Mostafa Ayaz
Mar 18 at 13:12




$begingroup$
Use Integration by parts en.wikipedia.org/wiki/Integration_by_parts
$endgroup$
– Mostafa Ayaz
Mar 18 at 13:12










1 Answer
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0












$begingroup$

Hint:



$$int P(t)sin t,dt=-P(t)cos t+int P'(t)cos t,dt=-P(t)cos t+ P'(t)sin t,dt-int P''(t)cos t,dt.$$



Also notice that $P$ and the sine vanish at the bounds, so that only the last term remains.






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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

    oldest

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    0












    $begingroup$

    Hint:



    $$int P(t)sin t,dt=-P(t)cos t+int P'(t)cos t,dt=-P(t)cos t+ P'(t)sin t,dt-int P''(t)cos t,dt.$$



    Also notice that $P$ and the sine vanish at the bounds, so that only the last term remains.






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      Hint:



      $$int P(t)sin t,dt=-P(t)cos t+int P'(t)cos t,dt=-P(t)cos t+ P'(t)sin t,dt-int P''(t)cos t,dt.$$



      Also notice that $P$ and the sine vanish at the bounds, so that only the last term remains.






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        Hint:



        $$int P(t)sin t,dt=-P(t)cos t+int P'(t)cos t,dt=-P(t)cos t+ P'(t)sin t,dt-int P''(t)cos t,dt.$$



        Also notice that $P$ and the sine vanish at the bounds, so that only the last term remains.






        share|cite|improve this answer









        $endgroup$



        Hint:



        $$int P(t)sin t,dt=-P(t)cos t+int P'(t)cos t,dt=-P(t)cos t+ P'(t)sin t,dt-int P''(t)cos t,dt.$$



        Also notice that $P$ and the sine vanish at the bounds, so that only the last term remains.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 18 at 13:30









        Yves DaoustYves Daoust

        131k676229




        131k676229



























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