Saturday, 13 May 2017

calculus - Alternative methods for finding antiderivative of fracx(1+sinx)2



This problem was assigned to a student I'm tutoring as extra credit:





x(1+sinx)2dx




I have two potential methods in mind, though I suspect the first, which employs the Weierstrass tangent half-angle substitution, goes beyond the student's curriculum, while the second probably falls within what he's learned (or at least in my opinion, what he should absolutely know). Both methods are rather lengthy, and, while I'm convinced the "extra credit" part about this problem is correlated with that, I can't help but wonder if there are more straightforward methods to do it.



I outline my suggestions below.



Method 1:





  • Integrate by parts to get
    x(1+sinx)2dx=uvvduwhere u=x and v=dx(1+sinx)2.


  • Compute v using the Weierstrass substitution t=tanx2, which yields
    2+3tanx2+3tan2x23(1+tanx2)3x+2+3tanx2+3tan2x23(1+tanx2)3dx


  • Compute the remaining integral by expanding like so:
    dx(1+tanx2)2+1+tan2x2(1+tanx2)2dx43dx(1+tanx2)3and with appropriate substitutions (Weierstrass and w=1+tanx2) rewrite as
    2(1+t)2(1+t2)dt+2w2dw432(1+t)3(1+t2)dt


  • Decompose into partial fractions, etc.





Method 2:




  • Let x=π2y and rewrite the integral as
    x(1+sinx)2dx=yπ2(1+cosy)2dy


  • Recalling the half-angle identity for cosine, we can rewrite the integrand as
    (y4π8)sec4y2dy=(yπ4)sec4ydy


  • Integrate by parts
    (yπ4)sec4ydy=uvvduthis time with u=yπ4 and v=sec4ydy.



  • Compute v with a reduction formula or the expansion sec4y=sec2y(1+tan2y), etc.



Answer



If you let u=x,v=1(1+sinx)2dx=1(sinx2+cosx2)2dx=1(2cos(x2π4))2dx



=12sec2(x2π4)dx=tan(x2π4)+C, this gives



1(1+sinx)2dx=uvvdu=xtan(x2π4)[2ln|cos(x2π4)|]+C



=xtan(x2π4)+2ln|cos(x2π4)|+C



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