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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: dx/((sin x)^5+(cos x)^5) integrate this
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jyotirmoy_mazumdar1 (0)

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dx/((sin x)^5+(cos x)^5) integrate this
    
dip_xaverian (117)

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ramkumar_november (1270)

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let I =     dx
          ----------------------------
           (sinx)^5 + (cosx)^5

  I  =      dx
       ------------------------------------------------------------------------------
        (sinx +cosx)(1 - sinxcosx -  (sinxcosx)^2)
 
I = (sinx + cosx)dx
    -------------------------------------------------------------------------
     ( 1 - 2sinxcosx) ( 1 - sinxcosx - (sinxcosx)^2)


now proceed by substituting    sinx - cosx = t
squaring you will get
  1 - 2sinxcosx=t^2
sinxcosx= 1/2*(1 - t^2)
substitute in the integral and proceed via the method of partial fractions ..............................
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sandeepramesh (1247)

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I dont understand how u multiplied by sinx+cosx in the Nr and sth else in the Dr, i.e how you got (sinx+cosx)^2 = 1 - sin2x???
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ramkumar_november (1270)

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here is a more detailed proof.....
 
I  =  int
 
 
x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)
 
sin^5x+cos^5x=(sin^2x+cos^2x)(sin^3x+cos^3x)-sin^2xcos^2x(sinx+cosx)
 
sin^5x+cos^5x=sin^3x+cos^3x-sin^2xcos^2x(sinx+cosx)
 
sin^5x+cos^5x=(sinx+cosx)(sin^2x+cos^2x-sinxcosx)-sin^2xcos^2x(sinx+cosx)
 
sin^5x+cos^5x=(sinx+cosx)(1-sinxcosx)-sin^2xcos^2x(sinx+cosx)
 
sin^5x+cos^5x=(sinx+cosx)(1-sinxcosx-sin^2xcos^2x)
 
so
 
I  =  int
 
multiply numerator and denominator by  sinx+cosx
 
I  =  int
 
I  =  int
 
now proceed by substituting , sinx-cosx=t......
 
clear????
 
sorry for typing mistakes....
 
 
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sandeepramesh (1247)

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I got that but last time you had made a typo by writing 1-sin2x in the Dr instead of 1+sin2x
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