Sunday, 29 March 2015

calculus - Evaluation of intfracsqrtcos2xsinx,dx





Compute the indefinite integral
cos2xsinxdx




My Attempt:




cos2xsinxdx=cos2xsin2xcos2xsinxdx=2cos2x1(1cos2x)2cos2x1sinxdx



Let cosx=t, so that sinxdx=dt. This changes the integral to



(2t21)(t21)2t21dt=(2t22)+1(t21)2t21dt=2dt2t21+dt(t21)2t21



How can I solve the integral from here?


Answer



cos2xsinx dx=cos2xsin2xsinx dx[1]=t46t2+1t3+t dt[2]=12u26u+1u2+u du[3]=(y26y+1)2(y1)(y3)(y+1)(y2+2t7) dy[4]=[1y1+1y31y+116y2+2y7] dt=[1y1+1y31y+116(y+1)28] dt
The rest is yours.







Notes :



[1]Use Weierstrass substitution, tan(x2)=t.



[2]Use substitution u=t2.



[3]Use Euler substitution, yu=u26u+1y=u212u6.



[4]Use partial fractions decomposition.


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