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Sinx/sinx+cosx for limit 0 to pi/2what is the solution of this integration of applying the limit on both the sided for the upper limit to the lower limit

Sinx/sinx+cosx for limit 0 to pi/2what is the solution of this integration of applying the limit on both the sided for the upper limit to the lower limit

Grade:12th pass

1 Answers

Chandan
121 Points
6 years ago
put
numerator=L(denominator)+m(derivative of denominator)
 ie.
sinx=L(sinx+cosx)+m(cosx-sinx).................. eq 1
sinx=Lsinx+Lcosx+mcosx-msinx.
sinx=(L-m)sinx+(L+m)cosx.
comparing coeffecients of sinx and cosx on LHS AND RHS.
l-m=1 and m+m=0.
solving we get L=1/2,m=-1/2.
put sinx if form of eq 1
question is:
∫(L(sinx+cosx)+m(cosx-sinx))/(cosx+sinx)
=∫Ldx+∫m(cosx-sinx)/(sinx+cosx)dx
=(1/2)x-(1/2)log(sinx+cosx)
Putting limits
ans is:pi/4 -(1/2)*0=pi/4                                  (take sinx+cosx as t so (cosx-sinx)=dt so ∫1/tdt=logt=log(sinx+cosx)

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