Rt R T T ie e ks s s Smz-. {an Ao यु दम न कला (डर

1.	Rt R T T ie e ks s s Smz-. {an Ao यु दम न कला (डर ;e
Answer» 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)=dtso∫1/tdt=logt=log(sinx+cosx)

Rt R T T ie e ks s s Smz-. {an Ao यु दम न कला (डर ;e

Answer»

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)=dtso∫1/tdt=logt=log(sinx+cosx)