Question

# ∫cosx+2sinx7sinx−5cosxdx=ax+bln∣7sinx−5cosx∣+cthena+bis.

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Solution

## We have, ∫cosx+2sinx7sinx−5cosxdx=ax+bln|7sinx−5cosx|+C Solve it:- ∫cosx+2sinx7sinx−5cosxdx We know that, Numerator = A(Denominator)+B(DerivativeofDenominator)……. (1) Where A and B are two constants. Now, Differentiation denominator with respect to x and we get Let, 7sinx−5cosx=t (7cosx+5sinx)=dtdx (7cosx+5sinx)dx=dt Then, By equation (1) and we get, cosx+2sinx=A(7sinx−5cosx)+B(7cosx+5sinx) =7Asinx−5Acosx+7Bcosx+5Bsinx =7Asinx+5Bsinx−5Acosx+7Bcosx cosx+2sinx=(7A+5B)sinx+(−5A+7B)cosx Comparing both side and we get, 7A+5B=1.......(2) −5A+7B=2.......(3) Solving equation (2) and (3) to, and we get, A=−374,B=1974 Put the value of A and B in equation (1) and we get Numerator = -374(Denominator)+1974(DerivativeofDenominator) cosx+2sinx=−374(7sinx−5cosx)+1974(7cosx+5sinx) Then, According to given question, ∫cosx+2sinx7sinx−5cosxdx=∫−374(7sinx−5cosx)+1974(7cosx+5sinx)7sinx−5cosxdx =∫−374(7sinx−5cosx)7sinx−5cosxdx+1974(7cosx+5sinx)7sinx−5cosxdx =−374∫(7sinx−5cosx)7sinx−5cosxdx+1974∫(7cosx+5sinx)7sinx−5cosxdx =−374∫1dx+1974∫dtt Onintegratingandweget, =−374x+1974lnt =−374x+1974ln(7sinx−5cosx)+C ∫cosx+2sinx7sinx−5cosxdx=−374x+1974ln(7sinx−5cosx)+C.......(4) Given that, ∫cosx+2sinx7sinx−5cosxdx=ax+bln(7sinx−5cosx)+C.......(5) On comparing equation (4) and (5) to and we get, a=−374,b=1974Then, a+b=−374+1974a+b=1674a+b=837 Hence, this is the answer.

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