On comparison of coefficients of sinx,cosx we get sinx coefficients →5=α+2β...(i) cosx coefficients →0=−2α+β On solving Eqs. (i) and (ii), weget α=1 and β=2
∴5sinx=1(sinx−2cosx)+2(cosx+2sinx)
Now, substitute 5sinx value in the given question. ∫
5sinx
(sinx−2cosx)
dx
=∫
1(sinx−2cosx)+2(cosx+2sinx)
(sinx−2cosx)
dx =∫
1(sinx−2cosx)
sinx−2cosx
dx+2∫
cosx+2sinx
(sinx−2cosx)
dx =∫1⋅dx+2log|sinx−2cosx|+c
|∵∫
f′(x)
f(x)
dx=log|f(x)|| =x+2log|sinx−2cosx|+c On comparison with given RHS, we get α=1,β=2⇒α−β=−1