if the integral ∫5tanxtanx−2dx=x+alog |sinx−2cosx|+k, then a is equal to
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a
-1
b
-2
c
1
d
2
answer is D.
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Detailed Solution
Given integral ∫5tanxtanx−2dxTo find The value of a ', if ∫5tanxtanx−2dx =x+alog|sinx−2cosx|+k ---iNow, let us assume that I=∫5tanxtanx−2dxOn multiplying by cos x in numerator anddenominator, we get I=∫5sinxsinx−2cosxdxThis special integration requires special substitutionof typeNr=ADr+BdDrdx,A and B are constant. Let 5sinx=A(sinx−2cosx)+B(cosx+2sinx)⇒0cosx+5sinx=(A+2B)sinx+(B−2A)cosxA+2B=5 and B−2A=0On solving the above two equations in A and B, we getA=1 and B=2 ⇒5sinx=(sinx−2cosx)+2(cosx+2sinx)⇒ I=∫5sinxsinx−2cosxdx=∫(sinx−2cosx)+2(cosx+2sinx)(sinx−2cosx)dx=∫sinx−2cosxsinx−2cosxdx+2∫(cosx+2sinx)(sinx−2cosx)dx=∫1dx+2∫d(sinx−2cosx)(sinx−2cosx)=x+2log|(sinx−2cosx)|+k ----iiwhere, k is the constant of integration.Now, by comparing the value of I in Eqs. (i) and (ii),we get .