Q.

if the integral ∫5tan⁡xtan⁡x−2dx=x+alog |sin⁡x−2cos⁡x|+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 ∫5tan⁡xtan⁡x−2dxTo find The value of a ', if ∫5tan⁡xtan⁡x−2dx =x+alog⁡|sin⁡x−2cos⁡x|+k ---iNow, let us assume that I=∫5tan⁡xtan⁡x−2dxOn multiplying by cos x in numerator anddenominator, we get I=∫5sin⁡xsin⁡x−2cos⁡xdxThis special integration requires special substitutionof typeNr=ADr+BdDrdx,A and B are constant. Let 5sin⁡x=A(sin⁡x−2cos⁡x)+B(cos⁡x+2sin⁡x)⇒0cos⁡x+5sin⁡x=(A+2B)sin⁡x+(B−2A)cos⁡xA+2B=5 and B−2A=0On solving the above two equations in A and B, we getA=1 and  B=2 ⇒5sin⁡x=(sin⁡x−2cos⁡x)+2(cos⁡x+2sin⁡x)⇒  I=∫5sin⁡xsin⁡x−2cos⁡xdx=∫(sin⁡x−2cos⁡x)+2(cos⁡x+2sin⁡x)(sin⁡x−2cos⁡x)dx=∫sin⁡x−2cos⁡xsin⁡x−2cos⁡xdx+2∫(cos⁡x+2sin⁡x)(sin⁡x−2cos⁡x)dx=∫1dx+2∫d(sin⁡x−2cos⁡x)(sin⁡x−2cos⁡x)=x+2log⁡|(sin⁡x−2cos⁡x)|+k   ----iiwhere, k is the constant of integration.Now, by comparing the value of I in Eqs. (i) and (ii),we get .
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