if the integral ∫$$5tan$$⁡xtan⁡x−$$2dx=x+alog$$ |$$sin$$⁡x−$$2cos$$⁡x|$$+k$$, then a is equal to

Question

if the integral ∫$$5tan$$⁡xtan⁡x−$$2dx=x+alog$$ |$$sin$$⁡x−$$2cos$$⁡x|$$+k$$, then a is equal to

Infinity Learn · Accepted Answer

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 .

if the integral $\int \frac{5 \tan x}{\tan x - 2} dx = x + alog$ $| \sin x - 2 \cos x | + k, then a is equal to$

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if the integral ∫5tan⁡xtan⁡x−2dx=x+alog |sin⁡x−2cos⁡x|+k, then a is equal to

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if the integral $\int \frac{5 \tan x}{\tan x - 2} dx = x + alog$ $| \sin x - 2 \cos x | + k, then a is equal to$