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if integral 1+tan(x-a)tan(x+a)dx=B ln(cos(x+a)/cos(x-a))+c then find B cot2a tan2a -cot2a -tan2a

if integral 1+tan(x-a)tan(x+a)dx=B ln(cos(x+a)/cos(x-a))+c then find B
  1. cot2a
  2. tan2a
  3. -cot2a
  4. -tan2a

Grade:12th pass

1 Answers

Samyak Jain
333 Points
5 years ago
1 + tan(x – a) tan(x + a)   =   1 + {sin(x – a)/cos(x – a)} {sin(x + a)/cos(x + a)}  
                                                       ...Using tanA = sinA / cosA
                   = [cos(x + a)cos(x – a) + sin(x + a)sin(x – a)] / [cos(x + a)cos(x – a)]           
                   = cos{(x + a) – (x – a)} / cos(x + a)cos(x – a)  =  cos(2a) / cos(x + a)cos(x – a)
    ...Using cosAcosB + sinAsinB = cos(A – B)
                   = sin(2a)/sin(2a) x cos(2a) / cos(x + a)cos(x – a)
                   = cot(2a) . sin2a / cos(x + a)cos(x – a)  =  cot(2a) sin{(x+a) – (x–a)} / cos(x + a)cos(x – a)
                   = cot(2a) . [sin(x+a)cos(x–a) – cos(x+a)sin(x–a)] / cos(x + a)cos(x – a)
   … Using sin(A – B) = sinAcosB – cosAsinB
                   = cot(2a) [tan(x+a) – tan(x–a)]
\therefore ∫ [1 + tan(x – a) tan(x + a)]dx  =  ∫cot(2a) [tan(x+a) – tan(x–a)]dx
              =  cot(2a) [∫ tan(x+a) dx – ∫ tan(x–a) dx]    =   cot(2a)[ln(sec(x+a) – ln(sec(x–a))] + c
              =  cot(2a) ln(sec(x + a) / sec(x – a)) + c    =  cot(2a) ln(cos(x – a) / cos(x + a)) + c
              =  cot(2a) [– ln(cos(x + a) / cos(x – a))] + c
              =  – cot(2a) ln(cos(x + a) / cos(x – a)) + c
\therefore B = – cot(2a)

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