if integral 1+tan(x-a)tan(x+a)dx=B ln(cos(x+a)/cos(x-a))+c then find B\tcot2a\ttan2a\t-cot2a\t-tan2a
chandrika , 5 Years ago
Grade 12th pass
Integral Calculus> if integral 1+tan(x-a)tan(x+a)dx=B ln(cos...
1 AnswersSamyak Jain
Last Activity: 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)]
= 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
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