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प्रश्न
Evaluate:
`int cos x/(3 cos x - 5) dx`
मूल्यांकन
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उत्तर
Given, `I = int cos x/(3 cos x - 5) dx`
= `1/3 int (3 cos x - 5 + 5)/(3 cos x - 5) dx`
= `1/3 int ((3 cos x - 5))/(3 cos x - 5) dx + 1/3 int 5/(3 cos x - 5) dx`
= `x/3 + 5/3 int ((1 + tan^2 x/2))/(3(1 - tan^2 x/2) - 5(1 + tan^2 x/2)) dx` ...`[∵ cos x = (1 - tan^2 x/2)/(1 + tan^2 x/2)]`
= `x/3 + 5/3 int (sec^2 x/2)/(3 - 3 tan^2 x/2 - 5 - 5 tan^2 x/2) dx`
= `x/3 + 5/3 int (sec^2 x/2)/(-8 tan^2 x/2 - 2)`
= `x/3 + (-5)/6 int (sec^2 x/2)/(4 tan^2 x/2 + 1) dx`
Let `2 tan x/2 = t sec^2 x/2 dx = dt`
= `x/3 - 5/6 int dt/(t^2 + 1^2)`
= `x/3 - 5/6 tan^-1 t + C`
Now, `I = x/3 - 5/6 tan^-1 (2 tan x/2) + C`.
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