If ∫(1+tan(x−α)tan(x+α)dx=λlncos(x+α)cos(x−α)+c, then λ=

∫(1+tan (x-α) tan(x+α))dx =∫1+tan2x-tan2α1-tan2x tan2αdx =∫1-tan2x tan2α+tan2x-tan2α1-tan2x tan2αdx =∫(1+tan2x)(1-tan2α)1-tan2x tan2αdx =(1-tan2α)tan2α∫sec2xcot2α-tan2xdx put tanx=t =cos2αSin2α.12cot α logcot α+tcot α-t =cos2αSin2α logcot α+tanxcot α-tanx =cot2α logcos(x-α)cos(x+α) =-cot2α logcos(x+α)cos(x-α) ∴λ=-cot2α

If ∫(1+tan(x−α)tan(x+α)dx=λlncos(x+α)cos(x−α)+c, then λ=

∫(1+tan (x-α) tan(x+α))dx =∫1+tan2x-tan2α1-tan2x tan2αdx =∫1-tan2x tan2α+tan2x-tan2α1-tan2x tan2αdx =∫(1+tan2x)(1-tan2α)1-tan2x tan2αdx =(1-tan2α)tan2α∫sec2xcot2α-tan2xdx put tanx=t =cos2αSin2α.12cot α logcot α+tcot α-t =cos2αSin2α logcot α+tanxcot α-tanx =cot2α logcos(x-α)cos(x+α) =-cot2α logcos(x+α)cos(x-α) ∴λ=-cot2α

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If $\int ((1 + \tan (x - α) \tan (x + α)) d x = λ \ln |\frac{\cos (x + α)}{\cos (x - α)}| + c, t h e n λ =$

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$- c o t 2 α$

$- \tan 2 α$

$c o t 2 α$

$\tan 2 α$

answer is C.

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Detailed Solution

$\int (1 + \tan (x - α) \tan (x + α)) dx = \int (1 + \frac{\tan^{2} x - \tan^{2} α}{1 - \tan^{2} x \tan^{2} α}) dx = \int \frac{1 - \tan^{2} x \tan^{2} α + \tan^{2} x - \tan^{2} α}{1 - \tan^{2} x \tan^{2} α} dx = \int \frac{(1 + \tan^{2} x) (1 - \tan^{2} α)}{1 - \tan^{2} x \tan^{2} α} dx = \frac{(1 - \tan^{2} α)}{\tan^{2} α} \int \frac{\sec^{2} x}{\cot^{2} α - \tan^{2} x} dx p u t \tan x = t = \frac{\cos 2 α}{S i n^{2} α} . \frac{1}{2 \cot α} \log |\frac{\cot α + t}{\cot α - t}| = \frac{\cos 2 α}{S i n 2 α} \log |\frac{\cot α + tanx}{\cot α - tanx}| = \cot 2 α \log |\frac{\cos (x - α)}{\cos (x + α)}| = - \cot 2 α \log |\frac{\cos (x + α)}{\cos (x - α)}| ∴ λ = - \cot 2 α$