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Exercise 11.13 · Q19

Q.∫sec⁡2xtan⁡2x−1 dx\displaystyle\int \dfrac{\sec^2x}{\tan^2x-1}\,dx

(1) 2log⁡∣1−tan⁡x1+tan⁡x∣+c2\log\left|\dfrac{1-\tan x}{1+\tan x}\right|+c
(2) log⁡∣1+tan⁡x1−tan⁡x∣+c\log\left|\dfrac{1+\tan x}{1-\tan x}\right|+c
(3) 12log⁡∣tan⁡x+1tan⁡x−1∣+c\dfrac12\log\left|\dfrac{\tan x+1}{\tan x-1}\right|+c
(4) 12log⁡∣tan⁡x−1tan⁡x+1∣+c\dfrac12\log\left|\dfrac{\tan x-1}{\tan x+1}\right|+c
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Substitute t=tan⁡xt=\tan x, then apply the Type I rational form.

Step 1. Put t=tan⁡xt=\tan x, dt=sec⁡2x dxdt=\sec^2x\,dx: the integral becomes ∫dtt2−1\displaystyle\int\frac{dt}{t^2-1}. …

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