∫_0 ^(π/4) ((sec^2 x)/(tan x−sec x)) dx =?


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Question Number 182648 by cortano1 last updated on 12/Dec/22

$\underset{0}{\int^{π / 4}} \frac{\sec^{2} x}{\tan x - \sec x} dx = ?$
	Answered by Acem last updated on 12/Dec/22
	$tan x − sec x= t ..... m sec x (sec x− tan x) dx= dt sec x dx= ((−dt)/t) ... i ∗ sec^2 x − tan^2 x= 1 ⇔ (sec x+ tan x) (sec x− tan x)=1 ⇒ sec x + tan x= ((−1)/t) ∗ & sec x − tan x= −t ∗∗^( From m) Sum stars: sec x= −(1/2)((1/t) + t) ...ii i, ii in integ. : I= ∫ ((sec x sec x dx)/(tan x− sec x))= (1/2)∫ ((((1/t) + t))/t^( 2) ) dt I= (1/2) ∫ ((1/t) + (1/t^3 )) dt, from m { ((m=0 : t= −1)),((m= (π/4) : t= 1−(√2) )) :} a= (1/2) ∫_(−1) ^( 1−(√2)) ((1/t) + (1/t^3 )) dt a= (1/2) [ln ∣t∣ − (1/(2 t^2 ))]∣_( −1) ^(1−(√2)) ≈ − 1. 648$
	$\tan x - \sec x = t . . . . . m$ $\sec x (\sec x - \tan x) d x = d t$ $\sec x d x = \frac{- d t}{t} . . . i$ $* \sec^{2} x - \tan^{2} x = 1 \Leftrightarrow (\sec x + \tan x) (\sec x - \tan x) = 1$ $\Rightarrow \sec x + \tan x = \frac{- 1}{t} $ $& \sec x - \tan x = - t *^{F r o m m}$ $S u m s t a r s : \sec x = - \frac{1}{2} (\frac{1}{t} + t) . . . i i$ $i, i i i n i n t e g . : I = \int \frac{\sec x \sec x d x}{\tan x - \sec x} = \frac{1}{2} \int \frac{(\frac{1}{t} + t)}{t^{2}} d t$ $I = \frac{1}{2} \int (\frac{1}{t} + \frac{1}{t^{3}}) d t, f r o m m {\begin{cases} m = 0 : t = - 1 \\ m = \frac{π}{4} : t = 1 - \sqrt{2} \end{cases}$ $a = \frac{1}{2} \int_{- 1}^{1 - \sqrt{2}} (\frac{1}{t} + \frac{1}{t^{3}}) d t$ $a = \frac{1}{2} [\ln ∣ t ∣ - \frac{1}{2 t^{2}}] ∣_{- 1}^{1 - \sqrt{2}} \approx - 1 . 648$
	Answered by FelipeLz last updated on 12/Dec/22

	$\tan (x) = u \to d u = \sec^{2} (x) d x$ $x = 0 \to u = 0$ $x = \frac{π}{4} \to u = 1$ $I = \underset{0}{\int^{π / 4}} \frac{\sec^{2} (x)}{\tan (x) - \sec (x)} d x = \underset{0}{\int^{1}} \frac{1}{u - \sqrt{u^{2} + 1}} d u = \underset{0}{\int^{1}} \frac{1}{u - \sqrt{u^{2} + 1}} \times \frac{u + \sqrt{u^{2} + 1}}{u + \sqrt{u^{2} + 1}} d u = \underset{0}{\int^{1}} \frac{u + \sqrt{u^{2} + 1}}{u^{2} - u^{2} - 1} d u$ $I = - \underset{0}{\int^{1}} u d u - \underset{0}{\int^{1}} \sqrt{u^{2} + 1} d u$ $I = - \underset{0}{\int^{π / 4}} \tan (x) \sec^{2} (x) d x - \underset{0}{\int^{π / 4}} \sec^{3} (x) d x$ $\int \tan (x) \sec^{2} (x) d x = \sec^{2} (x) - \int \tan (x) \sec^{2} (x) d x$ $2 \int \tan (x) \sec^{2} (x) d x = \sec^{2} (x)$ $\int \tan (x) \sec^{2} (x) d x = \frac{1}{2} \sec^{2} (x)$ $\int \sec^{3} (x) d x = \sec (x) \tan (x) - \int \sec (x) \tan^{2} (x) d x$ $\int \sec^{3} (x) d x = \sec (x) \tan (x) - \int \sec^{3} (x) d x + \int \sec (x) d x$ $2 \int \sec^{3} (x) d x = \sec (x) \tan (x) + \ln ∣ \sec (x) + \tan (x) ∣$ $\int \sec^{3} (x) d x = \frac{1}{2} \sec (x) \tan (x) + \frac{1}{2} \ln ∣ \sec (x) + \tan (x) ∣$ $I = - \frac{1}{2} {[\sec^{2} (x)]}_{0}^{π / 4} - \frac{1}{2} {[\sec (x) \tan (x) + \ln ∣ \sec (x) + \tan (x) ∣]}_{0}^{π / 4}$ $I = - \frac{1}{2} [{(\sqrt{2})}^{2} - 1^{2}] - \frac{1}{2} [\sqrt{2} \times 1 + \ln ∣ \sqrt{2} + 1 ∣ - 1 \times 0 - \ln ∣ 1 + 0 ∣]$ $I = - \frac{1}{2} [1 + \sqrt{2} + \ln (1 + \sqrt{2})]$
	Answered by MJS_new last updated on 12/Dec/22

	$\int \frac{\sec^{2} x}{\tan x - \sec x} d x = \int \frac{d x}{\sin 2 x - 2 \cos x} =$ $[t = \tan \frac{x}{2} \to d x = \frac{2 d t}{t^{2} + 1}]$ $= 2 \int \frac{t^{2} + 1}{{(t - 1)}^{3} (t + 1)} d t =$ $= \frac{1}{2} \int (\frac{1}{t - 1} - \frac{1}{t + 1}) d t + \int \frac{t + 1}{{(t - 1)}^{3}} d t =$ $= \frac{1}{2} \ln \frac{t - 1}{t + 1} - \frac{t}{{(t - 1)}^{2}} =$ $= \frac{1}{2} (\ln ∣ \tan x - \sec x ∣ - (\tan x + \sec x) \tan x) + C$ $\Rightarrow answer is \frac{1}{2} \ln (- 1 + \sqrt{2}) - \frac{1 + \sqrt{2}}{2}$
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