0-pi-4-sec-2-x-tan-x-sec-x-dx-

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 \dots . . m

\sec x (\sec x - \tan x) d x = d t

\sec x d x = \frac{- d t}{t} \dots 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) \dots 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}

Facebook Tweet Pin