hi-x-pi-4-pi-3-f-x-1-cos-x-primitive-of-f-x-

Question Number 168225 by henderson last updated on 06/Apr/22

hi!

x \in] \frac{π}{4}; \frac{π}{3} [

f (x) = \frac{1}{c o s x}

primitive of f (x) .

Answered by MJS_new last updated on 06/Apr/22

\underset{π / 4}{\int^{π / 3}} \frac{d x}{\cos x} =

[t = \tan \frac{x}{2} \to d x = \frac{2 d t}{t^{2} + 1}]

= - 2 \underset{- 1 + \sqrt{2}}{\int^{1 / \sqrt{3}}} \frac{d t}{t^{2} - 1} = \int (\frac{1}{t + 1} - \frac{1}{t + 1}) d t =

= {[\ln ∣ t + 1 ∣ - \ln ∣ t - 1 ∣]}_{- 1 + \sqrt{2}}^{1 / \sqrt{3}} =

= \ln (2 + \sqrt{3}) - \ln (1 + \sqrt{2})

\int \frac{d x}{\cos x} = \ln ∣ \tan (\frac{x}{2} + \frac{π}{4}) ∣ + C =

= \ln ∣ \frac{\cos x}{1 - \sin x} ∣ + C

Answered by floor(10²Eta[1]) last updated on 06/Apr/22

\int_{π / 4}^{π / 3} \frac{1}{cosx} dx = \int_{π / 4}^{π / 3} \frac{cosx}{\cos^{2} x} dx = \int_{π / 4}^{π / 3} \frac{cosx}{1 - \sin^{2} x} dx

u = sinx \Rightarrow du = cosxdx

\int_{\sqrt{2} / 2}^{\sqrt{3} / 2} \frac{du}{1 - u^{2}} = \frac{1}{2} \int_{\sqrt{2} / 2}^{\sqrt{3} / 2} (\frac{1}{1 - u} + \frac{1}{1 + u}) du

= - \frac{1}{2} {[\ln ∣ 1 - u ∣]}_{\sqrt{2} / 2}^{\sqrt{3} / 2} + \frac{1}{2} {[\ln ∣ 1 + u ∣]}_{\sqrt{2} / 2}^{\sqrt{3} / 2}

- \frac{1}{2} (\ln (1 - \frac{\sqrt{3}}{2}) - \ln (1 - \frac{\sqrt{2}}{2})) + \frac{1}{2} (\ln (1 + \frac{\sqrt{3}}{2}) - \ln (1 + \frac{\sqrt{2}}{2}))

= - \frac{1}{2} \ln (\frac{2 - \sqrt{3}}{2}) + \frac{1}{2} \ln (\frac{2 - \sqrt{2}}{2}) + \frac{1}{2} \ln (\frac{2 + \sqrt{3}}{2}) - \frac{1}{2} \ln (\frac{2 + \sqrt{2}}{2})

= \frac{1}{2} \ln (\frac{2 + \sqrt{3}}{2 - \sqrt{3}}) - \frac{1}{2} \ln (\frac{2 + \sqrt{2}}{2 - \sqrt{2}})

= \ln (2 + \sqrt{3}) - \ln (2 + \sqrt{2}) + \ln \sqrt{2}

= \ln (\frac{2 + \sqrt{3}}{\sqrt{2} + 1})

Commented by floor(10²Eta[1]) last updated on 06/Apr/22

thanks i corrected it

Commented by MJS_new last updated on 06/Apr/22

something went wrong in the last part after

inserting the values . the integral solution is

ok but the result must be

\ln ((2 + \sqrt{3}) (- 1 + \sqrt{2})) = \ln (2 + \sqrt{3}) - \ln (1 + \sqrt{2})

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