0-1-1-1-4x-dx-

Question Number 186263 by mokys last updated on 02/Feb/23

\int_{0}^{1} \sqrt{1 + \frac{1}{4 x}} d x

Answered by cortano1 last updated on 03/Feb/23

l e t \frac{1}{2 \sqrt{x}} = \tan t \Rightarrow 2 \sqrt{x} = \cot t

\Rightarrow 4 x = \cot^{2} t

\Rightarrow 4 d x = 2 \cot t (- \csc^{2} t) d t

I = \int \frac{1}{\sqrt{1 + \tan^{2} t}} (- 2 \cot t \csc^{2} t) d t

I = - 2 \int \cos t \frac{\cos t}{\sin t} . \frac{1}{\sin^{2} t} d t

I = - 2 \int \frac{1 - \sin^{2} t}{\sin^{3} t} d t

I = - 2 \int (\csc^{3} t - \csc t) d t

Answered by Mathspace last updated on 03/Feb/23

l e t \frac{1}{4 x} = t \Rightarrow x = \frac{1}{4 t} \Rightarrow

I = - \int_{\frac{1}{4}}^{+ \infty} \sqrt{1 + t} (\frac{- d t}{4 t^{2}})

4 I = \int_{\frac{1}{4}}^{+ \infty} \frac{\sqrt{1 + t}}{t^{2}} d t (1 + t = z^{2})

= \int_{\frac{\sqrt{5}}{2}}^{\infty} \frac{z}{{(z^{2} - 1)}^{2}} (2 z) d z

= \int_{\frac{\sqrt{5}}{2}} z (\frac{2 z}{{(z^{2} - 1)}^{2}}) d z (b y p a r t s)

= {[- \frac{z}{z^{2} - 1}]}_{\frac{\sqrt{5}}{2}}^{\infty} - \int_{\frac{\sqrt{5}}{2}}^{\infty} (- \frac{1}{z^{2} - 1}) d z

= \frac{\sqrt{5}}{2 (\frac{5}{4} - 1)} + \frac{1}{2} \int_{\frac{\sqrt{5}}{2}}^{\infty} (\frac{1}{z - 1} - \frac{1}{z + 1}) d z

= 2 \sqrt{5} + \frac{1}{2} {[l n ∣ \frac{z - 1}{z + 1} ∣]}_{\frac{\sqrt{5}}{2}}^{\infty}

= 2 \sqrt{5} - \frac{1}{2} l n ∣ \frac{\frac{\sqrt{5}}{2} - 1}{\frac{\sqrt{5}}{2} + 1} ∣

= 2 \sqrt{5} - \frac{1}{2} l n (\frac{\sqrt{5} - 2}{\sqrt{5} + 2})