evaluate-2-3-2-1-x-2-4-x-2-dx-using-the-substitution-x-2tan-

Question Number 98623 by hardylanes last updated on 15/Jun/20

e v a l u a t e

\int_{\frac{2}{\sqrt{3}}}^{2} \frac{1}{x^{2} \sqrt{4 + x^{2}}} d x u s i n g t h e s u b s t i t u t i o n x = 2 \tan θ

Answered by Kunal12588 last updated on 15/Jun/20

x = 2 t a n θ

\Rightarrow d x = 2 {s e c}^{2} θ d θ

θ = {t a n}^{- 1} (\frac{x}{2});

\int_{π / 6}^{π / 4} \frac{2 {s e c}^{2} θ}{4 {t a n}^{2} θ \sqrt{4 + 4 {t a n}^{2} θ}} d θ

\int_{π / 6}^{π / 4} \frac{s e c θ}{4 {t a n}^{2} θ} d θ

\frac{1}{4} \int_{π / 6}^{π / 4} \frac{c o s θ}{{s i n}^{2} θ} d θ

\frac{1}{4} \int_{π / 6}^{π / 4} \frac{d (s i n θ)}{{s i n}^{2} θ}

\frac{1}{4} {[- \frac{1}{s i n θ}]}_{π / 6}^{π / 4} = \frac{1}{4} (- \sqrt{2} + 2) = \frac{2 - \sqrt{2}}{4}

Commented by hardylanes last updated on 15/Jun/20

thanks

Commented by abdomathmax last updated on 15/Jun/20

your answer is correct miss kunal

Commented by Kunal12588 last updated on 16/Jun/20

{i t}^{'} s m i s t e r!

Answered by MWSuSon last updated on 15/Jun/20

x = 2 t a n θ, d x = 2 {s e c}^{2} x d θ

\frac{2}{\sqrt{3}} = 2 t a n θ, θ = a r c t a n (\frac{1}{\sqrt{3}}) = \frac{π}{6}

2 = 2 t a n θ, θ = a r c t a n (1) = \frac{π}{4}

\underset{\frac{π}{6}}{\int^{\frac{π}{4}}} \frac{2 {s e c}^{2} θ}{4 {t a n}^{2} θ \sqrt{4 + 4 {t a n}^{2} θ}} d θ = \underset{\frac{π}{6}}{\int^{\frac{π}{4}}} \frac{{s e c}^{2} θ}{2 \times 2 {t a n}^{2} θ \sqrt{1 + {t a n}^{2} θ}} d θ

= \underset{\frac{π}{6}}{\int^{\frac{π}{4}}} \frac{s e c θ}{4 {t a n}^{2} θ} d θ = \frac{1}{4} \underset{\frac{π}{6}}{\int^{\frac{π}{4}}} \frac{1}{c o s θ} \times \frac{{c o s}^{2} θ}{{s i n}^{2} θ} d θ = \frac{1}{4} \underset{\frac{π}{6}}{\int^{\frac{π}{4}}} \frac{c o s θ}{{s i n}^{2} θ} d θ

l e t t = s i n θ, d t = c o s θ d θ

= \frac{1}{4} \underset{\frac{π}{6}}{\int^{\frac{π}{4}}} \frac{d t}{t^{2}} = \frac{1}{4} (- \frac{1}{s i n (\frac{π}{4})} + \frac{1}{s i n (\frac{π}{6})}) = \frac{1}{4} (2 - \sqrt{2})

Commented by hardylanes last updated on 15/Jun/20

thankiu

Answered by abdomathmax last updated on 15/Jun/20

let I = \int_{\frac{2}{\sqrt{3}}}^{2} \frac{dx}{x^{2} \sqrt{4 + x^{2}}} let do the changement

x = 2 sht \Rightarrow t = argsh (\frac{x}{2}) \Rightarrow

I = \int_{\ln (\frac{1}{\sqrt{3}} + \sqrt{1 + \frac{1}{3}})}^{\ln (1 + \sqrt{2})} \frac{2 ch (t) dt}{{4 sh}^{2} t (2 cht)}

= \int_{\ln (\sqrt{3})}^{\ln (1 + \sqrt{2})} \frac{dt}{4 (\frac{ch (2 t) - 1}{2})}

= \frac{1}{2} \int_{\ln (\sqrt{3})}^{\ln (1 + \sqrt{2})} \frac{dt}{\frac{e^{2 t} + e^{- 2 t}}{2} - 1}

= \int_{\ln (\sqrt{3})}^{\ln (1 + \sqrt{2})} \frac{dt}{e^{2 t} + e^{- 2 t} - 2}

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

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

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

= \frac{1}{2} {\frac{1}{2} - \frac{1}{2 + 2 \sqrt{2}}} = \frac{1}{4} {1 - \frac{1}{1 + \sqrt{2}}}

= \frac{1}{4} {\frac{\sqrt{2}}{1 + \sqrt{2}}} = \frac{1}{4} \times \frac{\sqrt{2} (\sqrt{2} - 1)}{1} = \frac{1}{4} (2 - \sqrt{2})

= \frac{1}{2} - \frac{\sqrt{2}}{4}