find-Vn-1-n-an-1-n-x-a-x-x-dx-

Question Number 53470 by maxmathsup by imad last updated on 22/Jan/19

f i n d V n = \int_{\frac{1}{n}}^{\frac{a n - 1}{n}} \frac{\sqrt{x}}{\sqrt{a - \sqrt{x} + x}} d x

Commented by maxmathsup by imad last updated on 24/Jan/19

c h a n g e m e n t \sqrt{x} = t g i v e x = t^{2} \Rightarrow d x = 2 t d t a n d

V_{n} = \int_{\frac{1}{\sqrt{n}}}^{\sqrt{\frac{a n - 1}{n}}} \frac{t}{\sqrt{a - t + t^{2}}} (2 t) d t = 2 \int_{\frac{1}{\sqrt{n}}}^{\sqrt{\frac{a n - 1}{n}}} \frac{t^{2}}{\sqrt{t^{2} - 2 \frac{1}{2} t + \frac{1}{4} + a - \frac{1}{4}}} d t

= 2 \int_{\frac{1}{\sqrt{n}}}^{\sqrt{\frac{a n - 1}{n}}} \frac{t^{2}}{\sqrt{{(t - \frac{1}{2})}^{2} + \frac{4 a - 1}{4}}} l e t s u p p o s e a > \frac{1}{4} \Rightarrow

V_{n} =_{t - \frac{1}{2} = \frac{\sqrt{4 a - 1}}{2} u} 2 \int_{\frac{\frac{2}{\sqrt{n}} - 1}{\sqrt{4 a - 1}}}^{\frac{2 \sqrt{\frac{a n - 1}{n}} - 1}{\sqrt{4 a - 1}}} \frac{{(\frac{\sqrt{4 a - 1}}{2} u + \frac{1}{2})}^{2}}{\frac{\sqrt{4 a - 1}}{2} \sqrt{1 + u^{2}}} \frac{\sqrt{4 a - 1}}{2} d u

= 2 \int_{\frac{2 - \sqrt{n}}{\sqrt{n} \sqrt{4 a - 1}}}^{\frac{2 (\sqrt{4 a - 1} - \sqrt{n}}{\sqrt{n} \sqrt{4 a - 1}}} \frac{1}{4} \frac{(4 a - 1) u^{2} + 2 \sqrt{4 a - 1} u + 1}{\sqrt{1 + u^{2}}} d u

= \frac{1}{2} \int_{α_{n}}^{β_{n}} \frac{(4 a - 1) u^{2} + 2 \sqrt{4 a - 1} u + 1}{\sqrt{1 + u^{2}}} d u

= \frac{4 a - 1}{2} \int_{α_{n}}^{β_{n}} \frac{u^{2} + 1 - 1}{\sqrt{1 + u^{2}}} + \sqrt{4 a - 1} \int_{α_{n}}^{β_{n}} \frac{u}{\sqrt{1 + u^{2}}} d u + \frac{1}{2} \int_{α_{n}}^{β_{n}} \frac{d u}{\sqrt{1 + u^{2}}}

\int_{α_{n}}^{β_{n}} \frac{d u}{\sqrt{1 + u^{2}}} = [l n (1 + \sqrt{{1 + x^{2}]}_{α_{n}}^{β_{n}}} = l n (1 + \sqrt{1 + β_{n}^{2}}) - l n (1 + \sqrt{1 + α_{n}^{2}})

\int_{α_{n}}^{β_{n}} \frac{u}{\sqrt{1 + u^{2}}} d u = {[\sqrt{1 + u^{2}}]}_{α_{n}}^{β_{n}} = \sqrt{1 + β_{n}^{2}} - \sqrt{1 + α_{n}^{2}}

\int_{α_{n}}^{β_{n}} \frac{u^{2} + 1 - 1}{\sqrt{1 + u^{2}}} d u = \int_{α_{n}}^{β_{n}} \sqrt{1 + u^{2}} d u - \int_{α_{n}}^{β_{n}} \frac{d u}{\sqrt{1 + u^{2}}} = \dots . .