1-calculate-f-x-0-1-t-2-x-2-t-2-dt-with-x-gt-0-2-calculste-g-x-0-1-t-2-x-2-t-2-dt-

Question Number 74637 by mathmax by abdo last updated on 28/Nov/19

1) c a l c u l a t e f (x) = \int_{0}^{1} t^{2} \sqrt{x^{2} + t^{2}} d t w i t h x > 0

2) c a l c u l s t e g (x) = \int_{0}^{1} \frac{t^{2}}{\sqrt{x^{2} + t^{2}}} d t

Commented by mathmax by abdo last updated on 28/Nov/19

1) w e h a v e f (x) = \int_{0}^{1} t^{2} \sqrt{x^{2} + t^{2}} d t \Rightarrow f (x) =_{t = x s h (u)} \int_{0}^{a r g s h (\frac{1}{x})} x^{2} {s h}^{2} u (x c h (u)) x c h (u) d u

= x^{4} \int_{0}^{l n (\frac{1}{x} + \sqrt{1 + \frac{1}{x^{2}}})} {s h}^{2} (u) {c h}^{2} (u) d u

= \frac{x^{4}}{4} \int_{0}^{l n (\frac{1 + \sqrt{x^{2} + 1}}{x})} {s h}^{2} (2 u) d u = \frac{x^{4}}{8} \int_{0}^{l n (\frac{1 + \sqrt{x^{2} + 1}}{x})} (c h (4 u) - 1)) d u

= \frac{x^{4}}{32} {[s h (4 u)]}_{0}^{l n (\frac{1 + \sqrt{x^{2} + 1}}{x})} - \frac{x^{4}}{8} l n (\frac{1 + \sqrt{x^{2} + 1}}{x})

= \frac{x^{4}}{64} {[e^{4 u} - e^{- 4 u}]}_{0}^{l n (\frac{1 + \sqrt{x^{2} + 1}}{x})} - \frac{x^{4}}{8} l n (\frac{1 + \sqrt{x^{2} + 1}}{x})

= \frac{x^{4}}{64} {{(\frac{1 + \sqrt{x^{2} + 1}}{x})}^{4} - {(\frac{1 + \sqrt{x^{2} + 1}}{x})}^{- 4}} - \frac{x^{4}}{8} l n (\frac{1 + \sqrt{x^{2} + 1}}{x})

Answered by MJS last updated on 28/Nov/19

both with the same substitution

t = x \sinh \ln u = \frac{x (u^{2} - 1)}{2 u} \to d t = \frac{x (u^{2} + 1)}{2 u^{2}} d u

\int t^{2} \sqrt{x^{2} + t^{2}} d t = \frac{x^{4}}{16} \int \frac{{(u^{4} - 1)}^{2}}{u^{5}} d u =

= x^{4} (\frac{u^{8} - 1}{64 u^{4}} - \frac{1}{8} \ln u) = f (x)

\int \frac{t^{2}}{\sqrt{x^{2} + t^{2}}} d t = \frac{x^{2}}{4} \int \frac{{(u^{2} - 1)}^{2}}{u^{3}} d u =

= x^{2} (\frac{u^{4} - 1}{8 u^{2}} - \frac{1}{2} \ln u) = g (x)

the borders : 0 ⩽ t ⩽ 1 \Rightarrow 1 ⩽ u ⩽ \frac{1 + \sqrt{x^{2} + 1}}{x}

f (1) = g (1) = 0

f (x) = x^{4} (\frac{u^{8} - 1}{64 u^{4}} - \frac{1}{8} \ln u) with u = \frac{1 + \sqrt{x^{2} + 1}}{x}

g (x) = x^{2} (\frac{u^{4} - 1}{8 u^{2}} - \frac{1}{2} \ln u) with u = \frac{1 + \sqrt{x^{2} + 1}}{x}

sorry I^{'} ve got no time to insert these and

transform

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