If-I-0-1-dx-1-x-4-then-

Question Number 63065 by Enebeli Chinedu Vitalis last updated on 28/Jun/19

If I = \underset{0}{\int^{1}} \frac{d x}{\sqrt{1 + x^{4}}}, then

Commented by Enebeli Chinedu Vitalis last updated on 28/Jun/19

t h a n k s b o s s ★ ★

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

c h a n g e m e n t x^{2} = t a n θ g i v e x = {(t a n θ)}^{\frac{1}{2}} \Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{1}{2 \sqrt{1 + {t a n}^{2} θ}} (1 + {t a n}^{2} θ) {(t a n θ)}^{- \frac{1}{2}} d θ

= \frac{1}{2} \int_{0}^{\frac{π}{4}} \sqrt{1 + {t a n}^{2} θ} {(t a n θ)}^{- \frac{1}{2}} = \frac{1}{2} \int_{0}^{\frac{π}{4}} \frac{1}{c o s θ \sqrt{\frac{s i n θ}{c o s θ}}} = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{d θ}{\sqrt{c o s θ s i n θ}}

= \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{d θ}{\sqrt{s i n (2 θ)}} = \frac{1}{\sqrt{2}} \int_{0}^{π} \frac{d t}{2 \sqrt{s i n t}} (2 θ = t)

= \frac{1}{2 \sqrt{2}} \int_{0}^{π} \frac{d t}{\sqrt{s i n t}} c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e

\int_{0}^{π} \frac{d t}{\sqrt{s i n t}} = \int_{0}^{\infty} \frac{2 d u}{(1 + u^{2}) \sqrt{\frac{2 u}{1 + u^{2}}}} = \sqrt{2} \int_{0}^{\infty} \frac{d u}{\sqrt{u} \sqrt{1 + u^{2}}}

=_{\sqrt{u} = α} \sqrt{2} \int_{0}^{\infty} \frac{2 α d α}{α \sqrt{1 + α^{4}}} = 2 \sqrt{2} \int_{0}^{\infty} \frac{d α}{\sqrt{1 + α^{4}}} \Rightarrow I = \int_{0}^{\infty} \frac{d x}{\sqrt{1 + x^{4}}}

I = \int_{0}^{1} \frac{d x}{\sqrt{1 + x^{4}}} + \int_{1}^{+ \infty} \frac{d x}{\sqrt{1 + x^{4}}} \Rightarrow \int_{1}^{+ \infty} \frac{d x}{\sqrt{1 + x^{4}}} = 0 \dots . b e c o n t n u e d \dots .

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

y o u a r e w e l c o m e .