Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 81400 by mind is power last updated on 12/Feb/20

Hello Nice day im thinking of this one a close forme? ∫(√(1+x^p ))dx p∈R_+ , x∈[0,1[

$${Hello}\:\:{Nice}\:{day}\:{im}\:{thinking}\:{of}\:{this}\:{one}\:\:{a}\:{close}\:{forme}? \\ $$$$\int\sqrt{\mathrm{1}+{x}^{{p}} }{dx} \\ $$$${p}\in\mathbb{R}_{+} , \\ $$$${x}\in\left[\mathrm{0},\mathrm{1}\left[\right.\right. \\ $$

Commented by abdomathmax last updated on 13/Feb/20

at form of serie we have (1+u)^α =1+(α/(1!))u +((α(α−1))/(2!))u^2 +... =1+Σ_(n=1) ^∞ ((α(α−1)...(α−n+1))/(n!)) u^n ⇒ (√(1+u))=1+Σ_(n=1) ^∞ (((1/2)((1/2)−1)....((1/2)−n+1))/(n!))u^n ⇒ (√(1+x^p ))=1+Σ_(n=1) ^∞ (((1/2)((1/2)−1)...((1/2)−n+1))/(n!))x^(pn) ⇒ ∫(√(1+x^p ))dx= 1+Σ_(n=1) ^∞ (((1/2)((1/2)−1)...((1/2)−n+1))/(n!))×(1/(pn+1))x^(pn+1) +C

$${at}\:{form}\:{of}\:{serie}\:{we}\:{have}\: \\ $$$$\left(\mathrm{1}+{u}\right)^{\alpha} =\mathrm{1}+\frac{\alpha}{\mathrm{1}!}{u}\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}!}{u}^{\mathrm{2}} +... \\ $$$$=\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\alpha\left(\alpha−\mathrm{1}\right)...\left(\alpha−{n}+\mathrm{1}\right)}{{n}!}\:{u}^{{n}} \:\Rightarrow \\ $$$$\sqrt{\mathrm{1}+{u}}=\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)....\left(\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right)}{{n}!}{u}^{{n}} \:\Rightarrow \\ $$$$\sqrt{\mathrm{1}+{x}^{{p}} }=\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)...\left(\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right)}{{n}!}{x}^{{pn}} \:\Rightarrow \\ $$$$\int\sqrt{\mathrm{1}+{x}^{{p}} }{dx}= \\ $$$$\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)...\left(\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right)}{{n}!}×\frac{\mathrm{1}}{{pn}+\mathrm{1}}{x}^{{pn}+\mathrm{1}} \:+{C} \\ $$

Commented by mind is power last updated on 14/Feb/20

=x_2 F_1 (−(1/2),(1/p);(1/p)+1;x^p )+c

$$={x}_{\mathrm{2}} {F}_{\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{{p}};\frac{\mathrm{1}}{{p}}+\mathrm{1};{x}^{{p}} \right)+{c} \\ $$

Commented by mind is power last updated on 14/Feb/20

nice Sir

$${nice}\:{Sir} \\ $$

Answered by behi83417@gmail.com last updated on 12/Feb/20

x^p =tg^2 t⇒px^(p−1) dx=2tgt(1+tg^2 t)dt ⇒dx=(2/p)tg^(((2/p)−1)) t(1+tg^2 t)dt ⇒I=(2/p)∫tg^(((2/p)−1)) t.(1/(cos^3 t))dt= =^(m=((2/p)−1)) (m+1)∫((tg^m t)/(cos^3 t))dt= =(m+1)∫sin^m t.cos^(−(m+3)) tdt=...

$$\mathrm{x}^{\mathrm{p}} =\mathrm{tg}^{\mathrm{2}} \mathrm{t}\Rightarrow\mathrm{px}^{\mathrm{p}−\mathrm{1}} \mathrm{dx}=\mathrm{2tgt}\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \mathrm{t}\right)\mathrm{dt} \\ $$$$\Rightarrow\mathrm{dx}=\frac{\mathrm{2}}{\mathrm{p}}\mathrm{tg}^{\left(\frac{\mathrm{2}}{\mathrm{p}}−\mathrm{1}\right)} \mathrm{t}\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \mathrm{t}\right)\mathrm{dt} \\ $$$$\Rightarrow\mathrm{I}=\frac{\mathrm{2}}{\mathrm{p}}\int\mathrm{tg}^{\left(\frac{\mathrm{2}}{\mathrm{p}}−\mathrm{1}\right)} \mathrm{t}.\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{3}} \mathrm{t}}\mathrm{dt}= \\ $$$$\overset{\mathrm{m}=\left(\frac{\mathrm{2}}{\mathrm{p}}−\mathrm{1}\right)} {=}\left(\mathrm{m}+\mathrm{1}\right)\int\frac{\mathrm{tg}^{\mathrm{m}} \mathrm{t}}{\mathrm{cos}^{\mathrm{3}} \mathrm{t}}\mathrm{dt}= \\ $$$$=\left(\mathrm{m}+\mathrm{1}\right)\int\mathrm{sin}^{\mathrm{m}} \mathrm{t}.\mathrm{cos}^{−\left(\mathrm{m}+\mathrm{3}\right)} \mathrm{tdt}=... \\ $$

Commented by mind is power last updated on 12/Feb/20

nice sir Try too use Hypergeometric Function i think a close forme by this one may bee

$${nice}\:{sir}\:\:\: \\ $$$${Try}\:{too}\:{use}\:\:{Hypergeometric}\:{Function}\: \\ $$$${i}\:{think}\:{a}\:{close}\:{forme}\:{by}\:{this}\:{one}\:{may}\:{bee} \\ $$$$ \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com