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Question Number 68481 Answers: 0 Comments: 6

I=∫_0 ^( 1) (√((c−x^2 )/(x(1−x^2 ))))dx (c >1)

$${I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\frac{{c}−{x}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}}{dx}\:\:\:\:\:\:\left({c}\:>\mathrm{1}\right) \\ $$

Question Number 68576 Answers: 1 Comments: 1

Question Number 68470 Answers: 0 Comments: 3

find the value of ∫_0 ^∞ ((arctan(2x^2 ))/(x^2 +4))dx

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$

Question Number 68466 Answers: 1 Comments: 2

let f(x) =∫_x^ ^(x^2 −x) arctan(e^(−x−t) )dt calculate f^′ (x) and f^′ (0).

$${let}\:{f}\left({x}\right)\:=\int_{{x}^{} } ^{{x}^{\mathrm{2}} −{x}} \:{arctan}\left({e}^{−{x}−{t}} \right){dt} \\ $$$${calculate}\:{f}^{'} \left({x}\right)\:\:\:{and}\:{f}^{'} \left(\mathrm{0}\right). \\ $$

Question Number 68473 Answers: 0 Comments: 1

((sin 72°)/(sin 42°)) = p tan 12° = ?

$$\frac{\mathrm{sin}\:\mathrm{72}°}{\mathrm{sin}\:\mathrm{42}°}\:\:=\:\:{p} \\ $$$$\mathrm{tan}\:\mathrm{12}°\:\:=\:\:? \\ $$

Question Number 68451 Answers: 1 Comments: 7

Question Number 68446 Answers: 0 Comments: 1

cos (x−60)+cos (x−30)=sin x prove

$$\mathrm{cos}\:\left({x}−\mathrm{60}\right)+\mathrm{cos}\:\left({x}−\mathrm{30}\right)=\mathrm{sin}\:{x} \\ $$$${prove} \\ $$

Question Number 68434 Answers: 1 Comments: 0

Question Number 68433 Answers: 0 Comments: 0

hello i search som lectur about hypergeometric fonction2F_1 (a,b,c,x)=((Γ(c))/(Γ(a)Γ(b)))Σ_(n≥0) ((Γ(a+n)Γ(b+n))/(Γ(c+n)n!))x^n

$${hello} \\ $$$${i}\:{search}\:{som}\:{lectur}\:{about}\:{hypergeometric}\:{fonction}\mathrm{2}{F}_{\mathrm{1}} \left({a},{b},{c},{x}\right)=\frac{\Gamma\left({c}\right)}{\Gamma\left({a}\right)\Gamma\left({b}\right)}\sum_{{n}\geqslant\mathrm{0}} \frac{\Gamma\left({a}+{n}\right)\Gamma\left({b}+{n}\right)}{\Gamma\left({c}+{n}\right){n}!}{x}^{{n}} \\ $$$$ \\ $$

Question Number 68487 Answers: 1 Comments: 2

Question Number 68425 Answers: 1 Comments: 2

Question Number 68422 Answers: 1 Comments: 0

Question Number 68418 Answers: 1 Comments: 3

Question Number 68414 Answers: 2 Comments: 1

Is it possible to find any value for a,b,c from below system of equetions? { ((sina+sinb=sinc)),((cosa+cosb=cosc)) :}

$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{any}\:\mathrm{value}\:\mathrm{for} \\ $$$$\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\:\mathrm{from}\:\mathrm{below}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equetions}? \\ $$$$\begin{cases}{\boldsymbol{\mathrm{sina}}+\boldsymbol{\mathrm{sinb}}=\boldsymbol{\mathrm{sinc}}}\\{\boldsymbol{\mathrm{cosa}}+\boldsymbol{\mathrm{cosb}}=\boldsymbol{\mathrm{cosc}}}\end{cases} \\ $$

Question Number 68409 Answers: 1 Comments: 3

calculate ∫_0 ^(+∞) ((arctan(x^2 ))/(1+x^2 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 68405 Answers: 1 Comments: 3

There are a few problem reported. Notification: google discontinued Google Cloud Messaging so notifications are not working. Other Problems: There has been several new phone models and android version updates. And newer android version or phone models might have other issues. We are working on app updates for supporting these and also migrating to new messaging platform. We will address these problems as soon as we can.

$$\mathrm{There}\:\mathrm{are}\:\mathrm{a}\:\mathrm{few}\:\mathrm{problem}\:\mathrm{reported}. \\ $$$$\mathrm{Notification}:\:\mathrm{google}\:\mathrm{discontinued} \\ $$$$\mathrm{Google}\:\mathrm{Cloud}\:\mathrm{Messaging}\:\mathrm{so}\:\mathrm{notifications} \\ $$$$\mathrm{are}\:\mathrm{not}\:\mathrm{working}.\: \\ $$$$\mathrm{Other}\:\mathrm{Problems}: \\ $$$$\mathrm{There}\:\mathrm{has}\:\mathrm{been}\:\mathrm{several}\:\mathrm{new}\:\mathrm{phone} \\ $$$$\mathrm{models}\:\mathrm{and}\:\mathrm{android}\:\mathrm{version}\:\mathrm{updates}. \\ $$$$\mathrm{And}\:\mathrm{newer}\:\mathrm{android}\:\mathrm{version}\:\mathrm{or}\:\mathrm{phone} \\ $$$$\mathrm{models}\:\mathrm{might}\:\mathrm{have}\:\mathrm{other}\:\mathrm{issues}. \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{are}\:\mathrm{working}\:\mathrm{on}\:\mathrm{app}\:\mathrm{updates}\:\mathrm{for} \\ $$$$\mathrm{supporting}\:\mathrm{these}\:\mathrm{and}\:\mathrm{also}\:\mathrm{migrating} \\ $$$$\mathrm{to}\:\mathrm{new}\:\mathrm{messaging}\:\mathrm{platform}. \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{will}\:\mathrm{address}\:\mathrm{these}\:\mathrm{problems}\:\mathrm{as}\: \\ $$$$\mathrm{soon}\:\mathrm{as}\:\mathrm{we}\:\mathrm{can}. \\ $$

Question Number 68397 Answers: 2 Comments: 0