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Question Number 108145    Answers: 3   Comments: 2

((⊚BeMath⊚)/Π) (1) cos^2 ((x/4)) > ((√2)/2) + sin^2 ((x/4)) (2) ∫ (√((1+x)/(1−x))) dx (3) ∫_0 ^(π/2) ((√(tan x))/((cos x+sin x)^2 )) dx

$$\:\:\:\frac{\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc}{\Pi} \\ $$$$\:\left(\mathrm{1}\right)\:\:\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right)\:>\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right)\: \\ $$$$\:\left(\mathrm{2}\right)\:\int\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx} \\ $$$$\left(\mathrm{3}\right)\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\sqrt{\mathrm{tan}\:{x}}}{\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:{dx} \\ $$

Question Number 108144    Answers: 0   Comments: 0

∫ (√(ln(x)))dx

$$\int\:\sqrt{{ln}\left({x}\right)}{dx} \\ $$

Question Number 108143    Answers: 2   Comments: 0

cos^2 x+cos^2 2x=sin^2 3x Solve the equation. Please help ASAP

$$\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{2x}=\mathrm{sin}^{\mathrm{2}} \mathrm{3x} \\ $$$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{ASAP} \\ $$

Question Number 108133    Answers: 2   Comments: 1

Question Number 108131    Answers: 3   Comments: 1

Question Number 108118    Answers: 1   Comments: 0

Question Number 108114    Answers: 1   Comments: 0

Question Number 108107    Answers: 1   Comments: 2

Question Number 108104    Answers: 1   Comments: 0

Question Number 108099    Answers: 3   Comments: 1

((★BeMath⊚)/⊓) (1) ∫ x tan^(−1) (x) dx ? (2) Find the distance of the point (3,3,1) from the plane Π with equation (r^→ −i^→ −j^→ )•(i^→ −j^→ +k^→ ) = 0 , also find the point on the plane that is nearest to (3,3,1).

$$\:\:\:\:\frac{\bigstar\mathcal{B}{e}\mathcal{M}{ath}\circledcirc}{\sqcap} \\ $$$$\:\left(\mathrm{1}\right)\:\int\:{x}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:{dx}\:? \\ $$$$\left(\mathrm{2}\right)\:{Find}\:{the}\:{distance}\:{of}\:{the}\:{point}\: \\ $$$$\left(\mathrm{3},\mathrm{3},\mathrm{1}\right)\:{from}\:{the}\:{plane}\:\Pi\:{with}\:{equation} \\ $$$$\left(\overset{\rightarrow} {{r}}−\overset{\rightarrow} {{i}}−\overset{\rightarrow} {{j}}\right)\bullet\left(\overset{\rightarrow} {{i}}−\overset{\rightarrow} {{j}}+\overset{\rightarrow} {{k}}\right)\:=\:\mathrm{0}\:,\:{also}\: \\ $$$${find}\:{the}\:{point}\:{on}\:{the}\:{plane}\:{that}\:{is} \\ $$$${nearest}\:{to}\:\left(\mathrm{3},\mathrm{3},\mathrm{1}\right). \\ $$$$ \\ $$

Question Number 108095    Answers: 3   Comments: 0

((∞BeMath∞)/♠) ∫ 2x cos^(−1) (x) dx

$$\:\:\:\frac{\infty\mathcal{B}{e}\mathcal{M}{ath}\infty}{\spadesuit} \\ $$$$\:\:\:\int\:\mathrm{2}{x}\:\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\:{dx}\: \\ $$

Question Number 108085    Answers: 2   Comments: 1

((⋓BeMath⋓)/∞) sin^8 75°−cos^8 75° =

$$\:\:\:\:\:\frac{\Cup\mathcal{B}{e}\mathcal{M}{ath}\Cup}{\infty} \\ $$$$\:\:\:\mathrm{sin}\:^{\mathrm{8}} \mathrm{75}°−\mathrm{cos}\:^{\mathrm{8}} \mathrm{75}°\:= \\ $$

Question Number 108082    Answers: 1   Comments: 0

Question Number 108076    Answers: 0   Comments: 0

can someone please show how to get ∫_0 ^π sin (a sin (x)) dx=πH_0 (a) where H_0 (a) is the Struve−H−Function

$$\mathrm{can}\:\mathrm{someone}\:\mathrm{please}\:\mathrm{show}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get} \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{sin}\:\left({a}\:\mathrm{sin}\:\left({x}\right)\right)\:{dx}=\pi\mathrm{H}_{\mathrm{0}} \:\left({a}\right) \\ $$$$\mathrm{where}\:\mathrm{H}_{\mathrm{0}} \:\left({a}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{Struve}−\mathrm{H}−\mathrm{Function} \\ $$

Question Number 108073    Answers: 3   Comments: 0

((≜BeMath≜)/≺) ∫ ((x dx)/(x^8 −1)) ?

$$\:\:\:\:\frac{\triangleq\mathcal{B}{e}\mathcal{M}{ath}\triangleq}{\prec} \\ $$$$\:\:\int\:\frac{{x}\:{dx}}{{x}^{\mathrm{8}} −\mathrm{1}}\:? \\ $$

Question Number 108071    Answers: 1   Comments: 0

Question Number 108068    Answers: 4   Comments: 0

Question Number 108067    Answers: 2   Comments: 0

Question Number 108059    Answers: 1   Comments: 0

((♥JS♥)/(°js°)) Given a matrix A= ((( 3 2)),((−5 −4)) ) and A^2 +♭A−2I=0 where ♭ is a constant , I= (((1 0)),((0 1)) ). If B = (((−3♭ 2)),(( 5♭ −1)) ) , then A^(−1) B =

$$\:\:\:\frac{\heartsuit{JS}\heartsuit}{°{js}°} \\ $$$${Given}\:{a}\:{matrix}\:{A}=\begin{pmatrix}{\:\:\:\mathrm{3}\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{5}\:\:\:−\mathrm{4}}\end{pmatrix} \\ $$$${and}\:{A}^{\mathrm{2}} +\flat{A}−\mathrm{2}{I}=\mathrm{0}\:{where}\:\flat\:{is}\:{a} \\ $$$${constant}\:,\:{I}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}.\:{If}\:{B}\:= \\ $$$$\begin{pmatrix}{−\mathrm{3}\flat\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\mathrm{5}\flat\:\:\:\:−\mathrm{1}}\end{pmatrix}\:,\:{then}\:{A}^{−\mathrm{1}} {B}\:=\: \\ $$

Question Number 108053    Answers: 2   Comments: 0

((⋏J S⋏)/^⇉ ) ((y/x)+(√((x^2 +y^2 )/x^2 )))dx=dy

$$\:\:\:\:\:\:\:\:\:\:\:\frac{\curlywedge\mathcal{J}\:\mathbb{S}\curlywedge}{\:^{\rightrightarrows} } \\ $$$$\:\:\:\:\:\left(\frac{{y}}{{x}}+\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\right){dx}={dy} \\ $$

Question Number 108051    Answers: 1   Comments: 0

Question Number 108047    Answers: 1   Comments: 0

((°•BeMath•°)/Σ) y−x (dy/dx) = a (1+x^2 (dy/dx))

$$\:\:\:\:\:\:\frac{°\bullet\mathcal{B}{e}\mathcal{M}{ath}\bullet°}{\Sigma} \\ $$$$\:\:\:{y}−{x}\:\frac{{dy}}{{dx}}\:=\:{a}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\right)\: \\ $$

Question Number 108043    Answers: 2   Comments: 0

Given f(x)=x(x+1)(x+2)...(x+n) find the value of f′(0).

$$\mathrm{Given}\:\mathrm{f}\left({x}\right)={x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)...\left({x}+{n}\right) \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}'\left(\mathrm{0}\right). \\ $$

Question Number 108042    Answers: 0   Comments: 0

A particle in an electric and magnetic field is in motion. The time equations are in polar coordinates. r=r_0 e^(−(t/b)) and θ=(t/b) and b are positive constants. 1\Calculate the vector equation of the velocity of the particle. 2\Show that the angle (v_1 ^′ ,u_0 ′) is constant, and find the value. 3\Find the vector of acceleration of the particle. 4\Show that the angle (v_1 ^→ ,u_n ′) is constant, and find it. 5\Calculate the radius of this trajectory.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{in}\:\mathrm{an}\:\mathrm{electric}\:\mathrm{and}\:\mathrm{magnetic}\:\mathrm{field}\:\mathrm{is}\:\mathrm{in}\:\mathrm{motion}. \\ $$$$\mathrm{The}\:\mathrm{time}\:\mathrm{equations}\:\mathrm{are}\:\mathrm{in}\:\mathrm{polar}\:\mathrm{coordinates}. \\ $$$$\mathrm{r}=\mathrm{r}_{\mathrm{0}} \mathrm{e}^{−\frac{\mathrm{t}}{\mathrm{b}}} \:\mathrm{and}\:\theta=\frac{\mathrm{t}}{\mathrm{b}}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{constants}. \\ $$$$\mathrm{1}\backslash\mathrm{Calculate}\:\mathrm{the}\:\mathrm{vector}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}. \\ $$$$\mathrm{2}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{angle}\:\left(\mathrm{v}_{\mathrm{1}} ^{'} ,\mathrm{u}_{\mathrm{0}} '\right)\:\mathrm{is}\:\mathrm{constant},\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}. \\ $$$$\mathrm{3}\backslash\mathrm{Find}\:\mathrm{the}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}. \\ $$$$\mathrm{4}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{angle}\:\left(\overset{\rightarrow} {\mathrm{v}}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{n}} '\right)\:\mathrm{is}\:\mathrm{constant},\:\mathrm{and}\:\mathrm{find}\:\mathrm{it}. \\ $$$$\mathrm{5}\backslash\mathrm{Calculate}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{this}\:\mathrm{trajectory}. \\ $$

Question Number 108094    Answers: 1   Comments: 0

(d^2 Ψ/dt^2 )+(dΨ/dt)+Ψ=0

$$\frac{{d}^{\mathrm{2}} \Psi}{{dt}^{\mathrm{2}} }+\frac{{d}\Psi}{{dt}}+\Psi=\mathrm{0} \\ $$

Question Number 108034    Answers: 1   Comments: 0

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