[frac{xsqrt{4-x^2}}{2}+4/2sin^-1x/2]_{sqrt{3}}^2 eching

Omil

\frac{4\arcsin(\frac{x}{2})+x\sqrt{4-x^{2}}}{2}

Yechim qadamlari

[ \frac { x \sqrt { 4 - x ^ { 2 } } } { 2 } + \frac { 4 } { 2 } \sin ^ { - 1 } \frac { x } { 2 } ] _ { \sqrt { 3 } } ^ { 2 }

\frac{1}{2} omili.

\frac{x\sqrt{4-x^{2}}+4\arcsin(\frac{x}{2})}{2}

Baholash

\frac{4\arcsin(\frac{x}{2})+x\sqrt{4-x^{2}}}{2}

Grafik

Viktorina

Trigonometry

5xshash muammolar:

[ \frac { x \sqrt { 4 - x ^ { 2 } } } { 2 } + \frac { 4 } { 2 } \sin ^ { - 1 } \frac { x } { 2 } ] _ { \sqrt { 3 } } ^ { 2 }

Veb-qidiruvdagi o'xshash muammolar

Evaluate \int{\sqrt{a^2 - x^2}}dx

https://math.stackexchange.com/q/1789216

\cos(\pi/2-x)=\sin(x) and vice versa. If you haven't seen this before, the geometric explanation is that the sine of one of the acute angles in a right triangle is the cosine of the other. Therefore ...

Step by step solution to these derivations [closed]

https://math.stackexchange.com/questions/1551608/step-by-step-solution-to-these-derivations

HINT: you Need the power and the chain rule 2\,{1\arctan \left( 3\,{\frac {x}{2\,{x}^{2}-8}} \right) \left( 3\, \left( 2\,{x}^{2}-8 \right) ^{-1}-12\,{\frac {{x}^{2}}{ \left( 2\,{x} ^{2}-8 \right) ^{2}}} \right) \left( 9\,{\frac {{x}^{2}}{ \left( 2\,{ x}^{2}-8 \right) ^{2}}}+1 \right) ^{-1}} ...

How to \int \sqrt{10x-x^2}dx

https://math.stackexchange.com/q/136175

What you "did wrong" in making the subsitution was forget to take into account the \frac{dx}{dt} = 5\cos t. Continuing, \begin{align} \int \sqrt{10x-x^2}dx & = \int \sqrt{10(5 + 5\sin{t}) + (5+5\sin{t})^2}5\cos{t}dt\\ & = 5\int \sqrt{25 - 25\sin^2t}\cos{t}dt\\ & = 25\int \cos^2tdt\\ & = \frac{25}{2}\sin{t}\cos{t} + \frac{25}{2}t\\ & = \frac{25}{2}\frac{(x-5)}{5}\sqrt{1-(\frac{x-5}{5})^2} + \frac{25}{2}\sin^{-1}(\frac{x-5}{5})\\ & = \frac{x-5}{2}\sqrt{25 - (x-5)^2} + \frac{25}{2}\sin^{-1}(\frac{x-5}{5})\\ & = \frac{x-5}{2}\sqrt{10x-x^2} + \frac{25}{2}\sin^{-1}(\frac{x-5}{5}). \end{align} ...

Integrating \int \frac{\sin^{-1}(x)}{\sqrt{1+x}}dx by parts

https://math.stackexchange.com/q/1357901

First, you wrote -\int\frac{\sqrt{1+x}}{\sqrt{(1+x)(1-x)}}dx\tag 1 where (1) should have been -2\int\frac{\sqrt{1+x}}{\sqrt{(1+x)(1-x)}}dx \tag 2 This was likely a typographical error ...

Computing the inverse Jacobi function \mathrm{arccd} with elliptic integrals

https://math.stackexchange.com/questions/451708/computing-the-inverse-jacobi-function-mathrmarccd-with-elliptic-integrals

As you can see from your formulas, \mathrm{arccd}(x,k) is defined in [1] as an even function of x. This is not quite compatible with being the inverse function of \mathrm{cd}(x,k), which is also ...

If \frac{df}{dx} is continous on [a,b] is f continous on [a,b] as well?

https://math.stackexchange.com/questions/2032762/if-fracdfdx-is-continous-on-a-b-is-f-continous-on-a-b-as-well

On open sets it is true that if f is differentiable then it is continuous. Hence if we replace [a,b] with ]a,b[ in your original question everything works. However, for closed intervals of the ...

Baham ko'rish

Klipbordga nusxa olish

\frac{x\sqrt{4-x^{2}}+4\arcsin(\frac{x}{2})}{2}

\frac{1}{2} omili.

Misollar

Ikkilik tenglama

{ x } ^ { 2 } - 4 x - 5 = 0

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4 \sin \theta \cos \theta = 2 \sin \theta

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

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\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

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\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

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\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

Mavzular

Mavzular

Veb-qidiruvdagi o'xshash muammolar

Baham ko'rish

Misollar