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

Ffactor

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

Camau Ateb

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

Ffactora allan \frac{1}{2}.

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

Enrhifo

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

Graff

Cwis

Trigonometry

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

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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}} ...

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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 ...

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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 ...

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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 ...

Rhannu

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

Ffactora allan \frac{1}{2}.

Enghreifftiau

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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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