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

Tauwehe

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

Ngā Upane Otinga

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

Tauwehea te \frac{1}{2}.

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

Aromātai

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

Graph

Pātaitai

Trigonometry

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

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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 ...

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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}} ...

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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 ...

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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 ...

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Kua tāruatia ki te papatopenga

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

Tauwehea te \frac{1}{2}.

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