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

Lahuta teguriteks

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

Lahenduse etapid

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

Tooge \frac{1}{2} sulgude ette.

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

Arvuta

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

Graafik

Viktoriin

Trigonometry

5 probleemid, mis on sarnased:

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

Sarnased probleemid veebiotsingust

What am I doing wrong finding the derivative of \frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}?

https://math.stackexchange.com/questions/3009739/what-am-i-doing-wrong-finding-the-derivative-of-frac3-x2-sqrt1-2x-x22

You got A' wrong. A = \frac{3-x}{2}\cdot(\sqrt{1-2x-x^2}) = f\cdot g where: f =\frac{3-x}{2} , g = \sqrt{1-2x-x^2} A'= f'g + g'f f'g = (-\frac{1}{2})(\sqrt{1-2x-x^2}) is correct, ...

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} ...

Calculate \lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})

https://math.stackexchange.com/questions/3057710/calculate-lim-limits-x-to-inftyx-cdot-sin-sqrtx23-sqrtx22

Letting h=\frac1x : \begin{array}{cl} &\displaystyle \lim_{x \to \infty} x \sin \left( \sqrt{x^2+3} - \sqrt{x^2+2} \right) \\ =&\displaystyle \lim_{x \to \infty} x \sin \left( \frac 1 {\sqrt{x^2+3} + \sqrt{x^2+2} } \right) \\ =&\displaystyle \lim_{h \to 0^+} \frac1h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\ =&\displaystyle \lim_{h \to 0^+} \frac 1 {\sqrt{1+3h^2} + \sqrt{1+2h^2}} \frac {\sqrt{1+3h^2} + \sqrt{1+2h^2}} h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\ =&\displaystyle \frac12 \times \lim_{h \to 0^+} \frac {\sqrt{1+3h^2} + \sqrt{1+2h^2}} h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\ =&\displaystyle \frac12 \times 1 \\ =&\dfrac12 \end{array}

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 ...

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 ...

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}} ...

Jagama

Lõikelauale kopeeritud

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

Tooge \frac{1}{2} sulgude ette.

Näited

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{ 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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Teemad

Teemad

Sarnased probleemid veebiotsingust

Jagama

Näited