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Question Number 97368 by Riad last updated on 07/Jun/20

Commented by Tony Lin last updated on 07/Jun/20

$$\left(1\right)\left(a\right)\left(i\right)$$$$\int x^2{tan}^{-1}xdx$$$$=\frac{x^3}{3}{tan}^{-1}x-\int \frac{x^3}{3(1+x^2)}dx$$$$=\frac{x^3}{3}{tan}^{-1}x-\frac{1}{3}\int (x-\frac{x}{1+x^2})dx$$$$=\frac{x^3}{3}{tan}^{-1}x-\frac{x^2}{6}+\frac{1}{6}ln(x^2+1)+c$$

Commented by Riad last updated on 07/Jun/20

$$thanks$$

Commented by Tony Lin last updated on 07/Jun/20

$$\left(1\right)\left(a\right)\left(ii\right)$$$${\int }_0^1\frac{dx}{(1+x^2)\sqrt{1-x^2}}$$$$firstconsider$$$$\int \frac{dx}{(1+x^2)\sqrt{1-x^2}}$$$$letx=sin\theta ,dx=cos\theta d\theta$$$$\int \frac{d\theta }{1+{sin}^2\theta }$$$$=\int \frac{d\theta }{1+\frac{1-cos2\theta }{2}}$$$$=2\int \frac{d\theta }{3-cos2\theta }$$$$lettan\theta =t$$$$cos2\theta =\frac{1-t^2}{1+t^2},d\theta =\frac{1}{1+t^2}dt$$$$\int \frac{1}{2t^2+1}dt$$$$=\int \frac{1}{{\left(\sqrt{2}t\right)}^2+1}dt$$$$=\frac{1}{\sqrt{2}}{tan}^{-1}\left(\sqrt{2}tan\theta \right)+c$$$$=\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{\sqrt{2}x}{\sqrt{1-x^2}}\right)+c$$$$\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{\sqrt{2}x}{\sqrt{1-x^2}}\right){\mid }_0^1$$$$=\frac{1}{\sqrt{2}}(\underset{x\to \mathrm{\infty }}{\lim }{tan}^{-1}x-{tan}^{-1}0)$$$$=\frac{1}{\sqrt{2}}(\frac{\pi }{2}-0)$$$$=\frac{\pi \sqrt{2}}{4}$$

Commented by Tony Lin last updated on 07/Jun/20

$$\left(2\right)\left(a\right)$$$$u_x=\frac{y^2}{1+{xy}^2}$$$$u_y=\frac{2xy}{1+{xy}^2}$$$$u_{xx}=\frac{-y^2}{{(1+{xy}^2)}^2}$$$$u_{yy}=\frac{2x-2x^2{y}^2}{{(1+{xy}^2)}^2}$$$$u_{xy}=\frac{2y-2y^2{x}^2}{{(1+y^{2}x)}^2}$$$$2u_{xx}+u_{yy}+y^3{u}_{xy}$$$$=-2\left[\frac{y^2-x+x^2{y}^2-y^4+y^4{x}^2}{{(1+y^{2}x)}^2}\right]$$

Commented by Tony Lin last updated on 07/Jun/20

$$\left(1\right)\left(a\right)\left(iv\right)$$$${\int }_0^1{xe}^{-3x}dx$$$$=-\frac{1}{3}{xe}^{-3x}{\mid }_0^1+{\int }_0^1\frac{e^{-3x}}{3}dx$$$$=(-\frac{1}{3}{xe}^{-3x}-\frac{e^{-3x}}{9}){\mid }_0^1$$$$=\frac{1-4e^{-3}}{9}$$

Commented by Tony Lin last updated on 07/Jun/20

$$\left(1\right)\left(a\right)\left(iii\right)$$$$\int \sqrt{\frac{a+x}{a-x}}dx$$$$=\int \frac{a+x}{\sqrt{a^2-x^2}}dx$$$$letx=asin\theta ,dx=acos\theta d\theta$$$$a\int (1+sin\theta )d\theta$$$$=a\theta -acos\theta +c$$$$={asin}^{-1}\frac{x}{a}-\sqrt{a^2-x^2}+c$$

Commented by Tony Lin last updated on 07/Jun/20

$$\left(2\right)\left(b\right)$$$$f\left(x\right)=x^3-4x^2+5x-3$$$$f{}^{\prime }\left(x\right)=3x^2-8x+5=0$$$$(3x-5)(x-1)=0$$$$x=\frac{5}{3}orx=1$$$$f{}^{″}\left(x\right)=6x-8$$$$f{}^{″}\left(\frac{5}{3}\right)=2>0\to min$$$$f{}^{″}\left(1\right)=-2<0\to Max$$$$Therefore,$$$$f\left(\frac{5}{3}\right)=-\frac{31}{27}\to min$$$$f\left(1\right)=-1\to Max$$

Commented by Tony Lin last updated on 07/Jun/20

$$\left(2\right)\left(c\right)$$$${\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}{\int }_0^{sinx}{e}^{y}cosxdydx$$$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}(e^{sinx}-1)cosxdx$$$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}{e}^{sinx}\cdot cosx-{\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}cosxdx$$$$=e^{sin\frac{\pi }{4}}-e^{sin\frac{\pi }{6}}-sin\frac{\pi }{4}+sin\frac{\pi }{6}$$$$=e^{\frac{\sqrt{2}}{2}}-e^{\frac{1}{2}}-\frac{\sqrt{2}}{2}+\frac{1}{2}$$

Commented by Tony Lin last updated on 07/Jun/20

$$\left(3\right){\int }_0^{\frac{\pi }{8}}{\int }_0^2\sqrt{4-r^2}rdrd\theta$$$$=\frac{\pi }{8}{\int }_0^{2}r\sqrt{4-r^2}dr$$$$let4-r^2=u,dr=\frac{du}{-2r}$$$$\frac{1}{2}\times \frac{\pi }{8}{\int }_0^4\sqrt{u}du$$$$=\frac{\pi }{16}\times \frac{2}{3}{u}^{\frac{3}{2}}{\mid }_0^4$$$$=\frac{\pi }{3}$$

Commented by Tony Lin last updated on 07/Jun/20

$$Allaredone!$$

Commented by Riad last updated on 07/Jun/20

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}.\mathrm{love}\mathrm{you}$$

Answered by mathmax by abdo last updated on 07/Jun/20

$$3)\mathrm{I}=\int {\int }_{\mathrm{D}}\sqrt{4-{\mathrm{r}}^2}\mathrm{rdrd}\theta ={\int }_0^2\mathrm{r}\sqrt{4-{\mathrm{r}}^2}\mathrm{dr}{\int }_0^{\frac{\pi }{8}}\mathrm{d}\theta$$$$=\frac{\pi }{8}{\int }_0^2\mathrm{r}\sqrt{4-{\mathrm{r}}^2}\mathrm{dr}=\frac{\pi }{8}{\int }_0^2\mathrm{r}{(4-{\mathrm{r}}^2)}^{\frac{1}{2}}\mathrm{dr}=\frac{\pi }{8}{[-\frac{1}{3}{(4-{\mathrm{r}}^2)}^{\frac{3}{2}}]}_0^2$$$$=-\frac{\pi }{24}\{-4^{\frac{3}{2}}\}=\frac{\pi }{24}\times 8=\frac{\pi }{3}$$$$$$

Answered by mathmax by abdo last updated on 07/Jun/20

$$\mathrm{I}={\int }_0^1\frac{\mathrm{dx}}{(1+{\mathrm{x}}^2)\sqrt{1-{\mathrm{x}}^2}}\mathrm{changement}\mathrm{x}=\mathrm{sint}\mathrm{give}$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cost}}{(1+{\sin }^2\mathrm{t})\mathrm{cost}}\mathrm{dt}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{1+{\sin }^2\mathrm{t}}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{2-{\cos }^2\mathrm{t}}$$$$={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{2-\frac{1}{1+{\tan }^2\mathrm{t}}}={\int }_0^{\frac{\pi }{2}}\frac{1+{\tan }^2\mathrm{t}}{2+{2\tan }^2\mathrm{t}-1}\mathrm{dt}={\int }_0^{\frac{\pi }{2}}\frac{1+{\tan }^2\mathrm{t}}{1+{2\tan }^2\mathrm{t}}\mathrm{dt}$$$${=}_{\mathrm{tant}=\mathrm{u}}{\int }_0^{\mathrm{\infty }}\frac{1+{\mathrm{u}}^2}{1+{2\mathrm{u}}^2}\frac{\mathrm{du}}{1+{\mathrm{u}}^2}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{du}}{1+{2\mathrm{u}}^2}{=}_{\sqrt{2}\mathrm{u}=\mathrm{z}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dz}}{\sqrt{2}(1+{\mathrm{z}}^2)}$$$$=\frac{1}{\sqrt{2}}\times \frac{\pi }{2}=\frac{\pi }{2\sqrt{2}}$$

Answered by mathmax by abdo last updated on 07/Jun/20

$$2)\mathrm{u}(\mathrm{x},\mathrm{y})=\ln (1+{\mathrm{xy}}^2)\Rightarrow \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{{\mathrm{y}}^2}{1+{\mathrm{xy}}^2}\Rightarrow \frac{{\partial }^2}{\partial {\mathrm{x}}^2}\mathrm{u}={\mathrm{y}}^2(-\frac{{\mathrm{y}}^2}{{(1+{\mathrm{xy}}^2)}^2})$$$$=-\frac{{\mathrm{y}}^4}{{(1+{\mathrm{xy}}^2)}^2}$$$$\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\frac{2\mathrm{yx}}{1+{\mathrm{xy}}^2}\Rightarrow \frac{{\partial }^2\mathrm{u}}{{\partial }^2\mathrm{y}}=2\mathrm{x}\left(\frac{1+{\mathrm{xy}}^2-\mathrm{y}\left(2\mathrm{xy}\right)}{{(1+{\mathrm{xy}}^2)}^2}\right)=2\mathrm{x}\frac{1-{\mathrm{xy}}^2}{{(1+{\mathrm{xy}}^2)}^2}$$$$\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\frac{2\mathrm{yx}}{1+{\mathrm{xy}}^2}\Rightarrow \frac{\partial }{\partial \mathrm{x}}\left(\frac{\partial \mathrm{u}}{\partial \mathrm{y}}\right)=2\mathrm{y}\left(\frac{1+{\mathrm{xy}}^2-{\mathrm{xy}}^2}{{(1+{\mathrm{xy}}^2)}^2}\right)=\frac{2\mathrm{y}}{{(1+{\mathrm{xy}}^2)}^2}$$$${2\mathrm{u}}_{\mathrm{xx}}+{\mathrm{u}}_{\mathrm{yy}}+{\mathrm{y}}^3{\mathrm{u}}_{\mathrm{xy}}=-\frac{{2\mathrm{y}}^4}{{(1+{\mathrm{xy}}^2)}^2}+\frac{2\mathrm{x}-{2\mathrm{x}}^2{\mathrm{y}}^2}{{(1+{\mathrm{xy}}^2)}^2}+\frac{{2\mathrm{y}}^4}{{(1+{\mathrm{xy}}^2)}^2}$$$$=\frac{2\mathrm{x}-{2\mathrm{x}}^2{\mathrm{y}}^2}{{(1+{\mathrm{xy}}^2)}^2}$$

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