Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 131247 by bramlexs22 last updated on 03/Feb/21

$$\left(1\right)\psi =\int \frac{dx}{1-\sin x\cos x}=?$$$$\left(2\right){\int }_0^{\mathrm{\infty }}\frac{x}{x^3+1}dx=?$$$$\left(3\right){\int }_0^{\mathrm{\infty }}\frac{1}{x^{3/2}+1}dx=?$$

Answered by liberty last updated on 03/Feb/21

$$\mathrm{Evaluate}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{\mathrm{x}}^3+1}\mathrm{dx}.$$$$\left(2\right)\mathrm{L}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{\mathrm{x}}^3+1}\mathrm{dx};\mathrm{replace}\mathrm{x}\mathrm{with}\frac{1}{\mathrm{x}}$$$$\mathrm{L}={\int }_0^{\mathrm{\infty }}\frac{\frac{1}{\mathrm{x}}}{\frac{1}{{\mathrm{x}}^3}+1}.(-\frac{\mathrm{dx}}{{\mathrm{x}}^2})={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^3}$$$$\mathrm{now}\mathrm{we}\mathrm{find}\mathrm{that}:2\mathrm{L}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}+1}{{\mathrm{x}}^3+1}\mathrm{dx}$$$$2\mathrm{L}={\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{x}}^2-\mathrm{x}+1}\mathrm{dx};\mathrm{L}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{x}-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$\mathrm{L}=\frac{1}{2}.\frac{2}{\sqrt{3}}{\left[{\tan }^{-1}\left(\frac{\mathrm{x}-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]}_0^{\mathrm{\infty }}=\frac{1}{\sqrt{3}}{\left[{\tan }^{-1}\left(\frac{2\mathrm{x}-1}{\sqrt{3}}\right)\right]}_0^{\mathrm{\infty }}$$$$\mathrm{L}=\frac{1}{\sqrt{3}}(\frac{\pi }{2}-(-\frac{\pi }{6}))=\frac{2\pi \sqrt{3}}{9}.$$$$$$

Answered by liberty last updated on 03/Feb/21

$$\mathrm{Solve}\psi =\int \frac{\mathrm{dx}}{1-\sin \mathrm{x}\cos \mathrm{x}}$$$$\left(1\right)\psi =\int \frac{\mathrm{dx}}{1-\sin \mathrm{x}\cos \mathrm{x}}=\int \frac{\sec {}^2\mathrm{x}}{\sec {}^2\mathrm{x}-\tan \mathrm{x}}\mathrm{dx}$$$$\psi =\int \frac{\mathrm{d}\left(\tan \mathrm{x}\right)}{\tan {}^2\mathrm{x}-\tan \mathrm{x}+1};[\mathrm{change}\mathrm{variable}\mathrm{t}=\tan \mathrm{x}]$$$$\psi =\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-\mathrm{t}+1}=\int \frac{\mathrm{dt}}{{(\mathrm{t}-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$\psi =\frac{2}{\sqrt{3}}\arctan \left(\frac{\mathrm{t}-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+\mathrm{c}$$$$\psi =\frac{2}{\sqrt{3}}\arctan \left(\frac{2\mathrm{t}-1}{\sqrt{3}}\right)+\mathrm{c}=\frac{2}{\sqrt{3}}\arctan \left(\frac{2\tan \mathrm{x}-1}{\sqrt{3}}\right)+\mathrm{c}$$$$\psi =\frac{2\arctan \left(\frac{2\tan \mathrm{x}-1}{\sqrt{3}}\right)}{\sqrt{3}}+\mathrm{c}$$

Commented by malwan last updated on 03/Feb/21

$$\int \frac{dx}{1-sinxcosx}=2\int \frac{dx}{2-sin2x}$$$$y=tanx\Rightarrow dy={sec}^{2}xdx$$$$andwegetthesameresult$$$$thankyoumrliberty$$

Answered by liberty last updated on 03/Feb/21

$$\left(3\right)\mathrm{we}\mathrm{want}\mathrm{to}\mathrm{evaluate}\eta ={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^{3/2}+1}.$$$$\mathrm{by}\mathrm{rationalizing}\mathrm{replace}{\mathrm{x}}^{1/2}\mathrm{by}\mathrm{t}\mathrm{or}\mathrm{x}={\mathrm{t}}^2$$$$\eta =2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}}{{\mathrm{t}}^3+1}\mathrm{dt};\mathrm{where}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{\mathrm{x}}^3+1}\mathrm{dx}=\frac{2\pi \sqrt{3}}{9}$$$$\left(\mathrm{see}\mathrm{the}\mathrm{solution}\mathrm{in}\mathrm{part}\left(2\right)\right)$$$$\mathrm{Hence}\mathrm{we}\mathrm{find}\eta =\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\mathrm{dx}}{{\mathrm{x}}^{3/2}+1}=2\times \left(\frac{2\pi \sqrt{3}}{9}\right)=\frac{4\pi \sqrt{3}}{9}$$

Answered by Ar Brandon last updated on 03/Feb/21

$$2.\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{\mathrm{x}}^3+1}\mathrm{dx},{\mathrm{x}}^3=\mathrm{t}\Rightarrow {3\mathrm{x}}^2\mathrm{dx}=\mathrm{dt}$$$$=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{-\frac{1}{3}}}{1+\mathrm{t}}\mathrm{dt}=\frac{\beta (\frac{2}{3},\frac{1}{3})}{3}$$$$=\frac{\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{1}{3}\right)}{3}=\frac{\pi }{3\sin \left(\pi /3\right)}=\frac{2\pi }{3\sqrt{3}}$$

Commented by Engr_Jidda last updated on 03/Feb/21

$$sirplshowdo\frac{\pi }{3sin\left(\frac{\pi }{3}\right)}=\frac{2\pi }{3\sqrt{3}}?$$

Commented by malwan last updated on 03/Feb/21

$$\frac{\pi }{3sin\left(\frac{\pi }{3}\right)}=\frac{\pi }{3\times \left(\frac{\sqrt{3}}{2}\right)}=\frac{2\pi }{3\sqrt{3}}$$

Commented by Engr_Jidda last updated on 04/Feb/21

$$ohyeah,thanks$$

Answered by Dwaipayan Shikari last updated on 03/Feb/21

$${\int }_0^{\mathrm{\infty }}\frac{1}{x^{\frac{3}{2}}+1}dx=\frac{2}{3}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{2}{3}-1}}{{(1+u)}^{\frac{4}{3}+\frac{2}{3}-1}}du{x}^{\frac{3}{2}}=u\Rightarrow \frac{3}{2}\sqrt{x}=\frac{du}{dx}$$$$=\frac{2}{3}.\frac{\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{4}{3}\right)}{\mathrm{\Gamma }\left(2\right)}=\frac{2}{9}.\frac{\pi }{sin\left(\frac{\pi }{3}\right)}=\frac{4\pi }{9\sqrt{3}}$$

Answered by mathmax by abdo last updated on 03/Feb/21

$$1)\mathrm{\Phi }=\int \frac{\mathrm{dx}}{1-\mathrm{sinx}\mathrm{cosx}}\Rightarrow \mathrm{\Phi }=\int \frac{\mathrm{dx}}{1-\frac{\sin \left(2\mathrm{x}\right)}{2}}=2\int \frac{\mathrm{dx}}{2-\sin \left(2\mathrm{x}\right)}$$$${=}_{2\mathrm{x}=\mathrm{t}}\int \frac{\mathrm{dt}}{2-\mathrm{sint}}{=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{y}}\int \frac{2\mathrm{dy}}{(1+{\mathrm{y}}^2)(2-\frac{2\mathrm{y}}{1+{\mathrm{y}}^2})}$$$$=\int \frac{2\mathrm{dy}}{2+{2\mathrm{y}}^2-2\mathrm{y}}=\int \frac{\mathrm{dy}}{{\mathrm{y}}^2-\mathrm{y}+1}=\int \frac{\mathrm{dy}}{{(\mathrm{y}-\frac{1}{2})}^2+\frac{3}{4}}$$$${=}_{\mathrm{y}-\frac{1}{2}=\frac{\sqrt{3}}{2}\mathrm{u}}\frac{4}{3}\int \frac{1}{{\mathrm{u}}^2+1}\frac{\sqrt{3}}{2}\mathrm{du}=\frac{1}{\sqrt{3}}\mathrm{arctanu}+\mathrm{C}$$$$=\frac{1}{\sqrt{3}}\arctan \left(\frac{2\mathrm{y}-1}{\sqrt{3}}\right)+\mathrm{C}=\frac{1}{\sqrt{3}}\arctan (\frac{1}{\sqrt{3}}(2\tan \left(\mathrm{x}\right)-1)+\mathrm{C}$$

Answered by mathmax by abdo last updated on 03/Feb/21

$$2)\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{\mathrm{x}}^3+1}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{cha}7\mathrm{gement}{\mathrm{x}}^3=\mathrm{t}\Rightarrow$$$$\mathrm{I}=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{3}}}{1+\mathrm{t}}{\mathrm{t}}^{\frac{1}{3}-1}\mathrm{dt}=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{2}{3}-1}}{1+\mathrm{t}}\mathrm{dt}=\frac{1}{3}.\frac{\pi }{\sin \left(\frac{2\pi }{3}\right)}=\frac{\pi }{3.\frac{\sqrt{3}}{2}}\Rightarrow$$$$\mathrm{I}=\frac{2\pi }{3\sqrt{3}}$$

Answered by mathmax by abdo last updated on 03/Feb/21

$$3)\mathrm{J}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{\frac{3}{2}}}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{cha}7\mathrm{gement}{\mathrm{x}}^{\frac{3}{2}}=\mathrm{t}\Rightarrow \mathrm{x}={\mathrm{t}}^{\frac{2}{3}}\Rightarrow$$$$\mathrm{J}=\frac{2}{3}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{2}{3}-1}}{1+\mathrm{t}}\mathrm{dt}=\frac{2}{3}.\frac{\pi }{\sin \left(\frac{2\pi }{3}\right)}=\frac{2\pi }{3.\frac{\sqrt{3}}{2}}=\frac{4\pi }{3\sqrt{3}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com