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Question Number 196983 by universe last updated on 05/Sep/23

Commented by Frix last updated on 06/Sep/23

$$\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{solve}\mathrm{the}\mathrm{first}2\mathrm{but}\mathrm{wolframalpha}\mathrm{can}$$$$\left(\mathrm{C}\mathrm{is}\mathrm{the}\mathrm{Catalan}\mathrm{constant}\right)$$$$1.I_1=\underset{0}{\stackrel{\pi }{\int }}\frac{x^3}{...}dx=\frac{\pi (-1440\mathrm{C}+1144+15\pi (-41+20\pi +24\ln 2))}{3780}$$$$2.I_2=\underset{0}{\stackrel{\pi }{\int }}\frac{x^2}{...}dx=\frac{-1440\mathrm{C}+1144+15\pi (-41+30\pi +24\ln 2)}{5670}$$$$3.I_3=\underset{0}{\stackrel{\pi }{\int }}\frac{x}{...}dx=\frac{5\pi }{63}$$$$4.I_4=\underset{0}{\stackrel{\pi }{\int }}\frac{1}{...}dx=\frac{10}{63}$$

Answered by Blackpanther last updated on 05/Sep/23

$$7$$

Commented by Frix last updated on 05/Sep/23

$$\mathrm{I}\mathrm{also}\mathrm{get}\approx 7\mathrm{by}\mathrm{approximating}\mathrm{the}\mathrm{integral}$$

Commented by universe last updated on 05/Sep/23

$$sendyoursolutionsir$$

Commented by mokys last updated on 05/Sep/23

$$canyousolvebysteps$$

Answered by witcher3 last updated on 06/Sep/23

$${\int }_0^{\pi }\frac{{4\mathrm{x}}^3-6\pi {\mathrm{x}}^2+2\mathrm{x}+1}{{(\sin \left(\mathrm{x}\right)+1)}^5}\sin \left(\mathrm{x}\right)\mathrm{dx}=\mathrm{A}$$$$\mathrm{A}={\int }_0^{\pi }\frac{\sin \left(\mathrm{x}\right)}{{(\sin \left(\mathrm{x}\right)+1)}^5}(4{(\pi -\mathrm{x})}^3-6\pi {(\pi -\mathrm{x})}^2+2(\pi -\mathrm{x})+1))\mathrm{dx}$$$$2\mathrm{A}={\int }_0^{\pi }\frac{\mathrm{sim}\left(\mathrm{x}\right)(-2{\pi }^3+2\pi +2)}{{(\sin \left(\mathrm{x}\right)+1)}^5}$$$$\mathrm{A}=(-{\pi }^3+\pi +1){\int }_0^{\pi }\frac{\sin \left(\mathrm{x}\right)}{{(\sin \left(\mathrm{x}\right)+1)}^5}$$$${\int }_0^{\pi }\frac{\sin \left(\mathrm{x}\right)}{{(\sin \left(\mathrm{x}\right)+1)}^5}\mathrm{dx}=\mathrm{B}$$$${\int }_0^{\pi }\frac{2\mathrm{tg}\left(\frac{\mathrm{x}}{2}\right){(1+{\mathrm{tg}}^2\left(\frac{\mathrm{x}}{2}\right))}^4}{{(1+\mathrm{tg}\left(\frac{\mathrm{x}}{2}\right))}^{10}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{4\mathrm{x}{(1+{\mathrm{x}}^2)}^3}{{(1+\mathrm{x})}^{10}}\mathrm{dx}=4{\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{(1+\mathrm{x})}^{10}}+12{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^3}{{(1+\mathrm{x})}^{10}}+12{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^5}{{(1+\mathrm{x})}^{10}}$$$$+4{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^7}{{(1+\mathrm{x})}^{10}}$$$$‘‘\beta (\mathrm{x},\mathrm{y})={\int }_0^{\mathrm{\infty }}{\frac{{\mathrm{t}}^{\mathrm{x}-1}}{{(1+\mathrm{t})}^{\mathrm{x}+\mathrm{y}}}}^{″}$$$$\mathrm{B}=4\beta (2,8)+12\beta (4,6)+12\beta (6,4)+4\beta (8,2)$$$$=8\frac{\mathrm{\Gamma }\left(2\right)\mathrm{\Gamma }\left(8\right)}{\mathrm{\Gamma }\left(10\right)}+24\frac{\mathrm{\Gamma }\left(4\right)\mathrm{\Gamma }\left(6\right)}{\mathrm{\Gamma }\left(10\right)}$$$$=\frac{1}{9}+\frac{24.3.2}{9.8.7.6}=\frac{3}{63}+\frac{1}{9}=\frac{10}{63}$$$$\mathrm{A}=(-{\pi }^3+{\pi }^2+1)\mathrm{B}=\frac{10}{63}(-{\pi }^3+{\pi }^2+1)$$$$\frac{\mathrm{a}+3}{9\mathrm{a}}=\frac{10}{63},\Rightarrow \mathrm{a}=7$$$$$$$$$$

Commented by universe last updated on 06/Sep/23

$$thankyousir$$

Answered by lain_math last updated on 17/Sep/23

$$a=7$$$$:)$$

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