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Question Number 174194 by mokys last updated on 26/Jul/22

Answered by Mathspace last updated on 27/Jul/22

$$1){\int }_0^{\frac{\pi }{2}}{sin}^{10}x{cos}^{10}xdx$$$$wekmow{\int }_0^{\frac{\pi }{2}}{sin}^{2p-1}x{cos}^{2q-1}xdx$$$$=\frac{1}{2}B(p,q)=\frac{\mathrm{\Gamma }\left(p\right).\mathrm{\Gamma }\left(q\right)}{\mathrm{\Gamma }(p+q)}$$$$wegetp=\frac{11}{2}andq=\frac{11}{2}$$$$\Rightarrow {\int }_0^{\frac{\pi }{2}}{sin}^{10}x{cos}^{10}xdx$$$$=\frac{1}{2}\times \frac{{\mathrm{\Gamma }}^2\left(\frac{11}{2}\right)}{\mathrm{\Gamma }\left(11\right)}=\frac{1}{2.10!}{\mathrm{\Gamma }}^2\left(\frac{11}{2}\right)$$$$resttocalculate\mathrm{\Gamma }\left(\frac{11}{2}\right)$$$$...$$

Commented by aleks041103 last updated on 27/Jul/22

$$I(p,q)={\int }_0^{\pi /2}{sin}^{2p-1}x{cos}^{2q-1}xdx$$$$t={sin}^{2}x$$$$dt=2sinxcosxdx$$$$I(p,q)=\frac{1}{2}{\int }_0^{\pi /2}{\left({sin}^{2}x\right)}^{p-1}{\left({cos}^{2}x\right)}^{q-1}\left(2sinxcosxdx\right)=$$$$=\frac{1}{2}{\int }_0^1{t}^{p-1}{(1-t)}^{q-1}dt=\frac{1}{2}B(p,q)$$$$\Rightarrow {\int }_0^{\pi /2}{sin}^{\alpha }x{cos}^{\beta }xdx=\frac{1}{2}B(\frac{1+\alpha }{2},\frac{1+\beta }{2})$$$$\Rightarrow {\int }_0^{\pi /2}{sin}^{\alpha }x{cos}^{\beta }xdx=\frac{\mathrm{\Gamma }\left(\frac{1+\alpha }{2}\right)\mathrm{\Gamma }\left(\frac{1+\beta }{2}\right)}{2\mathrm{\Gamma }(1+\frac{\alpha +\beta }{2})}$$$$\Rightarrow {\int }_0^{\pi /2}{sin}^{10}x{cos}^{10}xdx=\frac{\mathrm{\Gamma }{\left(\frac{11}{2}\right)}^2}{2\mathrm{\Gamma }\left(11\right)}=\frac{{(\left(9/2\right)!)}^2}{2.10!}$$$$\left(\frac{9}{2}\right)!=\frac{9}{2}\left(\frac{7}{2}\right)!=...=\frac{9.7.5.3}{2.2.2.2}\left(\frac{1}{2}\right)!=\frac{9.7.5.3}{32}\sqrt{\pi }$$$$\Rightarrow {(\left(9/2\right)!)}^2/(2.10!)=\frac{893025}{2048.3628800}\pi =\frac{893025}{7431782400}\pi =$$$$=\frac{35721\pi }{297271296}=\frac{441\pi }{3670016}=\frac{63\pi }{524288}$$$$\Rightarrow {\int }_0^{\pi /2}{sin}^{\alpha }x{cos}^{\beta }xdx=\frac{63\pi }{524288}$$$$$$$$$$

Answered by Mathspace last updated on 27/Jul/22

$$3){\int }_{-5}^1\left[x\right]dx{=}_{x=-5+t}{\int }_0^6[t-5]dt$$$$={\int }_0^6(\left[t\right]-5)dt$$$$={\int }_0^6\left[t\right]dt-30$$$$=\sum _{k=0}^5{\int }_k^{k+1}kdt-30$$$$=\sum _{k=0}^{5}k-30$$$$=\frac{5.6}{2}-30=15-30=-15$$

Answered by Mathspace last updated on 27/Jul/22

$$4){\int }_{-2}^2\left[x\right]dx{=}_{x=-2+t}{\int }_0^4[-2+t]dt$$$$={\int }_0^4(-2+\left[t\right])dt$$$$=-8+{\int }_0^4\left[t\right]dt$$$$=-8+\sum _{k=0}^3{\int }_k^{k+1}kdt$$$$=-8+\sum _{k=1}^{3}k$$$$=-8+\frac{3.4}{2}=-8+6=2$$

Answered by Mathspace last updated on 27/Jul/22

$$2){\int }_{-3}^3\left\{x\right\}dx={\int }_{-3}^3(x-\left[x\right])dx$$$$={\int }_{-3}^{3}xdx(\to 0)-{\int }_{-3}^3\left[x\right]dx$$$${=}_{x=-3+t}-{\int }_o^6(-3+\left[t\right])dt$$$$=18-\sum _{k=0}^5{\int }_k^{k+1}kdt$$$$=18-\sum _{k=1}^{5}k$$$$=18-\frac{5.6}{2}=18-15=3$$

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