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Question Number 87581 by peter frank last updated on 05/Apr/20

Commented by TANMAY PANACEA. last updated on 05/Apr/20

$$i[shallsolveinpaper$$

Answered by TANMAY PANACEA. last updated on 05/Apr/20

$$1){\int }_0^1\frac{x^{2}dx}{\sqrt{1-x^4}}dx\times {\int }_0^1\frac{dx}{\sqrt{1+x^4}}$$$$I=I_1\times I_2$$$$I_1={\int }_0^1\frac{x^{2}dx}{\sqrt{1-x^4}}$$$$x^2=sina\to 2xdx=cosada$$$${\int }_0^{\frac{\pi }{2}}\frac{sina}{cosa}\times \frac{cosada}{2\sqrt{sina}}=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\left(sina\right)}^{2\times \frac{3}{4}-1}{\left(cosa\right)}^{2\times \frac{1}{2}-1}da$$$$usinggammafunction$$$$2{\int }_0^{\frac{\pi }{2}}{\left(sin\alpha \right)}^{2p-1}{\left(cos\alpha \right)}^{2q-1}d\alpha =\frac{\lceil \left(p\right)\lceil \left(q\right)}{\lceil (p+q)}$$$$I_1=\frac{1}{4}\times 2{\int }_0^{\frac{\pi }{2}}{\left(sina\right)}^{2\times \frac{3}{4}-1}{\left(cosa\right)}^{2\times \frac{1}{2}-1}da$$$$=\frac{1}{4}\times \frac{\lceil \left(\frac{3}{4}\right)\lceil \left(\frac{1}{2}\right)}{\lceil (\frac{3}{4}+\frac{1}{2})}$$$$I_2={\int }_0^1\frac{dx}{\sqrt{1+x^4}}$$$$x^2=tanb\to 2xdx={sec}^{2}bdb$$$$dx=\frac{{sec}^{2}b}{2\sqrt{tanb}}db$$$${\int }_0^{\frac{\pi }{4}}\frac{{sec}^{2}b}{2\sqrt{tanb}\times secb}db=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{db}{\sqrt{sinbcosb}}=\frac{1}{\sqrt{2}}{\int }_0^{\frac{\pi }{4}}\frac{db}{\sqrt{sin2b}}$$$$★★nowt=2bdt=2db$$$$I_2=\frac{1}{\sqrt{2}}{\int }_0^{\frac{\pi }{4}}\frac{db}{\sqrt{sin2b}}=\frac{1}{\sqrt{2}}{\int }_0^{\frac{\pi }{2}}\frac{dt}{2\sqrt{sint}}$$$$=\frac{1}{4\sqrt{2}}\times 2{\int }_0^{\frac{\pi }{2}}{\left(sint\right)}^{2\times \frac{1}{4}-1}{\left(cost\right)}^{2\times \frac{1}{2}-1}dt$$$$=\frac{1}{4\sqrt{2}}\times \frac{\lceil \left(\frac{1}{4}\right)\lceil \left(\frac{1}{2}\right)}{\lceil (\frac{1}{2}+\frac{1}{4})}=I_2$$$$I=I_1\times I_2$$$$=\frac{1}{4}\times \frac{\lceil \left(\frac{3}{4}\right)\lceil \left(\frac{1}{2}\right)}{\lceil \left(\frac{5}{4}\right)}\times \frac{1}{4\sqrt{2}}\times \frac{\lceil \left(\frac{1}{4}\right)\lceil \left(\frac{1}{2}\right)}{\lceil \left(\frac{3}{4}\right)}$$$$formula\lceil \left(\frac{1}{2}\right)=\sqrt{\pi }$$$$\lceil \left(p\right)\lceil (1-p)=\frac{p}{sinp\pi }$$$$\mathit{I}=\frac{1}{16\sqrt{2}}\times \pi \times \frac{\lceil \left(\frac{1}{4}\right)}{\lceil (1+\frac{1}{4})}=\frac{\pi }{16\sqrt{2}}\times \frac{\lceil \left(\frac{1}{4}\right)}{\frac{1}{4}\lceil \left(\frac{1}{4}\right)}$$$$=\frac{\pi }{4\sqrt{2}}proved$$$$$$$$note\lceil (n+1)=n\lceil \left(n\right)$$$$so\lceil \left(\frac{5}{4}\right)$$$$=\lceil (1+\frac{1}{4})=\frac{1}{4}\lceil \left(\frac{1}{4}\right)$$$$$$$$$$

Answered by TANMAY PANACEA. last updated on 05/Apr/20

$$2){\int }_0^{\frac{\pi }{2}}({tan}^2\theta +{tan}^5\theta )e^{-{tan}^2\theta }d\theta$$$$t={tan}^2\theta$$$$\frac{dt}{d\theta }=2tan\theta \times {sec}^2\theta =2tan\theta (1+{tan}^2\theta )=2\sqrt{t}(1+t)$$$${\int }_0^{\mathrm{\infty }}(t+t^{\frac{5}{2}})e^{-t}\times 2(\sqrt{t}+t^{\frac{3}{2}})dt$$$$I=2{\int }_0^{\mathrm{\infty }}{e}^{-t}(t^{\frac{3}{2}}+t^{\frac{5}{2}}+t^3+t^4)dt$$$$formula\lceil \left(n\right)={\int }_0^{\mathrm{\infty }}{e}^{-x}\times x^{n-1}dx$$$$\frac{I}{2}={\int }_0^{\mathrm{\infty }}{e}^{-t}.t^{\frac{5}{2}-1}+{\int }_0^{\mathrm{\infty }}{e}^{-t}.t^{\frac{7}{2}-1}+{\int }_0^{\mathrm{\infty }}{e}^{-t}.t^{4-1}+{\int }_0^{\mathrm{\infty }}{e}^{-t}.t^{5-1}$$$$=\lceil \left(\frac{5}{2}\right)+\lceil \left(\frac{7}{2}\right)+\lceil \left(4\right)+\lceil \left(5\right)$$$$\lceil \left(\frac{5}{2}\right)=\lceil (\frac{3}{2}+1)=\frac{3}{2}\times \frac{1}{2}\times \sqrt{\pi }=\frac{3\sqrt{\pi }}{4}$$$$\lceil \left(\frac{7}{2}\right)=\frac{5}{2}\times \frac{3}{2}\times \frac{1}{2}\times \sqrt{\pi }=\frac{15}{8}\sqrt{\pi }$$$$\lceil \left(4\right)=3\times 2\times 1=6$$$$\lceil \left(5\right)=4\times 3\times 2\times 1=24$$$$I=2[\frac{3\sqrt{\pi }}{4}+\frac{15\sqrt{\pi }}{8}+6+24]$$$$plscheck...$$$$$$

Commented by peter frank last updated on 15/Apr/20

$$correctsirthankyou$$

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