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Question Number 40823 by math khazana by abdo last updated on 28/Jul/18

$$calculate{\int }_0^{\frac{\pi }{2}}\sqrt{{cos}^{2}x+3{sin}^{2}x}dx$$

Commented by maxmathsup by imad last updated on 31/Jul/18

$$letI={\int }_0^{\frac{\pi }{2}}\sqrt{{cos}^{2}x+3{sin}^{2}x}dx$$$$I={\int }_0^{\frac{\pi }{2}}cosx\sqrt{1+3{tan}^{2}x}dx={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{1+3{tan}^{2}x}}{\sqrt{1+{tan}^{2}x}}dxch\sqrt{3}tanx=sh\left(u\right)$$$$={\int }_0^{\mathrm{\infty }}\frac{\sqrt{1+{sh}^{2}u}}{\sqrt{1+\frac{{sh}^{2}u}{3}}}\frac{\frac{1}{\sqrt{3}}chu}{(1+\frac{{sh}^{2}u}{3})}du(x=arctan\left(\frac{shu}{\sqrt{3}}\right))$$$$=\frac{1}{\sqrt{3}}{\int }_0^{\mathrm{\infty }}\frac{{ch}^{2}udu}{\sqrt{3+{sh}^{2}u}(3+{sh}^{2}u)}3\sqrt{3}du=3{\int }_0^{\mathrm{\infty }}\frac{{ch}^{2}u}{{(3+{sh}^{2}u)}^{\frac{3}{2}}}du$$$$=3{\int }_0^{\mathrm{\infty }}\frac{\frac{1+ch\left(2u\right)}{2}}{{(3+\frac{ch\left(2u\right)-1}{2})}^{\frac{3}{2}}}du=\frac{3}{2}{2}^{\frac{3}{2}}{\int }_0^{\mathrm{\infty }}\frac{1+ch\left(2u\right)}{{(5+ch\left(2u\right))}^{\frac{3}{2}}}du$$$$=3\sqrt{2}{\int }_0^{\mathrm{\infty }}\frac{1+\frac{e^{2u}+e^{-2u}}{2}}{{(5+\frac{e^{2u}+e^{-2u}}{2})}^{\frac{3}{2}}}du$$$$=3\sqrt{2}.\frac{2^{\frac{3}{2}}}{2}{\int }_0^{\mathrm{\infty }}\frac{2+e^{2u}+e^{-2u}}{{(10+e^{2u}+e^{-2u})}^{\frac{3}{2}}}du$$$${=}_{e^u=t}6{\int }_1^{+\mathrm{\infty }}\frac{2+t^2+t^{-2}}{{(10+t^2+t^{-2})}^{\frac{3}{2}}}\frac{dt}{t}$$$$=6{\int }_1^{+\mathrm{\infty }}\frac{2t^2+t^4+1}{t^3{\left(\frac{10t^2+t^4+1}{t^2}\right)}^{\frac{3}{2}}}dt=6{\int }_0^{\mathrm{\infty }}\frac{2t^2+t^4+1}{{(t^4+10t^2+1)}^{\frac{3}{2}}}dt...becontinued...$$$$$$

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