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Question Number 51551 by maxmathsup by imad last updated on 28/Dec/18

$$findf\left(\lambda \right)={\int }_0^{\frac{\pi }{4}}\sqrt{1+\lambda tant}dtwith\lambda >0$$ $$alsocalculate{\int }_0^{\frac{\pi }{4}}\frac{tant}{\sqrt{1+\lambda tant}}dt.$$

Commented byAbdo msup. last updated on 29/Dec/18

$$changementtant=xgive$$ $$f\left(\lambda \right)={\int }_0^1\sqrt{1+\lambda x}\frac{dx}{1+x^2}={\int }_0^1\frac{\sqrt{1+\lambda x}}{1+x^2}dx$$ $${=}_{\lambda x={sh}^{2}u}{\int }_0^{argsh\left(\sqrt{\lambda }\right)}\frac{chu}{1+\frac{1}{{\lambda }^2}{sh}^{4}u}\frac{1}{\lambda }2shuchudu$$ $$=\frac{2}{\lambda }{\int }_0^{ln(\sqrt{\lambda }+\sqrt{1+\lambda })}\frac{sh\left(u\right){ch}^2\left(u\right)}{{\lambda }^2+{sh}^{4}u}{\lambda }^{2}du$$ $$=2\lambda {\int }_0^{ln(\sqrt{\lambda }+\sqrt{1+\lambda })}\frac{sh\left(u\right)\frac{1+ch\left(2u\right)}{2}}{{\lambda }^2+{\left(\frac{ch\left(2u\right)-1}{2}\right)}^2}du$$ $$=4\lambda {\int }_0^{ln(\sqrt{\lambda }+\sqrt{1+\lambda })}\frac{(1+ch\left(2u\right))sh\left(u\right)}{4{\lambda }^2+{(ch\left(2u\right)-1)}^2}du$$ $$=4\lambda {\int }_0^{ln(\sqrt{\lambda }+\sqrt{1+\lambda })}\frac{(1+\frac{e^{2u}+e^{-2u}}{2})\left(\frac{e^u-e^{-u}}{2}\right)}{4{\lambda }^2+{(\frac{e^{2u}+e^{-2u}}{2}-1)}^2}du$$ $$=4\lambda {\int }_0^{ln(\sqrt{\lambda }+\sqrt{1+\lambda })}\frac{(2+e^{2u}+e^{-2u})(e^u-e^{-u})}{16{\lambda }^2+{(e^{2u}+e^{-2u}-2)}^2}du$$ $${=}_{e^u=t}4\lambda {\int }_1^{\sqrt{\lambda }+\sqrt{1+\lambda }}\frac{(2+t^2+t^{-2})(t-t^{-1})}{16{\lambda }^2+{(t^2+t^{-2}-2)}^2}\frac{dt}{t}$$ $$=4\lambda {\int }_1^{\sqrt{\lambda }+\sqrt{1+\lambda }}\frac{(t^4+2t^2+1)(t^2-1)}{t^4(16{\lambda }^2+{(t^2+t^{-2}-2)}^2)}dt$$ $$...becontinued...$$

Commented byAbdo msup. last updated on 29/Dec/18

$$letusethechangement\sqrt{1+\lambda tant}=u\Rightarrow$$ $$1+\lambda tant=u^2\Rightarrow \lambda tant=u^2-1\Rightarrow tant=\frac{u^2-1}{\lambda }\Rightarrow$$ $$t=arctan(\frac{1}{\lambda }{u}^2-\frac{1}{\lambda })\Rightarrow$$ $$f\left(\lambda \right)={\int }_1^{\sqrt{1+\lambda }}u.\frac{2}{\lambda }\frac{u}{1+{(\frac{1}{\lambda }{u}^2-\frac{1}{\lambda })}^2}du$$ $$=\frac{2}{\lambda }{\int }_1^{\sqrt{1+\lambda }}\frac{u^2}{{\lambda }^2+{(u^2-1)}^2}du$$ $${=}_{u=ch\left(t\right)}\frac{2}{\lambda }{\int }_{argch\left(1\right)}^{argch\left(\sqrt{1+\lambda }\right)}\frac{{ch}^{2}t}{{\lambda }^2+{sh}^{4}t}sh\left(t\right)dt$$ $$...becontinued...$$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18

$$g\left(t\right)=\sqrt{1+\lambda tant}$$ $$g\left(0\right)=1$$ $$g\left(\frac{\pi }{4}\right)=\sqrt{1+\lambda }$$ $${\int }_0^{\frac{\pi }{4}}\sqrt{1+\lambda tant}dt\approx \frac{1}{2}(1+\sqrt{1+\lambda })\times \frac{\pi }{4}$$ $${\int }_0^{\frac{\pi }{4}}\sqrt{1+\lambda tant}dt\approx \frac{\pi }{8}(1+\sqrt{1+\lambda })\left[approxmitation\right]$$ $$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18

$$g\left(t\right)=\sqrt{1+\lambda tant}$$ $$g\left(0\right)=1$$ $$g\left(\frac{\pi }{4}\right)=\sqrt{1+\lambda }$$ $${\int }_0^{\frac{\pi }{4}}\sqrt{1+\lambda tant}dt\approx \frac{1}{2}(1+\sqrt{1+\lambda })\times \frac{\pi }{4}$$ $${\int }_0^{\frac{\pi }{4}}\sqrt{1+\lambda tant}dt\approx \frac{\pi }{8}(1+\sqrt{1+\lambda })\left[approxmitation\right]$$ $$$$

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