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Question Number 122406 by mathdave last updated on 16/Nov/20

$$solve$$$${\int }_0^{\frac{\pi }{4}}\ln ({\tan }^{2}x+2)dx$$

Answered by TANMAY PANACEA last updated on 16/Nov/20

$$f\left(x\right)=ln(2+{tan}^{2}x)$$$$f\left(0\right)=ln2$$$$f\left(\frac{\pi }{4}\right)=ln3$$$${\int }_0^{\frac{\pi }{4}}ln3dx>{\int }_0^{\frac{\pi }{4}}f\left(x\right)dx>{\int }_0^{\frac{\pi }{4}}lm2dx$$$$\frac{\pi }{4}ln3>I>\frac{\pi }{4}ln2$$$$$$$$$$

Answered by mathmax by abdo last updated on 16/Nov/20

$$\mathrm{I}={\int }_0^{\frac{\pi }{4}}\ln (1+{\tan }^2\mathrm{x}+1)={\int }_0^{\frac{\pi }{4}}\ln (1+\frac{1}{{\cos }^2\mathrm{x}})\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{4}}\ln (1+{\cos }^2\mathrm{x})\mathrm{dx}-2{\int }_0^{\frac{\pi }{4}}\ln \left(\mathrm{cosx}\right)\mathrm{dx}$$$${\int }_0^{\frac{\pi }{4}}\ln (1+{\cos }^2\mathrm{x})\mathrm{dx}={\int }_0^{\frac{\pi }{4}}\ln (1+\frac{1+\cos \left(2\mathrm{x}\right)}{2})\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{4}}\ln (3+\cos \left(2\mathrm{x}\right))\mathrm{dx}-\frac{\pi }{4}\ln \left(2\right)$$$$=\frac{\pi }{4}\ln \left(3\right)+{\int }_0^{\frac{\pi }{4}}\ln (1+\frac{1}{3}\cos \left(2\mathrm{x}\right))\mathrm{dx}-\frac{\pi }{4}\ln \left(2\right)$$$${\int }_0^{\frac{\pi }{4}}\ln (1+\frac{1}{3}\cos \left(2\mathrm{x}\right))\mathrm{dx}{=}_{2\mathrm{x}=\mathrm{t}}\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln (1+\frac{1}{3}\mathrm{cost})\mathrm{dt}\mathrm{let}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}\ln (1+\mathrm{acost})\mathrm{dt}\mathrm{with}0<\mathrm{a}<1$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cost}}{1+\mathrm{acost}}\mathrm{dt}=\frac{1}{\mathrm{a}}{\int }_0^{\frac{\pi }{2}}\frac{1+\mathrm{acost}-1}{1+\mathrm{acost}}\mathrm{dt}$$$$=\frac{\pi }{2\mathrm{a}}-\frac{1}{\mathrm{a}}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{1+\mathrm{acost}}(\to \tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{u})$$$$=\frac{\pi }{2\mathrm{a}}-\frac{1}{\mathrm{a}}{\int }_0^1\frac{2\mathrm{du}}{(1+{\mathrm{u}}^2)(1+\mathrm{a}\frac{1-{\mathrm{u}}^2}{1+{\mathrm{u}}^2})}=\frac{\pi }{2\mathrm{a}}-\frac{2}{\mathrm{a}}{\int }_0^1\frac{\mathrm{du}}{1+{\mathrm{u}}^2+\mathrm{a}-{\mathrm{au}}^2}$$$$=\frac{\pi }{2\mathrm{a}}-\frac{2}{\mathrm{a}}{\int }_0^1\frac{\mathrm{du}}{1+\mathrm{a}+(1-\mathrm{a}){\mathrm{u}}^2}\mathrm{we}\mathrm{have}$$$$\frac{2}{\mathrm{a}}{\int }_0^1\frac{\mathrm{du}}{1+\mathrm{a}+(1-\mathrm{a}){\mathrm{u}}^2}=\frac{2}{\mathrm{a}(1+\mathrm{a})}{\int }_0^1\frac{\mathrm{du}}{1+\frac{1-\mathrm{a}}{1+\mathrm{a}}{\mathrm{u}}^2}$$$${=}_{\sqrt{\frac{1-\mathrm{a}}{1+\mathrm{a}}}\mathrm{u}=\mathrm{z}}\frac{2}{\mathrm{a}(1+\mathrm{a})}{\int }_0^{\sqrt{\frac{1-\mathrm{a}}{1+\mathrm{a}}}}\frac{1}{1+{\mathrm{z}}^2}\times \frac{\sqrt{1+\mathrm{a}}}{\sqrt{1-\mathrm{a}}}\mathrm{dz}$$$$=\frac{2}{\mathrm{a}\sqrt{1-{\mathrm{a}}^2}}\arctan \left(\sqrt{\frac{1-\mathrm{a}}{1+\mathrm{a}}}\right)\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2}\mathrm{lna}-2\int \frac{\arctan \left(\sqrt{\frac{1-\mathrm{a}}{1+\mathrm{a}}}\right)}{\mathrm{a}\sqrt{1-{\mathrm{a}}^2}}\mathrm{da}+\mathrm{c}$$$$\int \frac{\arctan \left(\sqrt{\frac{1-\mathrm{a}}{1+\mathrm{a}}}\right)}{\mathrm{a}\sqrt{1-{\mathrm{a}}^2}}\mathrm{da}{=}_{\mathrm{a}=\cos \theta }\int \frac{\arctan \left(\tan \left(\frac{\theta }{2}\right)\right)}{\cos \theta .\sin \theta }\sin \theta \mathrm{d}\theta$$$$=\frac{1}{2}\int \frac{\theta }{\cos \theta }\mathrm{d}\theta .....\mathrm{be}\mathrm{continued}....$$

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