Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 132124 by Chhing last updated on 11/Feb/21

$$$$$$\mathrm{Calculate}$$$${\int }_0^{\frac{\pi }{4}}\sqrt{\frac{\tan \left(\mathrm{x}\right)+{\tan }^2\left(\mathrm{x}\right)}{\tan \left(\mathrm{x}\right)-{\tan }^2\left(\mathrm{x}\right)}}\cos \left(\mathrm{x}\right)\mathrm{dx}$$$$$$

Commented by liberty last updated on 11/Feb/21

$$\sqrt{\frac{\tan \mathrm{x}(1+\tan \mathrm{x})}{\tan \mathrm{x}(1-\tan \mathrm{x})}}=\sqrt{\frac{\sin \mathrm{x}+\cos \mathrm{x}}{\sin \mathrm{x}-\cos \mathrm{x}}}$$$$\mathrm{but}\sin \mathrm{x}-\cos \mathrm{x}<0\mathrm{on}\mathrm{interval}$$$$0⩽\mathrm{x}⩽\frac{\pi }{4}.\mathrm{then}\mathrm{integral}\mathrm{diverges}$$

Commented by prakash jain last updated on 11/Feb/21

$$\sqrt{\frac{1+\tan x}{1-\tan x}}=\sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}$$

Answered by Ar Brandon last updated on 11/Feb/21

$$\mathcal{I}={\int }_0^{\frac{\pi }{4}}\sqrt{\frac{\mathrm{tanx}+{\tan }^2\mathrm{x}}{\mathrm{tanx}-{\tan }^2\mathrm{x}}}\cdot \mathrm{cosxdx}$$$$={\int }_0^{\frac{\pi }{4}}\sqrt{\frac{\mathrm{cosx}+\mathrm{sinx}}{\mathrm{cosx}-\mathrm{sinx}}}\cdot \mathrm{cosxdx}={\int }_0^{\frac{\pi }{4}}\frac{\mathrm{cosx}+\mathrm{sinx}}{\sqrt{\cos 2\mathrm{x}}}\cdot \mathrm{cosxdx}$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{1+\cos 2\mathrm{x}}{\sqrt{\cos 2\mathrm{x}}}\mathrm{dx}+\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{\sin 2\mathrm{x}}{\sqrt{\cos 2\mathrm{x}}}\mathrm{dx}$$$${\mathrm{u}}^2=\cos 2\mathrm{x},2\mathrm{udu}=-2\sqrt{1-{\mathrm{u}}^4}$$$$=\frac{1}{4}{\int }_0^1\frac{1+\mathrm{u}}{\sqrt{\mathrm{u}}}\cdot \frac{\mathrm{du}}{\sqrt{1-{\mathrm{u}}^2}}-{\left[\frac{\sqrt{\cos 2\mathrm{x}}}{2}\right]}_0^{\frac{\pi }{4}}=\frac{1}{4}{\int }_0^1\frac{1+\mathrm{u}}{\sqrt{\mathrm{u}-{\mathrm{u}}^3}}\mathrm{du}+\frac{1}{2}$$

Commented by Dwaipayan Shikari last updated on 17/Feb/21

$${\int }_0^1\frac{1}{\sqrt{u-u^3}}+\frac{u}{\sqrt{u-u^3}}du$$$$={\int }_0^1{u}^{-\frac{1}{2}}{(1-u^2)}^{-\frac{1}{2}}+u^{\frac{1}{2}}{(1-u^2)}^{-\frac{1}{2}}du$$$$=\frac{1}{2}{\int }_0^1{t}^{-\frac{3}{4}}{(1-t)}^{-\frac{1}{2}}+t^{-\frac{1}{4}}{(1-t)}^{-\frac{1}{2}}dt$$$$=\frac{1}{2}.\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\sqrt{\pi }}{\mathrm{\Gamma }\left(\frac{3}{4}\right)}+2\frac{\mathrm{\Gamma }\left(\frac{3}{4}\right)\sqrt{\pi }}{\mathrm{\Gamma }\left(\frac{1}{4}\right)}=\frac{\sqrt{\pi }}{2}\left(\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)+4{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right)}{\sqrt{2}\pi }\right)$$$$=\frac{1}{\sqrt{8\pi }}({\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)+4{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right))$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com