Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 201350 by sonukgindia last updated on 05/Dec/23

Answered by Calculusboy last updated on 05/Dec/23

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:\mathit{l}\mathit{e}\mathit{t}\mathit{y}=\frac{\mathit{\pi }}{4}-\mathit{x}\mathit{d}\mathit{y}=-\mathit{d}\mathit{x}$$$$\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathit{x}=\frac{\mathit{\pi }}{4}\mathit{y}=0\mathit{a}\mathit{n}\mathit{d}\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathit{x}=0\mathit{y}=\frac{\mathit{\pi }}{4}$$$$\mathit{I}={\int }_{\frac{\mathit{\pi }}{4}}^0\mathit{I}\mathit{n}[1+\mathit{t}\mathit{a}\mathit{n}(\frac{\mathit{\pi }}{4}-\mathit{y})]-\mathit{d}\mathit{y}\left(\mathit{c}\mathit{h}\mathit{a}\mathit{n}\mathit{g}\mathit{e}\mathit{o}\mathit{f}\mathit{v}\mathit{a}\mathit{r}\mathit{i}\mathit{a}\mathit{b}\mathit{l}\mathit{e}\right)$$$$\mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}[1+\mathit{t}\mathit{a}\mathit{n}(\frac{\mathit{\pi }}{4}-\mathit{x})]\mathit{d}\mathit{x}\iff \mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}[1+\frac{\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }}{4}\right)-\mathit{t}\mathit{a}\mathit{n}\mathit{x}}{1+\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }}{4}\right)\mathit{t}\mathit{a}\mathit{n}\mathit{x}}]\mathit{d}\mathit{x}$$$$\mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}[1+\frac{1-\mathit{t}\mathit{a}\mathit{n}\mathit{x}}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{x}}]\mathit{d}\mathit{x}\iff \mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}\left[\frac{1+\mathit{t}\mathit{a}\mathit{n}\mathit{x}+1-\mathit{t}\mathit{a}\mathit{n}\mathit{x}}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{x}}\right]\mathit{d}\mathit{x}$$$$\mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}\left[\frac{2}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{x}}\right]\mathit{d}\mathit{x}\iff \mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}[\mathit{I}\mathit{n}\left(2\right)-\mathit{I}\mathit{n}(1+\mathit{t}\mathit{a}\mathit{n}\mathit{x})]\mathit{d}\mathit{x}$$$$\mathit{I}=\mathit{I}\mathit{n}\left(2\right){\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{d}\mathit{x}-{\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}(1+\mathit{t}\mathit{a}\mathit{n}\mathit{x})\mathit{d}\mathit{x}$$$$\mathit{I}=\mathit{I}\mathit{n}\left(2\right){\left[\mathit{x}\right]}_0^{\frac{\mathit{\pi }}{4}}-\mathit{I}\iff 2\mathit{I}=\mathit{I}\mathit{n}\left(2\right)[\frac{\mathit{\pi }}{4}-0]$$$$\mathit{I}=\frac{\mathit{\pi }}{4}\times \frac{1}{2}\times \mathit{I}\mathit{n}\left(2\right)$$$$\mathit{I}=\frac{\mathit{\pi }\mathit{I}\mathit{n}\left(2\right)}{8}$$$$$$$$$$

Answered by Sutrisno last updated on 05/Dec/23

$$misal:$$$$I=\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}ln(1+tanx)dx$$$$({\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx)$$$$I=\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}ln(1+tan(\frac{\pi }{4}-x))dx$$$$I=\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}ln(1+\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}.tanx})dx$$$$I=\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}ln(1+\frac{1-tanx}{1+tanx})dx$$$$I=\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}ln\left(\frac{2}{1+tanx}\right)dx$$$$I=\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}ln\left(2\right)-\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}ln(1+tanx)dx$$$$I=\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}ln\left(2\right)-I$$$$2I=ln\left(2\right)x\underset{0}{\stackrel{\frac{\pi }{4}}{\mid }}$$$$I=\frac{\pi ln2}{8}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com