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Question Number 131211 by mnjuly1970 last updated on 02/Feb/21

$$...calculus...$$$$provethat::$$$$\mathbf{\Phi }={\int }_0^{\frac{\mathit{\pi }}{4}}\left(\frac{\sqrt{\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)+{\mathit{t}\mathit{a}\mathit{n}}^2\left(\mathit{x}\right)}}{\sqrt{\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)-{\mathit{t}\mathit{a}\mathit{n}}^2\left(\mathit{x}\right)}}\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}\right)\right)\mathit{d}\mathit{x}$$$$=\frac{1}{2}+\frac{\sqrt{\mathit{\pi }}}{8}(\frac{\mathbf{\Gamma }\left(\frac{1}{4}\right)}{\mathbf{\Gamma }\left(\frac{3}{4}\right)}-\frac{\mathbf{\Gamma }\left(\frac{3}{4}\right)}{\mathbf{\Gamma }\left(\frac{5}{4}\right)})$$

Answered by Ar Brandon last updated on 02/Feb/21

$$\mathrm{\Phi }={\int }_0^{\frac{\pi }{4}}\left(\frac{\sqrt{\mathrm{tanx}+{\tan }^2\mathrm{x}}}{\sqrt{\mathrm{tanx}-{\tan }^2\mathrm{x}}}\mathrm{sinx}\right)\mathrm{dx}={\int }_0^{\frac{\pi }{4}}\sqrt{\frac{1+\mathrm{tanx}}{1-\mathrm{tanx}}}\cdot \frac{\mathrm{tanx}}{\sqrt{1+{\tan }^2\mathrm{x}}}\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{4}}\frac{(1+\mathrm{tanx})\mathrm{tanx}}{\sqrt{(1-{\tan }^2\mathrm{x})(1+{\tan }^2\mathrm{x})}}\mathrm{dx},\mathrm{tanx}=\mathrm{t}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}$$$$={\int }_0^1\frac{\mathrm{t}+{\mathrm{t}}^2}{\sqrt{(1-{\mathrm{t}}^2)(1+{\mathrm{t}}^2)}}\cdot \frac{\mathrm{dt}}{1+{\mathrm{t}}^2}$$$$={\int }_0^1\{\frac{\mathrm{t}}{(1+{\mathrm{t}}^2)\sqrt{(1-{\mathrm{t}}^2)(1+{\mathrm{t}}^2)}}+\frac{{\mathrm{t}}^2}{(1+{\mathrm{t}}^2)\sqrt{(1-{\mathrm{t}}^2)(1+{\mathrm{t}}^2)}}\}\mathrm{dt}$$$$=\frac{1}{2}{\int }_0^1\frac{\mathrm{du}}{(1+\mathrm{u})\sqrt{1-{\mathrm{u}}^2}}+{\int }_0^1\frac{{\mathrm{u}}^{1/2}\mathrm{du}}{(1+\mathrm{u})\sqrt{1-{\mathrm{u}}^2}}$$$$\frac{1}{\mathrm{u}+1}=\mathrm{v}\Rightarrow -\frac{\mathrm{du}}{{(\mathrm{u}+1)}^2}=\mathrm{dv}\Rightarrow \mathrm{du}=-\frac{\mathrm{dv}}{{\mathrm{v}}^2}$$$$\mathrm{\Phi }=\frac{1}{2}{\int }_{1/2}^1\frac{\mathrm{v}}{\sqrt{\frac{2}{\mathrm{v}}-\frac{1}{{\mathrm{v}}^2}}}\cdot \frac{\mathrm{dv}}{{\mathrm{v}}^2}+{\int }_0^1\frac{{\mathrm{u}}^{1/2}\mathrm{du}}{(1+\mathrm{u})\sqrt{1-{\mathrm{u}}^2}}$$$$=\frac{1}{2}{\int }_{1/2}^1\frac{\mathrm{dv}}{\sqrt{2\mathrm{v}-1}}+\frac{1}{2}{\int }_0^1\frac{{\mathrm{p}}^{1/4}}{(1+{\mathrm{p}}^{1/2})\sqrt{1-\mathrm{p}}}\cdot \frac{\mathrm{dp}}{{\mathrm{p}}^{1/2}}$$$$=1+\frac{1}{2}{\int }_0^1\frac{{\mathrm{p}}^{-1/4}}{(1+{\mathrm{p}}^{1/2})\sqrt{1-\mathrm{p}}}\mathrm{dp}$$$$...$$

Commented by mnjuly1970 last updated on 02/Feb/21

$$grateful..foryoureffort$$$$mrbrandon...$$

Commented by Ar Brandon last updated on 02/Feb/21

Thank you Sir ���� Got stucked unfortunately

Commented by mnjuly1970 last updated on 02/Feb/21

Commented by Ar Brandon last updated on 02/Feb/21

Nice��

Commented by mnjuly1970 last updated on 02/Feb/21

$$thankyoumasterbrandon$$$$Godkeepyou...$$

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