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Question Number 63089 by mathmax by abdo last updated on 28/Jun/19

$$findthevalueof{\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\left(tanx\right)}^{\sqrt{2}}}.$$

Commented by edafe ovwie last updated on 29/Jun/19

$$\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}$$$${\int }_{\mathit{b}}^{\mathit{a}}\mathit{f}\left(\mathit{x}\right)\mathit{d}\mathit{x}={\int }_{\mathit{b}}^{\mathit{a}}\mathit{f}(\mathit{a}+\mathit{b}-\mathit{x})\mathit{d}\mathit{x}$$$$let$$$$\text{Missing left or extra right}$$$$\text{Missing left or extra right}$$$$tan(\frac{\pi }{2}-x)=cotx=\frac{1}{tanx}$$$$\text{Missing left or extra right}$$$$\text{Missing left or extra right}$$$$add1and2$$$$2\mathbf{\Omega }={\int }_0^{\frac{\pi }{2}}1dx$$$$2\mathrm{\Omega }=\frac{\pi }{2}$$$$\mathrm{\Omega }=\frac{\pi }{4}$$$$\text{Missing left or extra right}$$

Commented by mathmax by abdo last updated on 29/Jun/19

$$thankyousir.$$

Commented by mathmax by abdo last updated on 03/Jul/19

$$letA={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\left(tanx\right)}^{\sqrt{2}}}changementx=\frac{\pi }{2}-tgive$$$$A=-{\int }_0^{\frac{\pi }{2}}\frac{-dt}{1+{\left(\frac{1}{tant}\right)}^{\sqrt{2}}}={\int }_0^{\frac{\pi }{2}}\frac{{\left(tant\right)}^{\sqrt{2}}}{1+{\left(tant\right)}^{\sqrt{2}}}dt={\int }_0^{\frac{\pi }{2}}\frac{1+{\left(tant\right)}^{\sqrt{2}}-1}{1+{\left(tant\right)}^{\sqrt{2}}}dt$$$$=\frac{\pi }{2}-A\Rightarrow 2A=\frac{\pi }{2}\Rightarrow A=\frac{\pi }{4}.$$

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