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Question Number 121602 by benjo_mathlover last updated on 09/Nov/20

Answered by liberty last updated on 10/Nov/20

$$\mathrm{I}=\underset{0}{\stackrel{\pi /2}{\int }}\frac{\mathrm{dx}}{{\mathrm{b}}^2\tan {}^2\mathrm{x}+1}$$$$\mathrm{let}\varphi =\mathrm{b}\tan \mathrm{x}\Rightarrow \mathrm{d}\varphi =\mathrm{bsec}{}^2\mathrm{x}\mathrm{dx}$$$$\mathrm{dx}=\frac{\varphi }{{\varphi }^2+1}\mathrm{d}\varphi$$$$\mathrm{I}=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\varphi \mathrm{d}\varphi }{({\mathrm{b}}^2+{\varphi }^2)(1+{\varphi }^2)}=\frac{\mathrm{b}}{{\mathrm{b}}^2-1}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}[\frac{1}{1+{\varphi }^2}-\frac{1}{{\mathrm{b}}^2+{\varphi }^2}]\mathrm{d}\varphi$$$$\mathrm{I}=\frac{\mathrm{b}}{{\mathrm{b}}^2-1}{[{\tan }^{-1}\left(\varphi \right)-{\tan }^{-1}\left(\frac{\varphi }{\mathrm{b}}\right)]}_0^{\mathrm{\infty }}$$$$\mathrm{I}=\frac{\mathrm{b}}{{\mathrm{b}}^2-1}[\frac{\pi }{2}-\frac{\pi }{2\mathrm{b}}]$$$$\mathrm{I}=\frac{\pi }{2}(1-\frac{1}{\mathrm{b}})\left(\frac{\mathrm{b}}{{\mathrm{b}}^2-1}\right)=\frac{\pi }{2}\left(\frac{\mathrm{b}-1}{\mathrm{b}}\right)\left(\frac{\mathrm{b}}{(\mathrm{b}-1)(\mathrm{b}+1)}\right)$$$$\mathrm{I}=\frac{\pi }{2(\mathrm{b}+1)}.▴$$

Answered by Olaf last updated on 10/Nov/20

$$1)b=0$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}dx=\frac{\pi }{2}$$$$$$$$2)b=\pm 1$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\tan }^{2}x}$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{1+\cos \left(2x\right)}{2}dx$$$$\mathrm{I}={[\frac{1}{2}x+\frac{1}{4}\sin \left(2x\right)]}_0^{\frac{\pi }{2}}$$$$\mathrm{I}=\frac{\pi }{4}$$$$$$$$3)\mathrm{General}\mathrm{case}:b\ne 0,b\ne \pm 1$$$$u=\tan x$$$$du=(1+{\tan }^{2}x)dx=(1+u^2)dx$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{du}{(b^2{u}^2+1)(u^2+1)}$$$$\mathrm{I}=\frac{1}{b^2-1}{\int }_0^{\mathrm{\infty }}[\frac{b^2}{b^2{u}^2+1}-\frac{1}{1+u^2}]du$$$$\mathrm{I}=\frac{1}{b^2-1}{\int }_0^{\mathrm{\infty }}[\frac{1}{\frac{1}{b^2}+u^2}-\frac{1}{1+u^2}]du$$$$\mathrm{I}=\frac{1}{b^2-1}{[b\arctan \left(bu\right)-\arctan u]}_0^{\mathrm{\infty }}$$$$\mathrm{I}=\frac{1}{b^2-1}[(b\times \mathrm{sign}\left(b\right)\frac{\pi }{2}-\frac{\pi }{2})-(b\frac{\pi }{4}-\frac{\pi }{4})]$$$$\mathrm{I}=\frac{1}{b^2-1}[\frac{\pi }{2}(\mid b\mid -1)-\frac{\pi }{4}(b-1)]$$$$\mathrm{If}b>0:$$$$\mathrm{I}=\frac{1}{b^2-1}[\frac{\pi }{2}(b-1)-\frac{\pi }{4}(b-1)]$$$$\mathrm{I}=\frac{1}{b+1}[\frac{\pi }{2}-\frac{\pi }{4}]=\frac{\pi }{4(b+1)}$$$$\mathrm{If}b<0:$$$$\mathrm{I}=\frac{1}{b^2-1}[\frac{\pi }{2}(-b-1)-\frac{\pi }{4}(b-1)]$$$$\mathrm{I}=\frac{\pi }{4(b^2-1)}[2(-b-1)-(b-1)]$$$$\mathrm{I}=\frac{\pi }{4(b^2-1)}(-3b-3)$$$$\mathrm{I}=-\frac{3\pi }{4(b-1)}$$

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