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Question Number 186152 by normans last updated on 01/Feb/23

$$$$$$\mathit{I}=\underset{2}{\stackrel{\mathit{\pi }}{\int }}\frac{{\mathit{c}\mathit{o}\mathit{s}}^2\left(\mathit{x}\right)-1}{1+\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}\right)-\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)}\mathit{d}\mathit{x}$$$$$$

Answered by MJS_new last updated on 02/Feb/23

$$\int \frac{{\cos }^{2}x-1}{1+\sin x-\tan x}dx=-\int \frac{{\sin }^{2}x}{1+\sin x-\tan x}dx=$$$$[t=\tan (\frac{x}{2}+\frac{\pi }{8})\to dx=\frac{2dt}{t^2+1}]$$$$=-\frac{4-\sqrt{2}}{7}\int \frac{(t^2-2t-1){(t^2+2t-1)}^2}{{(t^2+1)}^2(t^2+1-2\sqrt{2})(t^2-\frac{1-2\sqrt{2}}{7})}dt$$$$\mathrm{now}\mathrm{decompose}\mathrm{\&}\mathrm{use}\mathrm{common}\mathrm{formulas}$$$$=-2\int \frac{t-1}{t^2+1}dt-$$$$-\sqrt{2}\int \frac{t^2+2t-1}{{(t^2+1)}^2}dt+$$$$+\frac{2-\sqrt{2}}{2}\int \frac{t-1+\sqrt{2}}{t^2+1-2\sqrt{2}}dt+$$$$+\frac{2+\sqrt{2}}{2}\int \frac{t-\frac{3+\sqrt{2}}{7}}{t^2-\frac{1-2\sqrt{2}}{7}}dt=$$$$=2\arctan t-\ln (t^2+1)+$$$$+\frac{\sqrt{2}(t+1)}{t^2+1}+$$$$+\frac{7(2-\sqrt{2})-(8-5\sqrt{2})\sqrt{-1+2\sqrt{2}}}{28}\ln \mid t+\sqrt{-1+2\sqrt{2}}\mid +$$$$+\frac{7(2-\sqrt{2})+(8-5\sqrt{2})\sqrt{-1+2\sqrt{2}}}{28}\ln \mid t-\sqrt{-1+2\sqrt{2}}\mid +$$$$+\frac{2+\sqrt{2}}{4}\ln \mid t^2-\frac{1-2\sqrt{2}}{7}\mid -$$$$-\frac{(4+3\sqrt{2})\sqrt{-1+2\sqrt{2}}}{2\sqrt{7}}\arctan \left(\sqrt{1+2\sqrt{2}}t\right)$$$$\mathrm{now}\mathrm{do}\mathrm{the}\mathrm{rest}\mathrm{if}\mathrm{you}\mathrm{want}.\mathrm{the}\mathrm{result}\mathrm{is}\mathrm{just}$$$$\mathrm{another}\mathrm{number}.$$

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