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Question Number 34862 by a.i msup by abdo last updated on 12/May/18

$$findthevalueof$$$$f\left(x\right)={\int }_0^{\pi }\frac{cosx}{1+2sin\left(2x\right)}dx$$

Commented by prof Abdo imad last updated on 14/May/18

$$thankyousirforthisremark..ihaveworking$$$$alotincomplexeanalysisbutifindoften$$$$somethingthatwecantunderstandand$$$$explainitsmoreandmoredifficult...$$

Commented by prof Abdo imad last updated on 12/May/18

$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=\frac{-4\pi }{\sqrt{3}}\{Im(\sqrt{z_1}+Im\left(\sqrt{{\stackrel{-}{z}}_1}\right)\}$$$$z_1=\frac{\sqrt{3}}{2}-\frac{i}{2}=e^{-i\frac{\pi }{6}}\Rightarrow \sqrt{z_1}=e^{-i\frac{\pi }{12}}\Rightarrow$$$$Im\left(\sqrt{z_1}\right)=-sin\left(\frac{\pi }{12}\right)andIm\left(\sqrt{{\stackrel{-}{z}}_1}\right)=sin\left(\frac{\pi }{12}\right)$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=0\Rightarrow I=0.$$

Commented by MJS last updated on 13/May/18

$$\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{follow}\mathrm{here}:$$$$\frac{2\pi \mathrm{i}}{\sqrt{3}}(z_1^{\frac{1}{2}}+z_1^{-\frac{1}{2}}-z_2^{\frac{1}{2}}-z_2^{-\frac{1}{2}})=...$$$$\mathrm{because}$$$$z_1=\frac{\sqrt{3}}{2}-\frac{1}{2}\mathrm{i}$$$$\sqrt{z_1}=\frac{\sqrt{6}+\sqrt{2}}{4}-\frac{\sqrt{6}-\sqrt{2}}{4}\mathrm{i}$$$$\frac{1}{\sqrt{z_1}}=\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}\mathrm{i}$$$$z_2=-\frac{\sqrt{3}}{2}-\frac{1}{2}\mathrm{i}$$$$\sqrt{z_2}=\frac{\sqrt{6}-\sqrt{2}}{4}-\frac{\sqrt{6}+\sqrt{2}}{4}\mathrm{i}$$$$\frac{1}{\sqrt{z_2}}=\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4}\mathrm{i}$$$$\mathrm{so}$$$$\frac{2\pi \mathrm{i}}{\sqrt{3}}(z_1^{\frac{1}{2}}+z_1^{-\frac{1}{2}}-z_2^{\frac{1}{2}}-z_2^{-\frac{1}{2}})=\frac{2\pi \mathrm{i}}{\sqrt{3}}\times \sqrt{2}$$$${\mathrm{where}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{bug}?$$

Commented by abdo mathsup 649 cc last updated on 13/May/18

$$sirMjsnoneadtousealgebricformbecause$$$$z_1=\frac{\sqrt{3}}{2}-\frac{i}{2}=e^{-i\frac{\pi }{6}}and\sqrt{z_1}=e^{-i\frac{\pi }{12}}$$$$z_1^{-\frac{1}{2}}={\stackrel{{-}^{\frac{1}{2}}}{z}}_1.....$$

Commented by MJS last updated on 14/May/18

$$z_1^{\frac{1}{2}}=\mathrm{conj}\left(z_1^{-\frac{1}{2}}\right)={(-z_2)}^{\frac{1}{2}}=\mathrm{conj}\left({(-z_2)}^{-\frac{1}{2}}\right)$$$$z_1={\mathrm{e}}^{-\mathrm{i}\frac{\pi }{6}}{z}_2={\mathrm{e}}^{-\mathrm{i}\frac{5\pi }{6}}$$$$\sqrt{z_1}={\mathrm{e}}^{-\mathrm{i}\frac{\pi }{12}}\sqrt{z_2}={\mathrm{e}}^{-\mathrm{i}\frac{5\pi }{12}}$$$$\frac{1}{\sqrt{z_1}}={\mathrm{e}}^{\mathrm{i}\frac{\pi }{12}}\frac{1}{\sqrt{z_2}}={\mathrm{e}}^{\mathrm{i}\frac{5\pi }{12}}$$

Commented by MJS last updated on 14/May/18

$$(z_1^{\frac{1}{2}}+z_1^{-\frac{1}{2}}-z_2^{\frac{1}{2}}-z_2^{-\frac{1}{2}})\ne (z_1^{\frac{1}{2}}-{\stackrel{{-}^{\frac{1}{2}}}{z}}_1+z_1^{-\frac{1}{2}}-{\stackrel{{-}^{-\frac{1}{2}}}{z}}_1)$$$$\mathrm{because}{z}_1\ne \mathrm{conj}\left(z_2\right)\mathrm{but}{z}_1=\mathrm{conj}(-z_2)$$$$\mathrm{sorry}\mathrm{but}\mathrm{I}\mathrm{really}\mathrm{want}\mathrm{to}\mathrm{understand}\mathrm{how}$$$$\mathrm{the}\mathrm{residue}\mathrm{method}\mathrm{works}...$$

Commented by abdo mathsup 649 cc last updated on 14/May/18

$$sirMjsiwilldeletmymethodusedinthisintegral$$$$andtryanotherbecausetheconseptofresidus$$$$ismoredifficultinsomecases....$$

Commented by MJS last updated on 14/May/18

$$\mathrm{Sir},\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{say}\mathrm{if}\mathrm{your}\mathrm{method}\mathrm{is}\mathrm{right}\mathrm{or}$$$$\mathrm{wrong},{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{just}\mathrm{trying}\mathrm{to}\mathrm{follow}\mathrm{and}$$$$\mathrm{hopefully}\mathrm{understand}\mathrm{what}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{doing}.$$$$\mathrm{I}\mathrm{believe}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{very}\mathrm{much}\mathrm{better}\mathrm{in}\mathrm{these}$$$$\mathrm{things}\mathrm{than}\mathrm{I}\mathrm{am},\mathrm{since}\mathrm{I}\mathrm{had}\mathrm{been}\mathrm{studying}$$$$\mathrm{maths}\mathrm{about}25\mathrm{years}\mathrm{ago}\mathrm{and}\mathrm{I}\mathrm{have}\mathrm{forgotten}$$$$\mathrm{many}\mathrm{things}...{\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{be}\mathrm{truly}\mathrm{thankful}\mathrm{for}\mathrm{any}$$$$\mathrm{explanation}\mathrm{you}\mathrm{give}\mathrm{in}\mathrm{this}\mathrm{or}\mathrm{another}\mathrm{case}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/May/18

$$letk=tanxdk={sec}^{2}xdx$$$$I^{\prime }=\int \frac{cosx}{1+4sinxcosx}$$$$=\int \frac{cosx}{1+4tanx.{cos}^{2}x}dx$$$$=\int \frac{cosx}{{cos}^{2}x({sec}^{2}x+4tanx)}dx$$$$=\int \frac{{sec}^{2}xdx}{secx(1+{tan}^{2}x+4tanx)}$$$$=\int \frac{dk}{(\sqrt{1+k^2)}(1+k^2+4k)}$$$$contd$$

Answered by math1967 last updated on 12/May/18

$$2I=\underset{0}{\stackrel{\pi }{\int }}\frac{cosx}{1+2sin\left(2x\right)}dx+\underset{0}{\stackrel{\pi }{\int }}\frac{cos(\pi -x)}{1+2sin(\pi -2x)}dx$$$$=\underset{0}{\stackrel{\pi }{\int }}\frac{cosx}{1+2sin\left(2x\right)}dx-\underset{0}{\stackrel{\pi }{\int }}\frac{cosx}{1+2sin\left(2x\right)}dx$$$$=0$$

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