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Question Number 141373 by SLVR last updated on 18/May/21

Commented by SLVR last updated on 18/May/21

$$kindlyhelpme$$

Answered by mathmax by abdo last updated on 19/May/21

$$\mathrm{I}={\int }_0^{4\pi }\log \mid 13\mathrm{sinx}+3\sqrt{3}\mathrm{cosx}\mid \mathrm{dx}=2{\int }_0^{2\pi }\log \mid 13\mathrm{sinx}+3\sqrt{3}\mathrm{cosx}\mid \mathrm{dx}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{2\pi }\log \mid 13\mathrm{sinx}+3\sqrt{3}\mathrm{cosx}+\mathrm{a}\mid \mathrm{dx}\Rightarrow \mathrm{f}\left(0\right)=\mathrm{I}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{2\pi }\frac{\mathrm{dx}}{13\mathrm{sinx}+3\sqrt{3}\mathrm{cosx}}{=}_{{\mathrm{e}}^{\mathrm{ix}}=\mathrm{z}}{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\mathrm{iz}(13\frac{\mathrm{z}-{\mathrm{z}}^{-1}}{2\mathrm{i}}+3\sqrt{3}.\frac{\mathrm{z}+{\mathrm{z}}^{-1}}{2})}$$$$={\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\mathrm{iz}(\frac{13(\mathrm{z}-{\mathrm{z}}^{-1})}{2\mathrm{i}}+\frac{\mathrm{i}3\sqrt{3}(\mathrm{z}+{\mathrm{z}}^{-1})}{2\mathrm{i}})}$$$$={\int }_{\mid \mathrm{z}\mid =1}\frac{2\mathrm{dz}}{13({\mathrm{z}}^2-1)+3\sqrt{3}\mathrm{i}({\mathrm{z}}^2+1)}$$$$={\int }_{\mid \mathrm{z}\mid =1}\frac{2\mathrm{dz}}{(13+3\sqrt{3}\mathrm{i}){\mathrm{z}}^2-13+3\sqrt{3}\mathrm{i}}$$$$=\frac{2}{13+3\sqrt{3}\mathrm{i}}{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{{\mathrm{z}}^2-\frac{13-3\sqrt{3}\mathrm{i}}{13+3\sqrt{3}\mathrm{i}}}$$$$=\frac{2}{13+3\sqrt{3}\mathrm{i}}{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{{\mathrm{z}}^2-\frac{{(13-3\sqrt{3}\mathrm{i})}^2}{{13}^2+27}}$$$$....\mathrm{be}\mathrm{continued}....$$

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