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Question Number 138691 by liberty last updated on 16/Apr/21

$$\int \cos 2x\sqrt{1+\sin {}^{2}x}dx=?$$

Answered by mathmax by abdo last updated on 17/Apr/21

$$\mathrm{I}=\int \cos \left(2\mathrm{x}\right)\sqrt{1+{\sin }^2\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{I}=\int \cos \left(2\mathrm{x}\right)\sqrt{1+\frac{1-\cos \left(2\mathrm{x}\right)}{2}}\mathrm{dx}$$$$=\int \cos \left(2\mathrm{x}\right)\sqrt{\frac{3-\cos \left(2\mathrm{x}\right)}{2}}\mathrm{dx}=\frac{1}{\sqrt{2}}\int \cos \left(2\mathrm{x}\right)\sqrt{3-\cos \left(2\mathrm{x}\right)}\mathrm{dx}$$$${=}_{2\mathrm{x}=\mathrm{t}}\frac{1}{\sqrt{2}}\int \mathrm{cost}\sqrt{3-\mathrm{cost}}\frac{\mathrm{dt}}{2}=\frac{1}{2\sqrt{2}}\int \mathrm{cost}\sqrt{3-\mathrm{cost}}\mathrm{dt}$$$${=}_{\mathrm{cost}=\mathrm{y}}\frac{1}{2\sqrt{2}}\int \mathrm{y}\sqrt{3-\mathrm{y}}(-\frac{\mathrm{dy}}{\sqrt{1-{\mathrm{y}}^2}})=-\frac{1}{2\sqrt{2}}\int \frac{\mathrm{y}}{\sqrt{1-{\mathrm{y}}^2}}\sqrt{3-\mathrm{y}}\mathrm{dy}$$$$\mathrm{by}\mathrm{parts}{\mathrm{f}}^{{}^{\prime }}=\frac{\mathrm{y}}{\sqrt{1-{\mathrm{y}}^2}}\mathrm{and}\mathrm{g}=\sqrt{3-\mathrm{y}}\Rightarrow$$$$\int \frac{\mathrm{y}}{\sqrt{1-{\mathrm{y}}^2}}\sqrt{3-\mathrm{y}}\mathrm{dy}=-\sqrt{1-{\mathrm{y}}^2}\sqrt{3-\mathrm{y}}-\int -\sqrt{1-{\mathrm{y}}^2}\left(\frac{-1}{2\sqrt{3-\mathrm{y}}}\right)\mathrm{dy}$$$$=-\sqrt{3-{\mathrm{y}}^2}\sqrt{1-{\mathrm{y}}^2}-\int \frac{\sqrt{1-{\mathrm{y}}^2}}{2\sqrt{3-\mathrm{y}}}\mathrm{dy}\mathrm{changement}\sqrt{3-\mathrm{y}}=\mathrm{u}\mathrm{give}$$$$3-\mathrm{y}={\mathrm{u}}^2\Rightarrow \mathrm{y}=3+{\mathrm{u}}^2\Rightarrow \int \frac{\sqrt{1-{\mathrm{y}}^2}}{2\sqrt{3-\mathrm{y}}}\mathrm{dy}$$$$=\int \frac{\sqrt{1-{(3+{\mathrm{u}}^2)}^2}}{2\mathrm{u}}\left(2\mathrm{u}\right)\mathrm{du}=\int \sqrt{1-{({\mathrm{u}}^2+3)}^2}\mathrm{du}$$$$....\mathrm{be}\mathrm{continued}...$$

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