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[TEX] \int_{}^{}\frac{sin2xdx}{cosx\sqrt{1+\sqrt{cosx}}} [/TEX]
[TEX] \int_{}^{}\frac{2.sinxdx}{\sqrt{1+\sqrt{cosx}}}[/TEX]
đặt
[TEX] \sqrt{1+\sqrt{cosx}} = u \Rightarrow u^2 = 1+\sqrt{cosx}\\ (u^2 -1)^2 = cosx \Rightarrow (4u^3 -4.u).du = -sinx.dx [/TEX]
[TEX] \int_{}^{}\frac{8.u-8.u^3}{u}.du = \int_{}^{}( 8 - 8.u^2) = 8u - \frac{8}{3}.u^3 \\ 8.\sqrt{1+\sqrt{cosx}} - \frac{8}{3}.(\sqrt{1+\sqrt{cosx}})^3 + C[/TEX]