Задача 59306 ...

5f54a07dd4ce5a71e171a096

19.04.2021 17:32:35

∫ cos x dx / √(sin x -4)^3 .

5f3ea7e3faf909182968ddd9

19.04.2021 19:16:12

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[m] ∫ \frac{du}{\sqrt{u^3}}= ∫ u^{-\frac{3}{2}}du=\frac{u^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}=\frac{u^{-\frac{1}{2}}}{-\frac{1}{2}}=-\frac{2}{\sqrt{u}} [/m]


u=sinx-4

du=(sinx-4)`dx

du=cosxdx

О т в е т.
[m]=-\frac{2}{\sqrt{(sinx-4)}}+С[/m]