Question

50 БАЛЛОВ СРОЧНО!!!!

Answer 1

Ответ:

$\displaystyle 4)\ \ \int \frac{sin^3x}{\sqrt{cosx}}\, dx=\int sin^2x\cdot sinx\cdot (cosx)^{-\frac{1}{2}}\, dx=\\\\\\=\int (\underbrace{1-cos^2x}_{sin^2x})\cdot (cosx)^{-\frac{1}{2}}\cdot \underbrace{sinx\, dx}_{-d(cosx)}=-\int (\, (cosx)^{-\frac{1}{2}}-(cosx)^{\frac{3}{2}}\, )\cdot d(cosx)=\\\\\\=-2(cosx)^{\frac{1}{2}}+\frac{2(cosx)^{\frac{5}{2}}}{5}+C=-2\sqrt{cosx}+\frac{2\sqrt{cos^5x}}{5}+C$

$\displaystyle 5)\ \ \int \frac{cosx\, dx}{sinx}=\int \frac{d(sinx)}{sinx}=ln|sinx|+C\\\\\\\star \ \ \frac{cosx}{sinx}=ctgx\ \ \ \Rightarrow \ \ \ \int ctgx=ln|sinx|+C\ \ \star$