Задача 54677 ...
Условие
математика ВУЗ
202
Решение
★
[m]\int \frac{cos3x}{sin^3x}dx=\int \frac{4cos^3x-3cosx}{sin^3x}dx=\int \frac{4cos^3x}{sin^3x}-\int \frac{3cosx}{sin^3x}dx=4\int \frac{cos^2x\cdot cosx }{sin^3x}-3\int \frac{cosx}{sin^3x}dx=4\int \frac{(1-sin^2x)\cdot cosx }{sin^3x}-3\int \frac{cosx}{sin^3x}dx=[/m]
[m]=4\int \frac{1}{sin^2x}dx-4\int cosx dx-3\int sin^{-3}xd(sinx)=4(-ctgx)-4sinx-3\frac{sin^{-2}x}{-2}+C=-4ctgx-4sinx+\frac{3}{2sin^2x}+C[/m]