Задача 43662 ...

sova

2020-01-25 10:16:05

sin^5x=sin^4x*sinx=(sin^2x)^2*sinx=(1-cos^2x)^2*sinx=(1-2cos^2x+cos^4x)*sinx

[m]\int \frac{sin^5x}{cos^3x}dx=\int \frac{1-2cos^2x+cos^4x}{cos^3x}sinxdx=[/m]

[m]=\int \frac{sinxdx}{cos^3x}-2\int \frac{sinxdx}{cosx}+\int cosx\cdot sinxdx=[/m]

[m]=-\int cos^{-3}d(cosx)+2\int \frac{d(cosx)}{cosx}-\int cosxd(cosx)=[/m]

[m]=-\frac{cos^{-3+1}}{-3+1}+2\cdot ln|cosx|-\frac{cos^2x}{2}+C=\frac{1}{3cos^2x}+2\cdot ln|cosx|-\frac{cos^2x}{2}+C[/m]