Интеграл sinxdx/(4+cosx)^5

4+cosx=t⇒dt=-sinxdx [latex] \int\limits {sinx/(4+cosx)^5} \, dx =- \int\limits {1/(4+t)^5} \, dt =1/4(1+y)^4=[/latex][latex]1/4(4+cosx)^4+C[/latex]

[latex]\displaystyle \int\frac{\sin(x)\text{d}x}{(4+\cos(x))^5}=\int\frac{-\text{d}(\cos(x))}{(4+\cos(x))^5}=-\int\frac{\text{d}(4+\cos(x))}{(4+\cos(x))^5}=[/latex] [latex]\displaystyle =-\int(4+\cos(x))^{-5}\text{d}(4+\cos(x))=-\frac{(4+\cos(x))^{-4}}{-4}+C=\boxed{\frac{1}{4(4+\cos(x))^4}+C}\phantom{.}.[/latex]