Question

Integration by Partial Fractions

∫dy/(5+4cosy)

Answer 1

$Given \ that \ \int(\frac{1}{5+4cosy})dy\\ Substituting \ cos(y)= \frac{1-tan^2{\frac{y}{2}}}{1+tan^2{\frac{y}{2}}}\\ we \ get \int(\frac{1}{5+4cosy})dy=\int(\frac{1}{5+4(\frac{1-tan^2{\frac{y}{2}}}{1+tan^2{\frac{y}{2}}})})dy\\ =\int(\frac{1+tan^2{\frac{y}{2}}}{5(1+tan^2{\frac{y}{2}})+4({1-tan^2{\frac{y}{2}}})})dy\\ =\int(\frac{1+tan^2{\frac{y}{2}}}{5+5tan^2{\frac{y}{2}}+4-4tan^2{\frac{y}{2}}})dy =\int(\frac{1+tan^2{\frac{y}{2}}}{9+tan^2{\frac{y}{2}}})dy\\ Substituting \ u=tan{\frac{y}{2}} \& \ (1+tan^2{\frac{y}{2}})dy=2du\\ we \ get \int(\frac{2}{9+u^2})du=\frac{2}{3}tan^{-1}(\frac{u}{3})+C,\\ \because \int(\frac{1}{a^2+x^2})dx=\frac{1}{a}tan^{-1}(\frac{x}{a})\\where \ C \ is \ a \ constant \ of \ integration.\\ Substituting \ back \ u=tan{\frac{y}{2}} \ , we \ get \\ \int(\frac{1}{5+4cosy})dy=\frac{2}{3}tan^{-1}(\frac{tan{\frac{y}{2}}}{3})+C\\ \therefore \int(\frac{1}{5+4cosy})dy=\frac{2}{3}tan^{-1}(\frac{1}{3}tan{\frac{y}{2}})+C$