Evaluating an indefinite cosine integral Evaluate $$\int\frac{1+2\cos x}{(\cos x+2)^2} dx$$ The problem looks easy but I don't think it is. Please just don't give hints for possible substitutions as I myself have tried quite many and all of them are just taking too long to solve. Thanks.
 A: $$\int  \frac { 1+2\cos  x }{ (\cos  x+2)^{ 2 } } dx=2\int { \frac { \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } +2\cos { x }  }{ { \left( \cos  x+2 \right)  }^{ 2 } }  } dx=\int { \frac { \cos { x } \left( \cos { x } +2 \right) +\sin ^{ 2 }{ x }  }{ { \left( \cos  x+2 \right)  }^{ 2 } }  } dx=\\ =\int { \frac { { { \left( \sin { x }  \right)  }^{ \prime  }\left( \cos { x } +2 \right) -\sin { x{ \left( \cos { x } +2 \right)  }^{ \prime  } }  } }{ { \left( \cos  x+2 \right)  }^{ 2 } }  } dx=\int { d\left( \frac { \sin { x }  }{ { \cos  x+2 } }  \right)  } =\frac { \sin { x }  }{ { \cos  x+2 } } +C$$
A: Just use the half angle formula...
$$\int \frac{2 \cos (x)+1}{(\cos (x)+2)^2} \, \mathrm{d}x = \frac{\sin (x)}{\cos (x)+2}$$
A: Hint use $cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ the integration reduces to evaluating $\int \frac{3-t^2}{(3+t^2)^2}dt=\frac{t}{3+t}+c$ after using Weierstrass substitution ie $\tan(x/2)=t$. Now making necessary substitutions get the integral
A: The one is set up for the eccentric anomaly:
$$\sin E=\frac{\sqrt{1-e^2}\sin x}{1+e\cos x},\,\,\,\cos E=\frac{\cos x+e}{1+e\cos x},\,\,\,dE=\frac{\sqrt{1-e^2}dx}{1+e\cos x}$$
Then with $e=\frac12$,
$$\begin{align}\int\frac{1+2\cos x}{(2+\cos x)^2}dx&=\frac12\int\frac{\cos x+e}{(1+e\cos x)^2}dx=\frac12\int\cos E\frac{dE}{\sqrt{1-e^2}}\\
&=\frac{\sin E}{2\sqrt{1-e^2}}+C=\frac1{2\sqrt{1-e^2}}\frac{\sqrt{1-e^2}\sin x}{1+e\cos x}+C\\
&=\frac{\sin x}{2+\cos x}+C\end{align}$$
A: Let $$\int\frac{1+2\cos x}{(\cos x+2)^2}dx $$
Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\sin^2 x$
$$= \int\frac{\csc^2 x+2\cot x\csc x}{(\cot x+2\csc x)^2}dx$$
Now Put $\cot x+2\csc x = t\;,$ Then $\left(\csc^2 x+2\csc x\cot x\right)dx = -dt$
So we get $$I = -\int\frac{1}{t^2}dt = \frac{1}{t}+\mathcal{C}=\frac{\sin x}{\cos x+2}+\mathcal{C}.$$
