Evaluate $\int \frac{1}{2+\sin x+\cos x}dx.$

Question

$$\int \frac{1}{2+\sin x+\cos x}dx.$$

My attempts:

1. Let $y = \sin x+\cos x.\implies \frac{dy}{dx}=\cos x-\sin x=y'.$

$$\int \frac{1}{2+\sin x+\cos x}dx=\int\frac{1}{2+y}\frac{dy}{y'}.$$

And I tried to use the fact $(\ln x)'=1/x,$ but $(\ln(2+y))'=\frac{y'}{2+y}$ : the form doesn't match. So I think I've failed at this moment.

2. Let $u=\sin x.\implies du=\cos x\ dx=\sqrt{1-u^2}dx.$

$$\int \frac{1}{2+\sin x+\cos x}dx=\int\frac{1}{2+u+\sqrt{1-u^2}}\frac{du}{\sqrt{1-u^2}}.$$

And it looks more uncomputable.

Both of my attempts are at a dead end. How to evaluate this integral? Any help would be appreciated.

Answer 1

Hint: Use the so-called Weierstrass substitution $$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ $$dx=\frac{2}{1+t^2}dt$$ A possible solution is given by $$\sqrt {2}\arctan \left( 1/4\, \left( 2\,\tan \left( x/2 \right) +2 \right) \sqrt {2} \right) +C$$

Answer 2

Hint:

$$\text{Let }\begin{bmatrix}\sin x \\ \cos x \\ \mathrm dx\end{bmatrix}=\begin{bmatrix}\dfrac{2t}{1+t^2}\\ \dfrac{1-t^2}{1+t^2}\\ \dfrac{2\mathrm dt}{1+t^2}\end{bmatrix}$$

This transforms $R(\sin x, \cos x)$ to a rational function in $t$ and you can proceed with Partial Fraction Decomposition.