# Integrate $\int\frac{dx}{7+5\cos x}$

$$\int\frac{dx}{7+5\cos x}$$

I'm stuck in this problem plzz give some clue to solve this problem

If I will write numerator as $$~\sin^2 x + \cos^2 x=1~$$

Then it become more complex

Please tell me an approach how to tackle this one

Integrals of this form are normally easily solved using the Tangent half-angle substitution: $$t=\tan\frac x2$$ $$dx=\frac 2{1+t^2}dt$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ so this would give you: $$\int\frac{dx}{7+5\cos(x)}dx=\int\frac{1}{7+5\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}dt=\int\frac{2}{7(1+t^2)+5(1-t^2)}dt$$ Now you should be able to solve it

• You say "easily" here. My experience with the Weierstrass substitution: if there is any other way to do the integral, it will be easier than the Weierstrass substutution. – GEdgar Jul 15 at 13:34
• @GEdgar As sad as it may be Weirerstrass substitution is a necessary reality for most non-elementary solvable integrals. – Sam Jul 17 at 17:37

$$I=\int\frac{dx}{7+5\cos x}=\int \frac{\sec^2\left(\frac{x}{2}\right)}{2\left(\tan^2\left(\frac{x}{2}\right)+6\right)}~dx$$

putting $$~u=\frac{1}{\sqrt{6}}~\tan\left(\frac{x}{2}\right)~$$, we have $$I=\frac{1}{\sqrt{6}}~\int \frac{du}{u^2+1}$$ $$=\frac{1}{\sqrt{6}}~\tan^{-1}u+C$$ $$=\frac{1}{\sqrt{6}}~\tan^{-1}\left(\frac{1}{\sqrt{6}}~\tan\left(\frac{x}{2}\right)\right)+C$$ where $$~C~$$ is constant.

we have $$\cos 2a=\frac{1-\tan^2\left(\frac{a}{2}\right)}{1+\tan^2\left(\frac{a}{2}\right)}$$

Use the so-called Weierstrass substitution: $$\cos(x)=\frac{1-t^2}{1+t^2}$$ $$dx=\frac{2dt}{1+t^2}$$