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Question Number 68868 by mathmax by abdo last updated on 16/Sep/19

$$find\int \frac{dx}{a+cosx}witha>0$$

Commented bymathmax by abdo last updated on 17/Sep/19

$$letI=\int \frac{dx}{a+cosx}changementtan\left(\frac{x}{2}\right)=tgive$$ $$I=\int \frac{1}{a+\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=\int \frac{2dt}{a+{at}^2+1-t^2}=\int \frac{2dt}{(a-1)t^2+a+1}$$ $$=\frac{2}{a-1}\int \frac{dt}{t^2+\frac{a+1}{a-1}}case10<a<1\Rightarrow I=\frac{2}{a-1}\int \frac{dt}{t^2-\frac{1+a}{1-a}}$$ $$=\frac{2}{a-1}\int \frac{dt}{t^2-{\left(\sqrt{\frac{1+a}{1-a}}\right)}^2}{=}_{t=\sqrt{\frac{1+a}{1-a}}u}\frac{2}{a-1}\frac{1-a}{1+a}\int \frac{1}{u^2-1}\sqrt{\frac{1+a}{1-a}}du$$ $$=\frac{2}{1+a}\times \frac{\sqrt{1+a}}{\sqrt{1-a}}\int (\frac{1}{u-1}-\frac{1}{u+1})du=\frac{1}{\sqrt{1-a^2}}ln\mid \frac{u-1}{u+1}\mid +c$$ $$=\frac{1}{\sqrt{1-a^2}}ln\mid \frac{\sqrt{\frac{1-a}{1+a}}tan\left(\frac{x}{2}\right)-1}{\sqrt{\frac{1-a}{1+a}}tan\left(\frac{x}{2}\right)+1}\mid +c$$ $$case2a>1\Rightarrow I{=}_{t=\sqrt{\frac{a+1}{a-1}}u}\frac{2}{a-1}\frac{a-1}{a+1}\int \frac{1}{u^2+1}\sqrt{\frac{a+1}{a-1}}du$$ $$=\frac{2}{\sqrt{a^2-1}}arctan\left(u\right)+c=\frac{2}{\sqrt{a^2-1}}arctan\left(\sqrt{\frac{a-1}{a+1}}t\right)+c.$$

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