calculate-cosx-cos-3x-dx-

Question Number 100088 by mathmax by abdo last updated on 24/Jun/20

calculate \int \frac{cosx}{\cos (3 x)} dx

Commented by Dwaipayan Shikari last updated on 24/Jun/20

- \frac{1}{2 \sqrt{3}} l o g (\frac{\sqrt{3} s i n θ - c o s θ}{\sqrt{3} s i n θ + c o s θ}) + C o n s t a n t

Answered by 1549442205 last updated on 25/Jun/20

\cos 3 x = {4 \cos}^{3} x - 3 \cos x, so F = \int \frac{dx}{{4 \cos}^{2} x - 3} = \int \frac{dx}{2 (1 + \cos 2 x) - 3}

= \int \frac{dx}{2 \cos 2 x - 1} = \frac{1}{2} \int \frac{dx}{\cos 2 x - \frac{1}{2}}

\underset{t = tanx}{=} \frac{1}{2} \int \frac{dt}{(1 + t^{2}) (\frac{1 - t^{2}}{1 + t^{2})} - \frac{1}{2})} = \frac{1}{2} \int \frac{dt}{1 - t^{2} - \frac{1 + t^{2}}{2}}

= \int \frac{dt}{1 - {3 t}^{2}} = \int \frac{dt}{(1 - \sqrt{3} t) (1 + \sqrt{3} t)} = \frac{1}{2} \int (\frac{1}{1 - \sqrt{3} t)} + \frac{1}{1 + \sqrt{3} t}) dt

= \frac{1}{2} \int \frac{dt}{1 + \sqrt{3} t} - \frac{1}{2} \int \frac{dt}{\sqrt{3} t - 1} = \frac{1}{2 \sqrt{3}} \int \frac{d (1 + \sqrt{3} t)}{1 + \sqrt{3} t} - \frac{1}{2 \sqrt{3}} \int \frac{d (\sqrt{3} t - 1)}{\sqrt{3} t - 1}

= \frac{1}{2 \sqrt{3}} \ln ∣ \frac{\sqrt{3} t + 1}{\sqrt{3} t - 1} ∣= \frac{1}{2 \sqrt{3}} \ln ∣ \frac{\sqrt{3} \arctan x + 1}{\sqrt{3} \arctan x - 1} ∣ + C