dx-3-tan-x-

Question Number 158159 by cortano last updated on 31/Oct/21

\int \frac{d x}{3 - \tan x} = ?

Answered by peter frank last updated on 31/Oct/21

\int \frac{\cos x}{3 \cos x - \sin x} dx

t - substitution

t = \tan \frac{x}{2}

\sin x = \frac{2 t}{1 + t^{2}}

\cos x = \frac{1 - t^{2}}{1 + t^{2}}

Answered by bobhans last updated on 31/Oct/21

let \tan x = h \Rightarrow dh = \sec^{2} x dx

\Rightarrow dx = \frac{dh}{1 + h^{2}}

I = \int \frac{dh}{(1 + h^{2}) (3 - h)}

(*) \frac{1}{(1 + h^{2}) (3 - h)} = \frac{A}{3 - h} + \frac{Bh + C}{1 + h^{2}}

(∙) A = {[\frac{1}{1 + h^{2}}]}_{h = 3} = \frac{1}{10}

(∙) \lim_{h \to \infty} hf (h) = \lim_{h \to \infty} (\frac{h}{10 (3 - h)} + \frac{{Bh}^{2} + Ch}{1 + h^{2}}) = 0

\Rightarrow - \frac{1}{10} + B = 0 \to B = \frac{1}{10}

F (0) = \frac{1}{3} = \frac{1}{30} + C \Rightarrow C = \frac{3}{10}

I = \int \frac{1}{10 (3 - h)} dh + \int \frac{h + 3}{10 (1 + h^{2})} dh

I = - \frac{1}{10} \ln ∣ 3 - h ∣ + \frac{1}{20} \int \frac{2 h + 6}{1 + h^{2}} dh

I = - \frac{1}{10} \ln ∣ 3 - h ∣ + \frac{1}{20} \int \frac{d (1 + h^{2})}{1 + h^{2}} + \frac{3}{10} \int \frac{dh}{1 + h^{2}}

I = - \frac{1}{10} \ln ∣ 3 - h ∣ + \frac{1}{20} \ln ∣ 1 + h^{2} ∣ + \frac{3}{10} \arctan (h) + c

I = - \frac{1}{10} \ln ∣ 3 - \tan x ∣ + \frac{1}{20} \ln ∣ \sec^{2} x ∣ + \frac{3}{10} \arctan (\tan x) + c

I = \frac{3 x}{10} - \frac{1}{10} \ln ∣ 3 - \tan x ∣ - \frac{1}{10} \ln ∣ \cos x ∣ + c

I = \frac{3 x}{10} - \frac{1}{10} \ln ∣ 3 \cos x - \sin x ∣ + c

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