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I'm am trying to evaluate $$\int \frac{1}{3+4\tan x} \, dx.$$

I tried to use the universal trig substitution, that is

$$t = \tan{x \over 2}, \quad \tan x = {2t \over 1-t^2}, \quad dx = {2 \over t^2 + 1}dt$$

and after manipulation I got the integral to be

$$\int \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3} \, dt.$$

Is there a more clever approach to this problem or do I have to continue and face partial fractions?

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    $\begingroup$ Try $t = \tan(x)$ instead. $\endgroup$ Jun 16, 2016 at 23:17
  • $\begingroup$ At least the denominator factors so that you can do the partial fractions. This is in fact a general property of $$\int \frac{dx}{a+b\tan x}$$ which comes out to be $$\frac{ax + b\log(a\cos x + b \sin x)}{a^2+b^2}$$. $\endgroup$ Jun 16, 2016 at 23:43

4 Answers 4

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Hint:

Let

\begin{equation} I=\int\frac{1}{a+b\tan x}\ dx=\int\frac{\cos x}{a\cos x+b\sin x}\ dx \end{equation}

and

\begin{equation} J=\int\frac{\sin x}{a\cos x+b\sin x}\ dx \end{equation}

then

\begin{equation} aI+bJ=x+C_1\tag1 \end{equation}

and

\begin{equation} bI-aJ=\log(a\cos x+b\sin x)+C_2\tag2 \end{equation}

Setting $a=3$ and $b=4$, then it's easy to determine $I$ by linear combinations $(1)$ and $(2)$.

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Note that $$\frac{1}{3 + 4 \tan x} = \frac{\cos x}{3 \cos x + 4 \sin x} = f(x).$$

Next, observe that $$\log (3 \cos x + 4 \sin x) + C = \int \frac{4 \cos x - 3 \sin x}{3 \cos x + 4 \sin x} \, dx = \int g(x) \, dx,$$ via the obvious substitution $u = 3 \cos x + 4 \sin x$. Now the goal is to find a linear combination of the integrands $f$ and $g$ such that for some real-valued scalar constants $A$, $B$, $$A f(x) + B g(x) = 1.$$ Well, this is simple: we have $$A \cos x + B(4 \cos x - 3 \sin x) = 3 \cos x + 4 \sin x,$$ and collecting like terms, we obtain $$A + 4B = 3, \quad -3B = 4,$$ or $$A = \frac{25}{3}, \quad B = -\frac{4}{3}.$$ Therefore, $$x = \int 1 \, dx = \int A f(x) + B g(x) \, dx = A \int f(x) \, dx + B \int g(x) \, dx,$$ or $$\int f(x) \, dx = \frac{3}{25} \left( x + \frac{4}{3} \log (3 \cos x + 4 \sin x) \right) + C,$$ and the rest is trivial.

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For sure, as shown in comments and answers, the tangent half-angle substitution is not the easiest way for this problem. However, computing $$I=\int \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3} \, dt$$ is doable noticing that the roots of the denominator are $-\frac 13$, $i$,$-i$ and $3$. Using it, by partial fraction decomposition, we can arrive to $$ \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3}=-\frac{2 (4 t-3)}{25 \left(t^2+1\right)}+\frac{4}{25 (t-3)}+\frac{12}{25 (3 t+1)}$$ and get $$I=\frac{6}{25} \tan ^{-1}(t)-\frac{4}{25} \log \left(t^2+1\right)+\frac{4}{25} \log (t-3)+\frac{4}{25} \log (75 t+25)$$ and simplifying $$I=\frac{2}{25} \left(3 \tan ^{-1}(t)+2\log \left(\frac{25 (t-3) (3 t+1)}{t^2+1}\right)\right)$$

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It is probably easier to use the substitution $x=\arctan(t)$ then find a partial fraction decomposition for $$ g(t) = \frac{1}{(3+4t)(1+t^2)}. $$ By computing the residue at the simple poles $t\in\left\{-\frac{3}{4},i,-i\right\}$ it follows that: $$ g(t) = \frac{4}{25}\cdot\frac{1}{x+\frac{3}{4}}+\frac{1}{25}\cdot\frac{3-4t}{1+t^2} $$ and:

$$ \int g(t)\,dt = C + \frac{1}{25}\left[4\log\left(t+\frac{3}{4}\right)+3\arctan(t)-2\log(1+t^2)\right].$$

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  • $\begingroup$ Should $x+\frac{3}{4}$ be replaced by $t+\frac{3}{4}$? $\endgroup$
    – user84413
    Jun 17, 2016 at 22:38
  • $\begingroup$ @user84413: sure, it was a typo, now fixed. Thanks. $\endgroup$ Jun 17, 2016 at 22:43

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