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$$\int \frac{1}{3\sin(x)+2\cos(x)+3}\ \text{d}x$$ This is one of the question which appeared in my exam today and I tried solving this but just couldn't solve it. I tried converting sine and cosine in terms of $\tan\left(\frac{x}{2}\right)$ but didn't help very well..

Does anyone have idea how to evaluate this integral?

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    $\begingroup$ use the tan half angle formulas $\endgroup$ Commented Oct 20, 2015 at 9:32
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    $\begingroup$ Please explain how the $u=\tan(x/2)$ "didn't help very well". It should. $\endgroup$
    – mickep
    Commented Oct 20, 2015 at 9:33
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    $\begingroup$ after two minutes you know that this will not help? $\endgroup$ Commented Oct 20, 2015 at 9:37
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    $\begingroup$ see here en.wikibooks.org/wiki/Calculus/Integration_techniques/… $\endgroup$ Commented Oct 20, 2015 at 9:37
  • $\begingroup$ @Dr.SonnhardGraubner. I know that we use this type of substitution for this form of integral, but isn't it too lengthy? I got the answer, but there has to be a shorter method also? Please can you point me towards it? $\endgroup$ Commented Oct 20, 2015 at 9:55

4 Answers 4

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Notice, $$\int \frac{1}{3\sin x+2\cos x+3}\ dx$$ $$=\int \frac{1}{3\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}+2\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+3}\ dx$$ $$=\int \frac{1+\tan^2\frac{x}{2}}{\tan^2\frac{x}{2}+6\tan\frac{x}{2}+5}\ dx$$ $$=\int \frac{\sec^2\frac{x}{2}}{\left(\tan\frac{x}{2}+3\right)^2-4}\ dx$$ Let $\tan\frac{x}{2}=t\implies \frac{1}{2}\sec^2\frac{x}{2}\ dx=dt$ $$=\int \frac{2\ dt}{t^2-4}$$ $$=\frac{2}{4}\ln\left|\frac{t-2}{t+2}\right|+C$$ $$=\color{blue}{\frac{1}{2}\ln\left|\frac{\tan\frac{x}{2}-2}{\tan\frac{x}{2}+2}\right|+C}$$

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HINT:

$$\int \frac{1}{3\sin(x)+2\cos(x)+3}\text{d}x =$$


Substitute $u=\tan\left(\frac{x}{2}\right)$ and $\text{d}u=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)\text{d}x$. Than transform the integrand using the substitutions $\sin(x)=\frac{2u}{u^2+1}$, $\cos(x)=\frac{1-u^2}{u^2+1}$ and $\text{d}x=\frac{2}{u^2+1}\text{d}u$:


$$\int \frac{2}{(u^2+1)\cdot\left(\frac{6u}{u^2+1}+\frac{2(1-u^2)}{u^2+1}+3\right)}\text{d}u =$$ $$\int \frac{2}{u^2+6u+5}\text{d}u =$$ $$2\int \frac{1}{(u+3)^2-4}\text{d}u=$$


Substitute $s=u+3$ and $\text{d}s=\text{d}u$:


$$2\int \frac{1}{s^2-4}\text{d}s=$$ $$2\int -\frac{1}{4\left(1-\frac{s^2}{4}\right)}\text{d}s=$$ $$-\frac{1}{2}\int \frac{1}{1-\frac{s^2}{4}}\text{d}s=$$


Substitute $p=\frac{s}{2}$ and $\text{d}p=\frac{1}{2}\text{d}s$:


$$-\int \frac{1}{1-p^2}\text{d}p=$$ $$-\tanh^{-1}(p)+C$$

Substitute everything back and you'll get the final answer!

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    $\begingroup$ Sure you've to substitute everything back $\endgroup$ Commented Oct 20, 2015 at 9:48
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Substitute $3\sin x+2\cos x=\sqrt{13}\sin t$ \begin{align} &\int \frac{1}{3\sin x+2\cos x+3}\ dx\\ =& \int \frac1{\sqrt{13}\sin t+3}dt = \int \frac{\sec^2t\ ( \sqrt{13}\sin t+3) }{\sec^2t\ (\sqrt{13}\sin t+3)^2}dt\\ =& \int \frac{d(\sqrt{13}\sec t+3\tan t)}{(\sqrt{13}\sec t+3\tan t)^2-4} =-\frac12\coth^{-1}\frac{\sqrt{13}\sec t+3\tan t}{2}\\ =& -\frac12\coth^{-1}\frac{{13}\sec x+9\tan x+6}{6-4\tan x}\\ \end{align}

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A slight twist on the tangent half-angle substitution produces an even simpler integral,

$$\begin{align*} \int\frac{dx}{3\sin x+2\cos x+3} &= \int \frac{-\frac2{1+y^2}\,dy}{3\frac{1-y^2}{1+y^2} + 2\frac{2y}{1+y^2}+3} & y=\tan\left(\frac\pi4-\frac x2\right) \\ &= -\int \frac{dy}{2y+3} \end{align*}$$

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