What went wrong in the evaluation of $\int \frac{1}{3-2\sin(x)}dx$?

Question

I tried to evaluate the following integral:

$$\int \frac{1}{3-2\sin(x)}dx\,\, $$ with universal substitution, using the fact that

$t(x):=\tan\left(\frac{x}{2}\right)$

$\sin(x)=\frac{2t(x)}{1+t(x)^2}$ and $t'(x)=\frac{1+t(x)^2}{2}$

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$\displaystyle\int \frac{1}{3-2\sin(x)}dx=$$\displaystyle\int \frac{1}{3-2(\frac{2t(x)}{1+t(x)^2})}dx=\displaystyle\int \frac{1}{3-\frac{4t(x)}{1+t(x)^2}}dx=\displaystyle\int \frac{1}{\frac{3+3t(x)^2-4t(x)}{1+t(x)^2}}dx=\displaystyle\int \frac{1+t(x)^2}{3+3t(x)^2-4t(x)}dx$

Now I substituted: $u:=t(x) \Longrightarrow dx= \frac{du}{t'(x)}=\frac{2du}{1+t(x)^2}$

So:

$\displaystyle\int \frac{1+u^2}{3+3u^2-4u}\frac{2}{1+u^2}du=\displaystyle\int \frac{2}{3+3u^2-4u}du=2\displaystyle\int \frac{1}{3u^2-4u+3}du=2\displaystyle\int \frac{1}{(\sqrt{3}u+\frac{2}{\sqrt{3}})^2+\frac{5}{3}}du$

Here I used the formular:

$\displaystyle\int \frac{1}{t^2+m^2}dt=\left[\frac{1}{m}\arctan\left(\frac{t}{m}\right)\right]$

So:

$2\displaystyle\int \frac{1}{(\sqrt{3}u+\frac{2}{\sqrt{3}})^2+\frac{5}{3}}du=2\left[\frac{\sqrt{3}}{\sqrt{5}}\arctan\left(\frac{(\sqrt{3}u-\frac{2}{\sqrt{3}})\sqrt{3}}{\sqrt{5}}\right)\right]=2\left[\frac{\sqrt{3}}{\sqrt{5}}\arctan\left(\frac{3u-2}{\sqrt{5}}\right)\right]$

Resubstituting:

$2\left[\frac{\sqrt{3}}{\sqrt{5}}\arctan\left(\frac{3\tan\left(\frac{x}{2}\right)-2}{\sqrt{5}}\right)\right]$

Here is the issue:

Wolfram tells:

$\displaystyle\int \frac{1}{3-2\sin(x)}dx=2\left[\frac{1}{\sqrt{5}}\arctan\left(\frac{3\tan\left(\frac{x}{2}\right)-2}{\sqrt{5}}\right)\right]$

I cannot figure out where I went wrong.. like.. its only one factor..

could someone maybe show me what went wrong? Thank you :)

Answer 1

You made a mistake when applying

$$\displaystyle\int \frac{1}{t^2+m^2}dt=\left[\frac{1}{m}\arctan\left(\frac{t}{m}\right)\right].$$

Define $t := (\sqrt{3}u + \frac{2}{\sqrt{3}})$ as you did implicitly, but make sure to rework $du$ into $c \cdot dt$.

Answer 2

Use the fact that, if $\color{blue}{\int f(x)dx = F(x)}$, then $\color{blue}{\int f(ax+b)dx = \frac{1}{a} F(x)}$

So, you'd have $$\frac{1}{\sqrt3}\cdot\left[\frac{2\sqrt3}{\sqrt5}\arctan\left(\frac{3u-2}{\sqrt5}\right)\right] =\frac{2}{\sqrt5}\arctan\left(\frac{3u-2}{\sqrt5}\right)$$