Inverse trig function integration by parts.

Question

1. $$\int \tan^{-1}{2y}dy$$

if I choose $u = \tan^{-1}{2y}$ then $du = \frac{2}{1 + (2y)^2} dy$

and

$dv = dy$ and $v = y$

But I have a more complicated du. What else can I do?

Answer 1

Another approach. Let $2y = \tan(\theta) \rightarrow 2\frac{dy}{d\theta} = \sec^2(\theta)$ and finally $dy = \frac{1}{2}\sec^2(\theta)d\theta$.

Thus,

\begin{align} I =\int \arctan(2y)\:dy = \int \arctan\left(\tan(\theta)\right) \cdot \frac{1}{2}\sec^2(\theta)\:d\theta = \frac{1}{2}\int \theta \sec^2(\theta)\:d\theta \end{align}

Now, integrate by parts:

\begin{align} v'(\theta) &= \sec^2(\theta) & u(\theta) &= \theta \\ v(\theta) &= \tan(\theta) & u'(\theta) &= 1 \end{align}

Thus:

\begin{align} I &= \frac{1}{2}\int \theta \sec^2(\theta)\:d\theta = \frac{1}{2}\left[\theta\tan(\theta) - \int \tan(\theta) \:d\theta \right] \\ &= \frac{1}{2}\left[\theta\tan(\theta) - \ln\left|\sec(\theta) \right| \right] + C \end{align}

Where $C$ is the constant of integration. Recall that $2y = \tan(\theta)$ and hence $\theta = \arctan(2y)$. For $\sec(\theta)$ we use the identity $\sec(\theta) = \sqrt{\tan^2(\theta)+ 1} = \sqrt{\left(2y\right)^2 + 1} = \sqrt{4y^2 + 1}$. Hence we arrive at

\begin{equation} I = \int \arctan(2y)\:dy = \frac{1}{2}\left[2y \arctan(2y) - \ln\left| \sqrt{4y^2 + 1} \right|\right] + C = y \arctan(2y) - \frac{1}{4}\ln\left| 4y^2 + 1 \right| + C \end{equation}

Answer 2

A neat identity.

Given some continuous inverse function $f^{-1}(x)$, we can find the antiderivative of this inverse function. Here's how.

Consider the integral $$I=\int f^{-1}(x)\mathrm dx$$ Assuming that we know what $f(x)$ is, we may preform the substitution $x=f(t)$ meaning that $\mathrm dx=f'(t)\mathrm dt$. Hence we have $$I=\int f^{-1}\left[f(t)\right]f'(t)\mathrm dt$$ $$I=\int tf'(t)\mathrm dt$$ We then integrate by parts: $$\mathrm dv=f'(t)\mathrm dt\Rightarrow v=f(t)\\ u=t\Rightarrow \mathrm du=\mathrm dt$$ Which gives $$I=tf(t)-\int f(t)\mathrm dt$$ And assuming that $F'(x)=f(x)$, $$I=tf(t)-F(t)+C$$ And since $x=f(t)$ we have that $t=f^{-1}(x)$ which gives $$I=xf^{-1}(x)-F\left[f^{-1}(x)\right]+C$$

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So, choosing $f^{-1}(x)=\arctan ax$ for some constant $a\neq 0$, we have that $f(x)=\frac1a\tan x$. Hence $$\begin{align} I&=x\arctan ax-\frac1a\int \tan t \,\mathrm dt\\ &=x\arctan ax+\frac1a\ln\left|\cos t\right|+C\\ &=x\arctan ax+\frac1a\ln\left|\cos[\arctan ax]\right|+C \end{align}$$ And from $$\cos[\arctan z]=\frac1{\sqrt{1+z^2}}$$ We have, after some simplification, $$\int\arctan(ax)\mathrm dx=x\arctan(ax)-\frac1{2a}\ln(1+a^2x^2)+C$$ And your integral is given by the case $a=2$.

Answer 3

$$\int\arctan2ydy=y\arctan2y-\int\frac{2y}{1+4y^2}dy=y\arctan2y-\frac{1}{4}\ln(1+4y^2)+C.$$

Answer 4

Integrating by parts, $\int tan^{-1}(2y)\, dy=y\, tan^{-1}(2y) -\int \frac {2y} {1+4y^{2}} \, dy+c$. Put $1+4y^{2}=u$ to evaluate this.