# Evaluate $\int_0^1x(\tan^{-1}x)^2~\textrm{d}x$

Evaluate $\int\limits_0^1x(\tan^{-1}x)^2~\textrm{d}x$

My Attempt

Let, $\tan^{-1}x=y\implies x=\tan y\implies dx=\sec^2y.dy=(1+\tan^2y)dy$ \begin{align} &\int\limits_0^1x(\tan^{-1}x)^2dx=\int\limits_0^{\pi/4}\tan y.y^2.(1+\tan^2y)dy\\ &=\int\limits_0^{\pi/4}\tan y.y^2dy+\int\limits_0^{\pi/4}\tan^3y.y^2dy\\ &=\bigg[y^2.\log|\sec y|-\int2y.\log|\sec y|dy\bigg]+\int_0^{\pi/4}\tan^3y.y^2dy\\ &=\bigg[y^2.\log|\sec y|-y^2.\log|\sec y|+\int\frac{\tan y\sec y}{\sec y}y^2dy\bigg]+\int_0^{\pi/4}\tan^3y.y^2dy\\ \end{align} How do I proceed further and solve the integration?

• Have you tried integration by parts? Commented May 24, 2018 at 16:24
• Try integration by parts with $\frac{x^2+1}2$ as antiderivative for $x$.
– A.Γ.
Commented May 24, 2018 at 16:25

Let's try integration by parts: \begin{align} \int x(\arctan x)^2\,dx &= \frac{x^2}{2}(\arctan x)^2- \int\frac{x^2}{2}2\arctan x\frac{1}{1+x^2}\,dx \\ &= \frac{x^2}{2}(\arctan x)^2- \int\frac{1+x^2-1}{1+x^2}\arctan x\,dx \\ &= \frac{x^2}{2}(\arctan x)^2- \int\arctan x\,dx+ \int\frac{1}{1+x^2}\arctan x\,dx \end{align} Now note that $$\int\arctan x\,dx=x\arctan x-\int\frac{x}{1+x^2}\,dx$$ and that $$\int\frac{1}{1+x^2}\arctan x\,dx= \frac{1}{2}(\arctan x)^2$$

The OP's initial change of variables will lead to an answer with a couple of integrations by parts and an intervening trig identity:

\begin{align} \int_0^1x(\arctan x)^2dx &=\int_0^{\pi/4}y^2\tan y\sec^2y\,dy\\ &={1\over2}y^2\tan^2y\,\Big|_0^{\pi/4}-\int_0^{\pi/4}y\tan^2y\,dy\quad(u=y^2,\,\, dv=\tan y\sec^2y\,dy)\\ &={1\over2}\left(\pi\over4\right)^2-\int_0^{\pi/4}y(\sec^2y-1)\,dy\quad\text{(trig identity)}\\ &={\pi^2\over32}+\int_0^{\pi/4}y\,dy-\int_0^{\pi/4}y\sec^2y\,dy\\ &={\pi^2\over32}+{\pi^2\over32}-\left(y\tan y\,\Big|_0^{\pi/4}-\int_0^{\pi/4}\tan y\,dy \right)\quad(u=y,\,\, dv=\sec^2y\,dy)\\ &={\pi^2\over16}-{\pi\over4}-\left(\ln\cos y\,\Big|_0^{\pi/4} \right)\\ &={\pi^2\over16}-{\pi\over4}-\ln(1/\sqrt2)+\ln(1)\\ &={\pi^2\over16}-{\pi\over4}+{1\over2}\ln2 \end{align}

Remarks: Once the change of variable left you integrating $y^2\times$trig($y$), you pretty much have to hammer away at the $y^2$ with integration by parts, first with $u=y^2$, then with $u=y$. The OP did a seemingly sensible thing, breaking $\int y^2(\tan y+\tan^3y)dy$ into separate pieces, $\int y^2\tan y\,dy$ and $\int y^2\tan^3y\,dy$. The problem is, neither of these pieces, by itself, has a nice antiderivative. Integration by parts on $\int y^2\tan y\,dy$, with $u=y^2$ and $dv=\tan y\,dy$, leaves $\int y\ln\sec y\,dy$, as the OP found, but at that point you're stuck. What the OP did was, in effect, to undo the first integration by parts, by letting $u=\ln\sec y$ and $dv=y\,dy$. The take-home lesson here is that trig identities and integration by parts are wonderful tools, but sometimes they'll lead you to dead ends, so you have to be ready, sometimes, to give up and start anew. Practice won't make perfect, but it will help you develop a nose for which pathways are likely to pay off.

Let $I$ be our integral and using integration by parts on $u=\arctan^2x$, we see that $I$ is equal as

\begin{align*}I & =\frac 12x^2\arctan^2x\,\Biggr\rvert_0^1-\int\limits_0^1dx\,\frac {x^2\arctan x}{1+x^2}\\ & =\frac {\pi^2}{32}-\int\limits_0^1dx\,\frac {x^2\arctan x}{1+x^2}\end{align*}

The second integral, again, can be solved using integration by parts with $u=\arctan x$. Note that for $dv$, by adding and subtracting one, we get$$v=x-\arctan x$$Hence\begin{align*}I & =\frac {\pi^2}{32}-\arctan x(x-\arctan x)\,\Biggr\rvert_0^1+\int\limits_0^1dx\,\frac x{1+x^2}-\int\limits_0^1dx\,\frac {\arctan x}{1+x^2}\\ & =\frac {\pi^2}{32}-\frac {\pi}4+\frac {\pi^2}{16}+\frac 12\log(1+x^2)\,\Biggr\rvert_0^1-\frac 12\arctan^2x\,\Biggr\rvert_0^1\end{align*}

Now it's just a matter of evaluating each result from its limits and using that $\arctan 1=\pi/4$ to get

$$\int\limits_0^1dx\, x\arctan^2x\color{blue}{=\frac {\pi^2}{16}-\frac {\pi}4+\frac 12\log2}$$

Don't rewrite the secant; instead keep the integral in the form $$\int_0^{\pi/4} y^2 \frac{\sin y}{\cos^3 y}\, dy$$ and apply integration by parts $\int u\, dv = uv - \int v\, du$ with $u = y^2$.

$$\int_ {0}^{1}x(arctan(x))^2dx$$ First solve the integral without boundaries $$\int x(arctan(x))^2dx$$ Apply Integration By Parts, where $u=\arctan^2(x),v^{\prime}=x$ $$=\arctan^2(x)\frac{x^2}{2}-\int \frac{2\arctan(x)}{1+x^2} (\frac{x^2}{2})dx$$ $$=\frac12 x^2\arctan^2(x)-\int \frac{x^2\arctan(x)}{x^2+1}dx$$ Note that $\int \frac{x^2\arctan(x)}{x^2+1}dx= x\arctan(x)-\arctan^2(x)+\ln|\frac{1}{\sqrt{1+x^2}}|+\frac12\arctan^2(x)$ So, $$=\frac12 x^2\arctan^2(x)-(x\arctan(x)-\arctan^2(x)+\ln|\frac{1}{\sqrt{1+x^2}}|+\frac12\arctan^2(x))$$ After simplifying we get, $$=\frac12 x^2\arctan^2(x)-x\arctan(x)+\frac12\arctan^2(x)-\ln|\frac{1}{\sqrt{1+x^2}}|$$ Now compute the boundaries and we get $$=\frac{\ln(2)}{2}+\frac{\pi^2}{16}-\frac{\pi}{4}$$ $$\int_ {0}^{1}x(arctan(x))^2dx=\frac{\ln(2)}{2}+\frac{\pi^2}{16}-\frac{\pi}{4}$$

A little trick on IBP \begin{aligned} \int_0^1 x(\arctan x)^2 d x = & \frac{1}{2} \int_0^1(\arctan x)^2 d\left(x^2+1\right) \\ = & \frac{1}{2}\left[\left(x^2+1\right)(\arctan x)^2\right]_0^1-\int_0^1 \arctan x d x \quad \textrm{( via IBP )}\\ = & \frac{1}{2}\left(2 \cdot \frac{\pi^2}{16}\right)-[x \arctan x]_0^1+\int_0^1 \frac{x d x}{1+x^2} \quad \textrm{( via IBP )} \\ = & \frac{\pi^2}{16}-\frac{\pi}{4}+\frac{1}{2} \ln 2 \end{aligned}