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Hint given: Write the integrand as an integral.

I'm supposed to do this as double integration.

My attempt:

$$\int_0^2 [\tan^{-1}y]^{\pi x}_{x}$$

$$= \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \mathrm{d}x} {y^2+1}$$

$$= \int_2^{2\pi} \int_{\frac{y}{\pi}}^2 \frac { \mathrm{d}x \mathrm{d}y} {y^2+1}$$

$$= \int_2^{2\pi} \frac { [x]^2_{\frac{y}{\pi}} \mathrm{d}y } { y^2+1}$$

$$= \int_2^{2\pi} \frac { 2- {\frac{y}{\pi}} \mathrm{d}y } { y^2+1}$$

Carrying out this integration, I got, $$2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]$$

But I'm supposed to get $$2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]+ [\frac {\pi-1}{2 \pi}] \ln 5$$

Can someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you.

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    $\begingroup$ At third line it seems to me the integral must be $$\int_0^{2\pi}\int_{\frac{y}{\pi}}^{y}\frac{\operatorname d\!x \operatorname d\!y}{y^2+1}$$ $\endgroup$ – Ángel Mario Gallegos Nov 4 '14 at 6:13
  • $\begingroup$ but that is taking my answer even further away from what I require. $\endgroup$ – Diya Nov 4 '14 at 6:50
  • $\begingroup$ I got it! You were right! The integral is divided into two regions. One is the one I used, and the other is the one you mentioned. Thank you so much! :) $\endgroup$ – Diya Nov 4 '14 at 6:57
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Use the change of variables $y = xt.$

$$I = \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \, \mathrm{d}x} {y^2+1}\\=\int_0^2 \int_1^{\pi } \frac { x} {x^2t^2+1}\mathrm{d}t \, \mathrm{d}x\\=\int_1^{\pi} \int_0^{2 } \frac { x} {x^2t^2+1}\mathrm{d}x \, \mathrm{d}t\\=\int_1^{\pi} \frac{\ln(1+4t^2)}{2t^2} \mathrm{d}t$$

Now use integration by parts.

$$I = -\left.\frac{\ln(1+4t^2)}{2t}\right|_1^{\pi}+4\int_1^{\pi}\frac1{1+4t^2} \, dt$$

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A more straight forward approach uses integration by parts.

Define: \begin{align} & I(c)=\int_{a}^{b}dx(1 \times \arctan{c x})=\int_{ac}^{bc}\frac{dy}{c} (1 \times\arctan{ y})=\\&\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\int_{ac}^{bc}\frac{y}{1+y^2} \end{align}

using partial fraction this reads: \begin{align} I(c)=\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\log(1+y^2)|_{ac}^{bc} \end{align}

taking $I(\pi)-I(0)$ with $a=0$ and $b=2$ we are done

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For the sake of an alternative approach, recall the formula for the integral of an inverse function $$\int f^{-1}(x)\mathrm{d}x=xf^{-1}(x)-F\circ f^{-1}(x)$$ Where $F'(x)=f(x)$. Plugging in $f(x)=\tan(x)$, $$I=\int\arctan(x)\mathrm{d}x=x\arctan(x)-\int_0^{\arctan(x)}\tan(t)\mathrm{d}t$$ Then recall that $$(-\log|\cos(x)|)'=\tan(x)$$ So we have $$I=x\arctan(x)+\log|\cos(\arctan(x))|$$ then using trig, $$I=x\arctan(x)-\frac12\log(x^2+1)$$ So $$I_1=\int_0^2\arctan(x)\mathrm{d}x=2\arctan2-\frac12\log5$$ And $$ \begin{align} I_2=&\int_0^2\arctan(\pi x)\mathrm{d}x\\ =&\frac1\pi\int_0^{2\pi}\arctan(x)\mathrm{d}x\\ =&\frac1\pi\bigg(2\pi\arctan2\pi-\frac12\log(4\pi^2+1)\bigg)\\ =&2\arctan2\pi-\frac1{2\pi}\log(4\pi^2+1)\\ \end{align} $$ So $$ \begin{align} \int_0^2(\arctan\pi x-\arctan x)\mathrm{d}x=&I_2-I_1\\ =&2\arctan2\pi-\frac1{2\pi}\log(4\pi^2+1)-2\arctan2+\frac12\log5\\ =&2\arctan2\pi-\frac1{2}\log\sqrt[\pi]{4\pi^2+1}-2\arctan2+\frac12\log5\\ =&2(\arctan2\pi-\arctan2)+\frac12\log\frac{5}{\sqrt[\pi]{4\pi^2+1}}\\ \end{align} $$

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