Board index **‹** Computation of Integrals **‹** Arctan integral
## Arctan integral

I tried to take a crack at it (letting Wolfram Alpha take care of all the nasty parts), but I didn't get a nice short solution that I felt like typing out. So instead I'll show the work I did so someone else can either work off it or find a better method, or maybe even show that there is a nicer form than what I will suggest. If I've made any errors please let me know.

To warm up I first did a simpler integral:

$$J=\int_0^1 arctan\left(\frac{88\sqrt{21}}{36x^2+215}\right)dx=arctan\left ( \frac{88\sqrt{21}}{251} \right )+\frac{u+\overline{u}}{2742}$$

with

$$u=w_1 w_2 arctan\left(\frac{6}{\overline{w_1}}\right)$$

where $w_1=\sqrt{215+i88\sqrt{21}}$ and $w_2=i(\overline{w_1})^2$.

Method of attack for J:

I let $f(x,\;t)=\frac{\partial}{\partial t}arctan\left(\frac{t88\sqrt{21}}{36x^2+215}\right)$, then considered

$$U_1=\int\int f(x,\;t) dx dt.$$

Setting all constants of integration in $U_1$ to $0$ and $t=1$, I call this new formula $U_2$. We can then calculate that

$$U_2| _{x=0}^{x=1}=J.$$

Attacking FDP's integral:

$$I=\int_0^1\frac{arctan\left(\frac{88\sqrt{21}}{36x^2+215}\right)}{\sqrt{1-x^2}}dx$$

I let

$$I'(t)=\int_0^1\frac{f(x,\;t)}{\sqrt{1-x^2}}dx$$

where $f(x,\;t)$ is the same as it was for $J$.

The roots of the denominator of $f(x,\;t)$ are $\pm\alpha(t)=\pm\sqrt{-215+i88\sqrt{21}}$ and $\pm\beta(t)=\pm\sqrt{-215-i88\sqrt{21}}$ so I went with partial fraction expansion as the next step. This led to me consider double integrals of the form

$$F=\int\int\frac{a}{\left(x-h\sqrt{u+vt}\right)\sqrt{1-x^2}}dxdt.$$

Typing ∫∫a/((x-h√(u+v*t))*√(1-x^2)) dx dt into Wolfram Alpha gives a messy result. But since four integrals of this form are being added together I wouldn't be surprised if some stuff cancelled out, and some of the logs could probably be combined into arctans.

I'd assume that using the $F$ integrals with proper values of $h,\;u,\;v$, and $a$ to account for the values of the roots and the coefficient obtained by the partial fraction expansion I could proceed by doing

$$H_1=\int\int\frac{f(x,\;t)}{\sqrt{1-x^2}}dxdt$$

then setting $t=1$ and all constants of integration to $0$ and calling this new formula $H_2$. After that we should have

$$H_2|_{x=0}^{x=1}=I.$$

I stopped at this point because the number of terms was making my head spin.

Note: For $F$, $a$ may have been a function of $t$. I didn't spend the time to find the coefficients $a$ from the partial fraction expansion, but I'm really thinking that all $a$ are indeed functions of $t$. If this is the case then everything after my defintion of $F$ is wrong!

Edit: Yeah, $a$ is a function of $t$ so the result Wolfram gave for $F$ is useless and my method is flawed.

**Moderators:** galactus, Random Variable, sos440

2 posts
• Page **1** of **1**

$\displaystyle \int_0^1 \dfrac{\arctan\left(\dfrac{88\sqrt{21}}{36x^2+215}\right)}{\sqrt{1-x^2}}dx$

To warm up I first did a simpler integral:

$$J=\int_0^1 arctan\left(\frac{88\sqrt{21}}{36x^2+215}\right)dx=arctan\left ( \frac{88\sqrt{21}}{251} \right )+\frac{u+\overline{u}}{2742}$$

with

$$u=w_1 w_2 arctan\left(\frac{6}{\overline{w_1}}\right)$$

where $w_1=\sqrt{215+i88\sqrt{21}}$ and $w_2=i(\overline{w_1})^2$.

Method of attack for J:

I let $f(x,\;t)=\frac{\partial}{\partial t}arctan\left(\frac{t88\sqrt{21}}{36x^2+215}\right)$, then considered

$$U_1=\int\int f(x,\;t) dx dt.$$

Setting all constants of integration in $U_1$ to $0$ and $t=1$, I call this new formula $U_2$. We can then calculate that

$$U_2| _{x=0}^{x=1}=J.$$

Attacking FDP's integral:

$$I=\int_0^1\frac{arctan\left(\frac{88\sqrt{21}}{36x^2+215}\right)}{\sqrt{1-x^2}}dx$$

I let

$$I'(t)=\int_0^1\frac{f(x,\;t)}{\sqrt{1-x^2}}dx$$

where $f(x,\;t)$ is the same as it was for $J$.

The roots of the denominator of $f(x,\;t)$ are $\pm\alpha(t)=\pm\sqrt{-215+i88\sqrt{21}}$ and $\pm\beta(t)=\pm\sqrt{-215-i88\sqrt{21}}$ so I went with partial fraction expansion as the next step. This led to me consider double integrals of the form

$$F=\int\int\frac{a}{\left(x-h\sqrt{u+vt}\right)\sqrt{1-x^2}}dxdt.$$

Typing ∫∫a/((x-h√(u+v*t))*√(1-x^2)) dx dt into Wolfram Alpha gives a messy result. But since four integrals of this form are being added together I wouldn't be surprised if some stuff cancelled out, and some of the logs could probably be combined into arctans.

I'd assume that using the $F$ integrals with proper values of $h,\;u,\;v$, and $a$ to account for the values of the roots and the coefficient obtained by the partial fraction expansion I could proceed by doing

$$H_1=\int\int\frac{f(x,\;t)}{\sqrt{1-x^2}}dxdt$$

then setting $t=1$ and all constants of integration to $0$ and calling this new formula $H_2$. After that we should have

$$H_2|_{x=0}^{x=1}=I.$$

I stopped at this point because the number of terms was making my head spin.

Note: For $F$, $a$ may have been a function of $t$. I didn't spend the time to find the coefficients $a$ from the partial fraction expansion, but I'm really thinking that all $a$ are indeed functions of $t$. If this is the case then everything after my defintion of $F$ is wrong!

Edit: Yeah, $a$ is a function of $t$ so the result Wolfram gave for $F$ is useless and my method is flawed.

2 posts
• Page **1** of **1**