Inverse trigonometric integral

Put $t = \ln x$ $\Rightarrow \frac{1}{x} \text{dx} = \text{dt}$

$\Rightarrow I(a) = \displaystyle \int_{- \ln a}^{\ln a}\tan^{-1} (e^t) \text{dt}$

= $\displaystyle \int_{0}^{\ln a} (\tan^{-1} e^t + \tan^{-1} e^{-t}) \text{dt}$

= $\displaystyle \int_{0}^{\ln a} (\tan^{-1} e^t + \cot^{-1} e^{t}) \text{dt}$ ( $e^t > 0$ )

= $\displaystyle \int_{0}^{\ln a} \frac{\pi}{2} \text{dt}$ = $\boxed{\frac{\pi}{2} \ln a}$

Rewrite $I(a) = \int_{1/a}^1 \frac{\arctan x}{x} \, dx + \int_1^a \frac{\arctan x}{x} \, dx$ . Set $y = 1/x$ . Then $dy/y = -dx/x$ and $\arctan y = \pi/2 - \arctan x$ . Putting it all together, we get $I(a) = \int_1^a \frac{\pi/2-\arctan y}{y} \, dy + \int_1^a \frac{\arctan x}{x} \, dx,$ which simplifies to $\int_1^a \frac{\pi/2}{x} \, dx = \frac{\pi}2 \ln(a)$ . Plugging in $3$ gives an approximate answer of $\fbox{1.726}$ .

Use the substitution $\ln x=t \Rightarrow dx=e^t\,dt$ .

Hence,

$\displaystyle I(a)=\int_{-\ln a}^{\ln a} \frac{\arctan(e^t)}{e^t}e^t\,dt=\int_{-\ln a}^{\ln a} \arctan(e^t)\,dt \,\, (*)$

From the property,

$\displaystyle \int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$

We have,

$\displaystyle I(a)=\int_{-\ln a}^{\ln a} \arctan(e^{-t})\,dt=\int_{-\ln a}^{\ln a} \text{arccot}(e^t)\,dt \,\,\, (**)$

Add $(*)$ and $(**)$ to get:

$\displaystyle 2I(a)=\int_{-\ln a}^{\ln a} \frac{\pi}{2}\,dt=\pi\ln a \Rightarrow \boxed {I(a)=\dfrac{\pi \ln a}{2}}$

Hence, $I(3)=1.725 \Rightarrow \boxed{\lceil{I(3)\rceil}=2}$

Answer: $\boxed{2}$ .

Let use the substitution $x=\frac{1}{u}$ . Then we'll have $dx=-\frac{1}{u^2}\,du$ and

$I(a)=\int^{\frac{1}{a}}_{a}\frac{\arctan{\frac{1}{u}}}{\frac{1}{u}}\cdot\left(-\frac{1}{u^2}\right)\,du=\int^{a}_{\frac{1}{a}}\frac{\arctan{\frac{1}{u}}}{u}\,du.$

Now we can write

$2I(a)=\int^{a}_{\frac{1}{a}}\frac{\arctan x+\arctan\frac{1}{x}}{x}\,dx=\frac{\pi}{2}\int^{a}_{\frac{1}{a}}\frac{dx}{x}=\pi\ln a$

Finally we obtain $I(a)=\dfrac{\pi \ln a}{2}$ and $\lceil I(3)\rceil=\lceil \dfrac{\pi \ln 3}{2}\rceil=\boxed{2}$ .

By newton's leibnitz rule we have ${ I }^{ ' }(a)=\frac { \tan ^{ -1 }{ a } }{ a } -(\frac { -1 }{ { a }^{ 2 } } )\frac { \tan ^{ -1 }{ \frac { 1 }{ a } } }{ \frac { 1 }{ a } } \\ =\frac { 1 }{ a } (\tan ^{ -1 }{ a } +\tan ^{ -1 }{ \frac { 1 }{ a } } )=\frac { \pi }{ 2a }$

$\Rightarrow I(a)=\frac { \pi }{ 2 } ln(a)+C$

Now $I(1)=0,\Rightarrow C=0\\ \Rightarrow I(a)=\frac { \pi }{ 2 } ln(a)$

The solution with C Programming!

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#include <stdio.h>
#include <math.h>
#include <string.h>
#define pi acos(-1)
int main()
{

double s=0,x=0.333333333,dx=pow(10,-2),a,b;
while(x<=3)

{
s+=(atan(x)*dx)/x;  
x+=dx;
}
printf("%lf",s);
printf("\n\n");

return 0;
}