Csdrhrt

Hint given: Write the integrand as an integral.

I'm supposed to do this as double integration.

My attempt:

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

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

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

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

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

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.

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.

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

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 timesarctan{ 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

For the sake of an alternative approach, recall the formula for the integral of an inverse function
$$int f^{-1}(x)mathrm{d}x=xf^{-1}(x)-Fcirc f^{-1}(x)$$
Where $F'(x)=f(x)$. Plugging in $f(x)=tan(x)$,
$$I=intarctan(x)mathrm{d}x=xarctan(x)-int_0^{arctan(x)}tan(t)mathrm{d}t$$
Then recall that $$(-log|cos(x)|)'=tan(x)$$
So we have
$$I=xarctan(x)+log|cos(arctan(x))|$$
then using trig,
$$I=xarctan(x)-frac12log(x^2+1)$$
So
$$I_1=int_0^2arctan(x)mathrm{d}x=2arctan2-frac12log5$$
And
$$
begin{align}
I_2=&int_0^2arctan(pi x)mathrm{d}x\
=&frac1piint_0^{2pi}arctan(x)mathrm{d}x\
=&frac1pibigg(2piarctan2pi-frac12log(4pi^2+1)bigg)\
=&2arctan2pi-frac1{2pi}log(4pi^2+1)\
end{align}
$$
So
$$
begin{align}
int_0^2(arctanpi x-arctan x)mathrm{d}x=&I_2-I_1\
=&2arctan2pi-frac1{2pi}log(4pi^2+1)-2arctan2+frac12log5\
=&2arctan2pi-frac1{2}logsqrt[pi]{4pi^2+1}-2arctan2+frac12log5\
=&2(arctan2pi-arctan2)+frac12logfrac{5}{sqrt[pi]{4pi^2+1}}\
end{align}
$$