Integration Bee 2022

UNSW Maths Society

Round 1

Question 16

Fun question, I don’t think my method was the intended!

∫1xx−x2 dxegin{align*} int rac{1}{ sqrt{xsqrt{x}-x^2} } dx end{align*}

Start by considering domain – the function is only defined if the numerator is nonzero, and this only happens if

xx−x2>0xx−x2>0xx>x2x3/2>x4/2egin{align*} sqrt{xsqrt{x}-x^2} &> 0 \ xsqrt{x}-x^2 &> 0 \ xsqrt{x} > x^2 \ x^{3/2} > x^{4/2} end{align*}

The only region where xn>xn+kx^n > x^{n+k} is true (i.e. a lower exponent of xx results in a larger value) is 0<x<10 < x < 1, where exponentiating xx makes it smaller instead of larger. So our integrand is only defined for 0<x<10 < x < 1.

We’ll change the fraction into an exponent for clarity:

 ∫1xx−x2 dx= ∫[xx−x2]−1 dxegin{align*} & int rac{1}{ sqrt{xsqrt{x}-x^2} } dx \ =& int left[ sqrt{xsqrt{x}-x^2} ight]^{-1} dx end{align*}

Now factorise out x\sqrt{x}:

=∫[x(x−x)]−1 dx=∫[xx−x]−1 dx=∫1xx−x dxegin{align*} &= int left[ sqrt{xleft(sqrt{x}-x ight)} ight]^{-1} dx \ &= int left[ sqrt{x}sqrt{sqrt{x}-x} ight]^{-1} dx \ &= int rac{1}{sqrt{x}sqrt{sqrt{x}-x}} dx end{align*}

We need to check domain when we do this – the leftover x−x\sqrt{\sqrt{x}-x} must still be valid. Since

x>xfor 0<x<1x−x>0⇒x−x is definedegin{align*} sqrt{x} &> x quad ext{for} 0 < x < 1 \ sqrt{x} - x &> 0 \ Rightarrow sqrt{sqrt{x} - x} & ext{ is defined} end{align*}

So we’re good to go. We can now substitute x=t\sqrt{x} = t:

 ∫1xx−x dx= 2∫1x−x⋅12x dx= 2∫1t−t2 dtegin{align*} & int rac{1}{sqrt{x}sqrt{sqrt{x}-x}} dx \ =& 2 int rac{1}{sqrt{sqrt{x}-x}} cdot rac{1}{2sqrt{x}} dx \ =& 2 int rac{1}{sqrt{t-t^2}} dt end{align*}

The rest is just trig sub WORK:

=2∫1−(t2−t) dx=2∫1−((t−12)2−14) dx=2∫114−(t−12)2 dxegin{align*} &= 2 int rac{1}{sqrt{-(t^2-t)}} dx \ &= 2 int rac{1}{sqrt{-((t- rac{1}{2})^2- rac{1}{4})}} dx \ &= 2 int rac{1}{sqrt{ rac{1}{4}-(t- rac{1}{2})^2}} dx end{align*} (t−12)2=14sin⁡2vt−12=12sin⁡vdt=12cos⁡v dvegin{align*} left( t- rac{1}{2} ight)^2 &= rac{1}{4} sin^2{v} \ t- rac{1}{2} &= rac{1}{2} sin{v} \ dt &= rac{1}{2} cos{v} dv end{align*} =2∫114−14sin⁡2v⋅12cos⁡v dv=2∫ dv=2v=2sin⁡−1(2t−1)=2sin⁡−1(2x−1)−cegin{align*} &= 2 int rac{1}{sqrt{ rac{1}{4}- rac{1}{4}sin^2{v}}} cdot rac{1}{2} cos{v} dv \ &= 2 int dv \ &= 2v \ &= 2sin^{-1}left( 2t-1 ight) \ &= 2sin^{-1}left( 2sqrt{x}-1 ight) - c end{align*}

Interesting that this is equivalent to the canonical answer of 4sin⁡−1(x1/4)4\sin^{-1}\left( x^{1/4} \right), but differs by π\pi.

Final

Question 1

Damn, all my recreational integration really paid off for this one.

 ∫0∞1(x+1/x)2 dx= ∫0∞1(x+1/x)2⋅x2x2 dx= ∫0∞x2(x2+1)2 dxegin{align*} & int_0^{infin} rac{1}{left( x+1/x ight)^2} dx \ =& int_0^{infin} rac{1}{left( x+1/x ight)^2} cdot rac{x^2}{x^2} dx \ =& int_0^{infin} rac{x^2}{left( x^2+1 ight)^2} dx end{align*}

Then we pull the trick similar to products of xnx^n and e−x2e^{-x^2} (like in this question):

=∫0∞x⋅x(x2+1)2 dx=12∫0∞x⋅2x(x2+1)2 dxegin{align*} &= int_0^{infin} rac{x cdot x}{left( x^2+1 ight)^2} dx \ &= rac{1}{2} int_0^{infin} rac{x cdot 2x}{left( x^2+1 ight)^2} dx end{align*}

So now we can parts:

f=xg′=2x(x2+1)2f′=1g=−1x2+1egin{align*} f &= x quad&quad g' &= rac{2x}{left(x^2+1 ight)^2} \ f' &= 1 quad&quad g &= - rac{1}{x^2+1} end{align*}  12[fg−∫f′g dx]0∞= 12[[−xx2+1]0∞−∫0∞−1x2+1 dx]= 12[(0−0)+[tan⁡−1x]0∞]= 12(π2−0)= π4egin{align*} & rac{1}{2} left[ fg - int f'g dx ight]_0^{infin} \ =& rac{1}{2} left[ left[ - rac{x}{x^2+1} ight]_0^{infin} - int_0^{infin} - rac{1}{x^2+1} dx ight] \ =& rac{1}{2} left[ (0 - 0) + left[ an^{-1}{x} ight]_0^{infin} ight] \ =& rac{1}{2} left( rac{pi}{2} - 0 ight) \ =& rac{pi}{4} end{align*}