Round 1 Question 16 Fun question, I don’t think my method was the intended!
∫ 1 x x − x 2 d x egin{align*}
int rac{1}{ sqrt{xsqrt{x}-x^2} } dx
end{align*} ∫ x x − x 2 1 d x Start by considering domain – the function is only defined if the numerator is nonzero, and this only happens if
x x − x 2 > 0 x x − x 2 > 0 x x > x 2 x 3 / 2 > x 4 / 2 egin{align*}
sqrt{xsqrt{x}-x^2} &> 0
\ xsqrt{x}-x^2 &> 0
\ xsqrt{x} > x^2
\ x^{3/2} > x^{4/2}
end{align*} x x − x 2 x x − x 2 x x > x 2 x 3/2 > x 4/2 > 0 > 0 The only region where x n > x n + k x^n > x^{n+k} x n > x n + k x x x 0 < x < 1 0 < x < 1 0 < x < 1 x x x 0 < x < 1 0 < x < 1 0 < x < 1
We’ll change the fraction into an exponent for clarity:
∫ 1 x x − x 2 d x = ∫ [ x x − x 2 ] − 1 d x egin{align*}
& int rac{1}{ sqrt{xsqrt{x}-x^2} } dx
\ =& int left[ sqrt{xsqrt{x}-x^2}
ight]^{-1} dx
end{align*} = ∫ x x − x 2 1 d x ∫ [ x x − x 2 ] − 1 d x Now factorise out x \sqrt{x} x
= ∫ [ x ( x − x ) ] − 1 d x = ∫ [ x x − x ] − 1 d x = ∫ 1 x x − x d x egin{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*} = ∫ [ x ( x − x ) ] − 1 d x = ∫ [ x x − x ] − 1 d x = ∫ x x − x 1 d x We need to check domain when we do this – the leftover x − x \sqrt{\sqrt{x}-x} x − x
x > x for 0 < x < 1 x − x > 0 ⇒ x − x is defined egin{align*}
sqrt{x} &> x quad ext{for} 0 < x < 1
\ sqrt{x} - x &> 0
\ Rightarrow sqrt{sqrt{x} - x} & ext{ is defined}
end{align*} x x − x ⇒ x − x > x for 0 < x < 1 > 0 is defined So we’re good to go. We can now substitute x = t \sqrt{x} = t x = t
∫ 1 x x − x d x = 2 ∫ 1 x − x ⋅ 1 2 x d x = 2 ∫ 1 t − t 2 d t egin{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*} = = ∫ x x − x 1 d x 2 ∫ x − x 1 ⋅ 2 x 1 d x 2 ∫ t − t 2 1 d t The rest is just trig sub WORK :
= 2 ∫ 1 − ( t 2 − t ) d x = 2 ∫ 1 − ( ( t − 1 2 ) 2 − 1 4 ) d x = 2 ∫ 1 1 4 − ( t − 1 2 ) 2 d x egin{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*} = 2 ∫ − ( t 2 − t ) 1 d x = 2 ∫ − (( t − 2 1 ) 2 − 4 1 ) 1 d x = 2 ∫ 4 1 − ( t − 2 1 ) 2 1 d x ( t − 1 2 ) 2 = 1 4 sin 2 v t − 1 2 = 1 2 sin v d t = 1 2 cos v d v egin{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*} ( t − 2 1 ) 2 t − 2 1 d t = 4 1 sin 2 v = 2 1 sin v = 2 1 cos v d v = 2 ∫ 1 1 4 − 1 4 sin 2 v ⋅ 1 2 cos v d v = 2 ∫ d v = 2 v = 2 sin − 1 ( 2 t − 1 ) = 2 sin − 1 ( 2 x − 1 ) − c egin{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*} = 2 ∫ 4 1 − 4 1 sin 2 v 1 ⋅ 2 1 cos v d v = 2 ∫ d v = 2 v = 2 sin − 1 ( 2 t − 1 ) = 2 sin − 1 ( 2 x − 1 ) − c Interesting that this is equivalent to the canonical answer of 4 sin − 1 ( x 1 / 4 ) 4\sin^{-1}\left( x^{1/4} \right) 4 sin − 1 ( x 1/4 ) π \pi π
Final Question 1 Damn, all my recreational integration really paid off for this one.
∫ 0 ∞ 1 ( x + 1 / x ) 2 d x = ∫ 0 ∞ 1 ( x + 1 / x ) 2 ⋅ x 2 x 2 d x = ∫ 0 ∞ x 2 ( x 2 + 1 ) 2 d x egin{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*} = = ∫ 0 ∞ ( x + 1/ x ) 2 1 d x ∫ 0 ∞ ( x + 1/ x ) 2 1 ⋅ x 2 x 2 d x ∫ 0 ∞ ( x 2 + 1 ) 2 x 2 d x Then we pull the trick similar to products of x n x^n x n e − x 2 e^{-x^2} e − x 2 in this question ):
= ∫ 0 ∞ x ⋅ x ( x 2 + 1 ) 2 d x = 1 2 ∫ 0 ∞ x ⋅ 2 x ( x 2 + 1 ) 2 d x egin{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*} = ∫ 0 ∞ ( x 2 + 1 ) 2 x ⋅ x d x = 2 1 ∫ 0 ∞ ( x 2 + 1 ) 2 x ⋅ 2 x d x So now we can parts:
f = x g ′ = 2 x ( x 2 + 1 ) 2 f ′ = 1 g = − 1 x 2 + 1 egin{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*} f f ′ = x = 1 g ′ g = ( x 2 + 1 ) 2 2 x = − x 2 + 1 1 1 2 [ f g − ∫ f ′ g d x ] 0 ∞ = 1 2 [ [ − x x 2 + 1 ] 0 ∞ − ∫ 0 ∞ − 1 x 2 + 1 d x ] = 1 2 [ ( 0 − 0 ) + [ tan − 1 x ] 0 ∞ ] = 1 2 ( π 2 − 0 ) = π 4 egin{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*} = = = = 2 1 [ f g − ∫ f ′ g d x ] 0 ∞ 2 1 [ [ − x 2 + 1 x ] 0 ∞ − ∫ 0 ∞ − x 2 + 1 1 d x ] 2 1 [ ( 0 − 0 ) + [ tan − 1 x ] 0 ∞ ] 2 1 ( 2 π − 0 ) 4 π