Breather problem

Calculus Level 4

∫ 0 4 x 2 + x 2 1 + x 2 d x = 1 2 [ 3 a + ln ⁡ ( b 2 + 17 ) − 2 − ln ⁡ ( c + c 2 ) ] \int_0^4 \sqrt{x^2+\frac{x^2}{1+x^2}}dx=\frac{1}{2}\left[3\sqrt{a}+\ln(b\sqrt{2}+\sqrt{17})-\sqrt{2}-\ln \left(\sqrt{c}+\frac{c}{2} \right)\right]

where a a , b b , and c c are integers and a a and c c are square-free. Submit a + b + 2 c a+b+2c .


The answer is 41.

James Wilson by James Wilson , 28, USA

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2 solutions

Chew-Seong Cheong
Jan 17, 2021

I = ∫ 0 4 x 2 + x 2 1 + x 2 d x Let x = sinh ⁡ θ ⟹ d x = cosh ⁡ θ d θ = ∫ 0 sinh ⁡ − 1 4 sinh ⁡ 2 θ + sinh ⁡ 2 θ cosh ⁡ 2 θ ⋅ cosh ⁡ θ d θ = ∫ 0 sinh ⁡ − 1 4 cosh ⁡ 2 θ + 1 sinh ⁡ θ d θ Let u = cosh ⁡ θ ⟹ d u = sinh ⁡ θ d θ = ∫ 1 17 u 2 + 1 d u Let u = sinh ⁡ ϕ ⟹ d u = cosh ⁡ ϕ d ϕ = ∫ sinh ⁡ − 1 1 sinh ⁡ − 1 17 cosh ⁡ 2 ϕ d ϕ = ∫ sinh ⁡ − 1 1 sinh ⁡ − 1 17 cosh ⁡ 2 ϕ + 1 2 d ϕ = sinh ⁡ 2 ϕ 4 + ϕ 2 ∣ sinh ⁡ − 1 1 sinh ⁡ − 1 17 Note that sinh ⁡ 2 ϕ = 2 sinh ⁡ ϕ cosh ⁡ ϕ = 3 34 2 + sinh ⁡ − 1 17 2 − 2 2 − sinh ⁡ − 1 1 2 and sinh ⁡ − 1 x = ln ⁡ ( x + x 2 + 1 ) = 3 34 + ln ⁡ ( 3 2 + 17 ) − 2 − ln ⁡ ( 2 + 1 ) 2 \begin{aligned} I & = \int_0^4 \sqrt{x^2+\frac {x^2}{1+x^2}}\ dx & \small \blue{\text{Let }x = \sinh \theta \implies dx = \cosh \theta\ d\theta} \\ & = \int_0^{\sinh^{-1} 4} \sqrt{\sinh^2 \theta +\frac {\sinh^2 \theta}{\cosh^2 \theta}}\cdot \cosh \theta \ d\theta \\ & = \int_0^{\sinh^{-1} 4} \sqrt{\cosh^2 \theta + 1} \sinh \theta \ d\theta & \small \blue{\text{Let }u = \cosh \theta \implies du = \sinh \theta \ d\theta} \\ & = \int_1^{\sqrt{17}} \sqrt{u^2+1} \ du & \small \blue{\text{Let }u = \sinh \phi \implies du = \cosh \phi\ d\phi} \\ & = \int_{\sinh^{-1}1}^{\sinh^{-1} \sqrt{17}} \cosh^2 \phi \ d \phi \\ & = \int_{\sinh^{-1}1}^{\sinh^{-1} \sqrt{17}} \frac {\cosh 2\phi +1}2\ d \phi \\ & = \frac \blue{\sinh 2\phi}4 + \frac \phi 2 \ \bigg|_{\sinh^{-1}1}^{\sinh^{-1} \sqrt{17}} & \small \blue{\text{Note that }\sinh 2\phi = 2\sinh \phi \cosh \phi} \\ & = \frac {3\sqrt{34}}2 + \frac \blue{\sinh^{-1} \sqrt{17}}2 - \frac {\sqrt 2}2 - \frac \blue{\sinh^{-1} 1}2 & \small \blue{\text{and }\sinh^{-1} x = \ln (x + \sqrt{x^2+1})} \\ & = \frac {3\sqrt{34} +\ln(3\sqrt 2 + \sqrt{17}) - \sqrt 2 - \ln (\sqrt 2 +1)}2 \end{aligned}

Therefore a + b + 2 c = 34 + 3 + 2 ⋅ 2 = 41 a+b+2c = 34 + 3 + 2\cdot 2 = \boxed{41} .

Reference: Hyperbolic trigonometric functions

2 Helpful 1 Interesting 2 Brilliant 0 Confused

Very interesting solution. Good job.

James Wilson - 4 months, 3 weeks ago

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Yes. Substituting sinh ⁡ u \sinh u is easier than tan ⁡ u \tan u . Nice problem.

Chew-Seong Cheong - 4 months, 3 weeks ago

Indeed an elegant solution! I made 2 trig substitutions as well but not hyperbolic...and my solution is much lengthier.

Veselin Dimov - 4 months, 3 weeks ago

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Yes, next time use sinh ⁡ u \sinh u instead of tan ⁡ u \tan u . It is easier.

Chew-Seong Cheong - 4 months, 3 weeks ago

Chew-Seong Cheong is very wise in his solutions. Gotta love Osbourne's Rule.

James Wilson - 4 months, 3 weeks ago

Hi, Chew-Seong Cheong. I found a typo in your solution. It should say 34 \sqrt{34} at the end, but it says 32 \sqrt{32} . Also, my latest problem, "Colliding Spring System," should be labeled as "Classical Mechanics," but I labeled it "Calculus." You can change it if you think "Classical Mechanics" is more suitable.

James Wilson - 4 months, 3 weeks ago

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Thanks, I have amended it. I thought you can do the changes on your own problems. Let me check. You can provide a link to your problem by using [problem title] {the link} (no space between [] and {}).

Chew-Seong Cheong - 4 months, 3 weeks ago

I don't see any way that lets me change it myself. You're right. I should've provided a link. I don't know what I was thinking. I appreciate your kindness.

James Wilson - 4 months, 3 weeks ago
Sathvik Acharya
Jan 17, 2021

Claim: ∫ x 2 + x 2 1 + x 2 d x = ln ⁡ ( x 2 + 1 + x 2 + 2 ) + ( x 2 + 1 ) ( x 2 + 2 ) 2 + C \int \sqrt{x^2+\frac{x^2}{1+x^2}} \,\mathrm{d}x=\frac{\ln\left(\sqrt{x^2+1}+\sqrt{x^2+2}\right)+\sqrt{(x^2+1)(x^2+2)}}{2}+C Proof: x 2 + x 2 1 + x 2 = x 4 + 2 x 2 x 2 + 1 = x x 2 + 2 x 2 + 1 \sqrt{x^2+\frac{x^2}{1+x^2}} = \sqrt{\frac{x^4+2x^2}{x^2+1}}=\frac{x\sqrt{x^2+2}}{\sqrt{x^2+1}} Let a = x 2 + 1 ⟹ d a d x = x x 2 + 1 ⟹ d x = a a 2 − 1 d a a=\sqrt{x^2+1}\implies \dfrac{\mathrm{d}a}{\mathrm{d}x}=\dfrac{x}{\sqrt{x^2+1}}\implies \mathrm{d}x=\dfrac{a}{\sqrt{a^2-1}}\mathrm{d}a . So we have, ∫ x x 2 + 2 x 2 + 1 d x = ∫ a 2 − 1 ⋅ a 2 + 1 a ⋅ a a 2 − 1 d a = ∫ a 2 + 1 d a = ln ⁡ ( a 2 + 1 + a ) + a a 2 + 1 2 + C = ln ⁡ ( x 2 + 1 + x 2 + 2 ) + ( x 2 + 1 ) ( x 2 + 2 ) 2 + C \begin{aligned} \int \frac{x\sqrt{x^2+2}}{\sqrt{x^2+1}} \,\mathrm{d}x&= \int \frac{\sqrt{a^2-1}\cdot \sqrt{a^2+1}}{a}\cdot \dfrac{a}{\sqrt{a^2-1}}\mathrm{d}a \\ &=\int \sqrt{a^2+1} \,\mathrm{d}a \\ &=\frac{\ln\left(\sqrt{a^2+1}+a\right)+a\sqrt{a^2+1}}{2}+C \\ &=\frac{\ln\left(\sqrt{x^2+1}+\sqrt{x^2+2}\right)+\sqrt{(x^2+1)(x^2+2)}}{2}+C \end{aligned}

Therefore, ∫ 0 4 x 2 + x 2 1 + x 2 = ln ⁡ ( 4 2 + 1 + 4 2 + 2 ) + ( 4 2 + 1 ) ( 4 2 + 2 ) 2 − ln ⁡ ( 0 2 + 1 + 0 2 + 2 ) + ( 0 2 + 1 ) ( 0 2 + 2 ) 2 = 1 2 ( 3 34 + ln ⁡ ( 3 2 + 17 ) − 2 − ln ⁡ ( 2 + 1 ) ) \begin{aligned} \int_{0}^4 \sqrt{x^2+\frac{x^2}{1+x^2}}&=\frac{\ln\left(\sqrt{4^2+1}+\sqrt{4^2+2}\right)+\sqrt{(4^2+1)(4^2+2)}}{2}-\frac{\ln\left(\sqrt{0^2+1}+\sqrt{0^2+2}\right)+\sqrt{(0^2+1)(0^2+2)}}{2} \\ &=\frac{1}{2} \left(3\sqrt{34}+\ln (3\sqrt{2}+\sqrt{17})-\sqrt{2}-\ln (\sqrt{2}+1)\right) \end{aligned} ⟹ a = 34 , b = 3 , c = 2 ⟹ a + b + 2 c = 41 \implies a=34,\;b=3,\;c=2 \implies a+b+2c=\boxed{41}

1 Helpful 1 Interesting 1 Brilliant 0 Confused

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Galdly. Let a = 1 2 ( t − 1 t ) . a=\frac{1}{2}\Big(t-\frac{1}{t}\Big). ⇒ a 2 + 1 = 1 4 ( t − 1 t ) 2 + 1 = 1 4 ( t 2 − 2 + 1 t 2 ) + 1 = 1 4 ( t 2 + 2 + 1 t 2 ) = 1 2 ( t + 1 t ) . \Rightarrow \sqrt{a^2+1}=\sqrt{\frac{1}{4}\Big(t-\frac{1}{t}\Big)^2+1}=\sqrt{\frac{1}{4}\Big(t^2-2+\frac{1}{t^2}\Big)+1}=\sqrt{\frac{1}{4}\Big(t^2+2+\frac{1}{t^2}\Big)}=\frac{1}{2}\Big(t+\frac{1}{t}\Big). Then we have d a = 1 2 ( 1 + 1 t 2 ) d t . da=\frac{1}{2}\Big(1+\frac{1}{t^2}\Big)dt. So, ∫ a 2 + 1 d a = ∫ 1 2 ( t + 1 t ) 1 2 ( 1 + 1 t 2 ) d t = ∫ 1 4 [ t + 2 t + 1 t 3 ] d t = 1 8 t 2 + 1 2 ln ⁡ t − 1 8 t 2 + C = 1 8 ( a + a 2 + 1 ) 2 + 1 2 ln ⁡ ( a + a 2 + 1 ) − 1 8 ( a + a 2 + 1 ) 2 + C \int \sqrt{a^2+1}da = \int \frac{1}{2}\Big(t+\frac{1}{t}\Big)\frac{1}{2}\Big(1+\frac{1}{t^2}\Big)dt = \int \frac{1}{4}\Big[t+\frac{2}{t}+\frac{1}{t^3}\Big]dt=\frac{1}{8}t^2+\frac{1}{2}\ln{t}-\frac{1}{8t^2} +C = \frac{1}{8}\Big(a+\sqrt{a^2+1}\Big)^2+\frac{1}{2}\ln\Big(a+\sqrt{a^2+1}\Big)-\frac{1}{8\Big(a+\sqrt{a^2+1}\Big)^2}+C . (You should check my work.) Very efficient method, by the way. Good job. (By the way, you typed the answer as 21 instead of 41.)

James Wilson - 4 months, 3 weeks ago

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Thanks., I have corrected it. Also, nice problem :)

Sathvik Acharya - 4 months, 3 weeks ago

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