Generalising a Special Case

Calculus Level 5

It is possible to find positive integers A , B , C , D , E A,B,C,D,E such that ∫ 0 2 a a 2 + 1 sin ⁡ − 1 ( ∣ 1 − a x ∣ 1 − x 2 ) d x = A a 2 + 1 sin ⁡ − 1 ( 1 a B ) − C sin ⁡ − 1 ( 1 a D ) + E a π a 2 + 1 \int_0^{\frac{2a}{a^2+1}} \sin^{-1}\left(\frac{|1-ax|}{\sqrt{1-x^2}}\right)\,dx \; = \; \frac{A}{\sqrt{a^2+1}}\sin^{-1}\left(\frac{1}{a^B}\right) - C\sin^{-1}\left(\frac{1}{a^D}\right) + \frac{Ea\pi}{a^2+1} for all real numbers a ≥ 3 a \ge 3 . What is the value of A + B + C + D + E A+B+C+D+E ?


The answer is 8.

Mark Hennings by Mark Hennings , 61, United Kingdom

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1 solution

Mark Hennings
May 10, 2019

First integrating by parts, then applying the substitution ( a 2 + 1 ) x − a = a sin ⁡ θ (a^2+1)x -a = a\sin\theta , and then applying the substitution t = tan ⁡ 1 2 θ t = \tan\tfrac12\theta , gives F ( x ) = ∫ sin ⁡ − 1 ( 1 − a x 1 − x 2 ) d x = x sin ⁡ − 1 ( 1 − a x 1 − x 2 ) + ∫ x ( a − x ) ( 1 − x 2 ) 2 a x − ( a 2 + 1 ) x 2 d x = x sin ⁡ − 1 ( 1 − a x 1 − x 2 ) + a 2 + 1 ∫ x ( a − x ) ( 1 − x 2 ) a 2 − ( ( a 2 + 1 ) x − a ) 2 d x = x sin ⁡ − 1 ( 1 − a x 1 − x 2 ) + a 2 a 2 + 1 ∫ ( 1 + sin ⁡ θ ) ( a 2 − sin ⁡ θ ) ( a 2 − a + 1 − a sin ⁡ θ ) ( a 2 + a + 1 + a sin ⁡ θ ) d θ = x sin ⁡ − 1 ( 1 − a x 1 − x 2 ) + a 2 a 2 + 1 ∫ ( 1 a 2 + ( a 2 + 1 ) ( a − 1 ) 2 a 2 ( a 2 − a + 1 − a sin ⁡ θ ) − ( a 2 + 1 ) ( a + 1 ) 2 a 2 ( a 2 + a + 1 + a sin ⁡ θ ) ) d t = x sin ⁡ − 1 ( 1 − a x 1 − x 2 ) + θ a 2 + 1 + a 2 + 1 ∫ ( a − 1 ( a 2 − a + 1 ) ( t 2 + 1 ) − 2 a t − a + 1 ( a 2 + a + 1 ) ( t 2 + 1 ) + 2 a t ) d t = x sin ⁡ − 1 ( 1 − a x 1 − x 2 ) + θ a 2 + 1 + tan ⁡ − 1 ( ( a 2 − a + 1 ) t − a ( a − 1 ) a 2 + 1 ) − tan ⁡ − 1 ( ( a 2 + a + 1 ) t + a ( a + 1 ) a 2 + 1 ) \begin{aligned} F(x)& = \; \int \sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right)\,dx \\ & = \; x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right) + \int \frac{x(a-x)}{(1-x^2)\sqrt{2ax - (a^2+1)x^2}}\,dx \\ & = \; x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right) + \sqrt{a^2+1}\int \frac{x(a-x)}{(1-x^2)\sqrt{a^2 - \big((a^2+1)x-a\big)^2}}\,dx \\ & = \; x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right) + \frac{a^2}{\sqrt{a^2+1}}\int \frac{(1+\sin\theta)(a^2-\sin\theta)}{(a^2-a+1-a\sin\theta)(a^2+a+1+a\sin\theta)}\,d\theta \\ & = \; x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right) + \frac{a^2}{\sqrt{a^2+1}}\int\left( \frac{1}{a^2} + \frac{(a^2+1)(a-1)}{2a^2(a^2-a+1-a\sin\theta)} - \frac{(a^2+1)(a+1)}{2a^2(a^2+a+1+a\sin\theta)}\right)\,dt \\ & = \; \begin{array}{l}\displaystyle x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right) + \frac{\theta}{\sqrt{a^2+1}} \\ \displaystyle + \sqrt{a^2+1}\int\left( \frac{a-1}{(a^2-a+1)(t^2+1)-2at} - \frac{a+1}{(a^2+a+1)(t^2+1)+2at}\right)\,dt \end{array} \\ & = \; x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right) + \frac{\theta}{\sqrt{a^2+1}} + \tan^{-1}\left(\frac{(a^2-a+1)t-a}{(a-1)\sqrt{a^2+1}}\right) - \tan^{-1}\left(\frac{(a^2+a+1)t+a}{(a+1)\sqrt{a^2+1}}\right) \end{aligned} for 0 ≤ x ≤ 2 a a 2 + 1 0 \le x \le \tfrac{2a}{a^2+1} , ignoring the constant of integration. Thus ∫ 0 2 a a 2 + 1 sin ⁡ − 1 ( ∣ 1 − a x ∣ 1 − x 2 ) d x = 2 F ( a − 1 ) − F ( 0 ) − F ( 2 a a 2 + 1 ) \int_0^{\frac{2a}{a^2+1}} \sin^{-1}\left(\frac{|1-ax|}{\sqrt{1-x^2}}\right)\,dx \; = \; 2F\big(a^{-1}\big) - F(0) - F\big(\tfrac{2a}{a^2+1}\big)

If x = 0 x=0 we have θ = − 1 2 π \theta = -\tfrac12\pi and t = − 1 t=-1 , so that F ( 0 ) = − π 2 a 2 + 1 + tan ⁡ − 1 ( − a 2 + 1 a − 1 ) − tan ⁡ − 1 ( − a 2 + 1 a + 1 ) = − π 2 a 2 + 1 − tan ⁡ − 1 ( a 2 + 1 a 2 ) F(0) \; = \; -\frac{\pi}{2\sqrt{a^2+1}} + \tan^{-1}\left(-\frac{\sqrt{a^2+1}}{a-1}\right) - \tan^{-1}\left(-\frac{\sqrt{a^2+1}}{a+1}\right) \; = \; -\frac{\pi}{2\sqrt{a^2+1}} - \tan^{-1}\left(\frac{\sqrt{a^2+1}}{a^2}\right) If x = 2 a a 2 + 1 x=\tfrac{2a}{a^2+1} we have θ = 1 2 π \theta = \tfrac12\pi and t = 1 t=1 , so that F ( 2 a a 2 + 1 ) = − a π a 2 + 1 + π 2 a 2 + 1 + tan ⁡ − 1 ( a − 1 a 2 + 1 ) − tan ⁡ − 1 ( a + 1 a 2 + 1 ) = − a π a 2 + 1 + π 2 a 2 + 1 − tan ⁡ − 1 ( a 2 + 1 a 2 ) \begin{aligned} F\big(\tfrac{2a}{a^2+1}\big) & = \; -\frac{a\pi}{a^2+1} + \frac{\pi}{2\sqrt{a^2+1}} + \tan^{-1}\left(\frac{a-1}{\sqrt{a^2+1}}\right) - \tan^{-1}\left(\frac{a+1}{\sqrt{a^2+1}}\right) \\ & = \; -\frac{a\pi}{a^2+1} + \frac{\pi}{2\sqrt{a^2+1}} - \tan^{-1}\left(\frac{\sqrt{a^2+1}}{a^2}\right) \end{aligned} Finally, if x = a − 1 x = a^{-1} then θ = sin ⁡ − 1 ( a − 2 ) \theta = \sin^{-1}\big(a^{-2}\big) and hence t = a 2 − a 4 − 1 t = a^2 - \sqrt{a^4-1} . If a ≥ 3 a \ge 3 then ( a 4 + a 2 + 1 ) a 2 − 1 − a 4 a 2 + 1 > 0 (a^4+a^2+1)\sqrt{a^2-1} - a^4\sqrt{a^2+1} > 0 and we can show that F ( a − 1 ) = 1 a 2 + 1 sin ⁡ − 1 ( a − 2 ) + tan ⁡ − 1 ( a a 4 − 1 − ( a 3 + 1 ) a 2 − 1 ) − tan ⁡ − 1 ( a a 4 − 1 − ( a 3 − 1 ) a 2 − 1 ) = 1 a 2 + 1 sin ⁡ − 1 ( a − 2 ) − tan ⁡ − 1 ( 1 ( a 4 + a 2 + 1 ) a 2 − 1 − a 4 a 2 + 1 ) = 1 a 2 + 1 sin ⁡ − 1 ( a − 2 ) − 1 2 π + tan ⁡ − 1 ( ( a 4 + a 2 + 1 ) a 2 − 1 − a 4 a 2 + 1 ) \begin{aligned} F\big(a^{-1}\big) & = \; \frac{1}{\sqrt{a^2+1}}\sin^{-1}\big(a^{-2}\big) + \tan^{-1}\left(\frac{a\sqrt{a^4-1}-(a^3+1)}{\sqrt{a^2-1}}\right) - \tan^{-1}\left(\frac{a\sqrt{a^4-1} - (a^3-1)}{\sqrt{a^2-1}}\right)\\ & = \; \frac{1}{\sqrt{a^2+1}}\sin^{-1}\big(a^{-2}\big) - \tan^{-1}\left(\frac{1}{(a^4+a^2+1)\sqrt{a^2-1} - a^4\sqrt{a^2+1}}\right) \\ & = \; \frac{1}{\sqrt{a^2+1}}\sin^{-1}\big(a^{-2}\big) - \tfrac12\pi + \tan^{-1}\left((a^4+a^2+1)\sqrt{a^2-1} - a^4\sqrt{a^2+1}\right) \end{aligned} and hence ∫ 0 2 a a 2 + 1 sin ⁡ − 1 ( ∣ 1 − a x ∣ 1 − x 2 ) d x = 2 a 2 + 1 sin ⁡ − 1 ( a − 2 ) − π + a π a 2 + 1 + 2 tan ⁡ − 1 ( ( a 4 + a 2 + 1 ) a 2 − 1 − a 4 a 2 + 1 ) + 2 tan ⁡ − 1 ( a 2 + 1 a 2 ) = 2 a 2 + 1 sin ⁡ − 1 ( a − 2 ) − π + a π a 2 + 1 + 2 tan ⁡ − 1 ( a 2 − 1 ) = 2 a 2 + 1 sin ⁡ − 1 ( a − 2 ) − π + a π a 2 + 1 + 2 cos ⁡ − 1 ( a − 1 ) = 2 a 2 + 1 sin ⁡ − 1 ( a − 2 ) − 2 sin ⁡ − 1 ( a − 1 ) + a π a 2 + 1 \begin{aligned} \int_0^{\frac{2a}{a^2+1}} \sin^{-1}\left(\frac{|1-ax|}{\sqrt{1-x^2}}\right)\,dx & = \; \begin{array}{l} \displaystyle \frac{2}{\sqrt{a^2+1}}\sin^{-1}\big(a^{-2}\big) - \pi + \frac{a\pi}{a^2+1} \\ \displaystyle + 2\tan^{-1}\left((a^4+a^2+1)\sqrt{a^2-1} - a^4\sqrt{a^2+1}\right) + 2\tan^{-1}\left(\frac{\sqrt{a^2+1}}{a^2}\right) \end{array} \\ & = \; \frac{2}{\sqrt{a^2+1}}\sin^{-1}\big(a^{-2}\big) - \pi + \frac{a\pi}{a^2+1} + 2\tan^{-1}\left(\sqrt{a^2-1}\right) \\ & = \; \frac{2}{\sqrt{a^2+1}}\sin^{-1}\big(a^{-2}\big) - \pi + \frac{a\pi}{a^2+1} + 2\cos^{-1}\big(a^{-1}\big) \\ & = \; \frac{2}{\sqrt{a^2+1}}\sin^{-1}\big(a^{-2}\big) - 2\sin^{-1}\big(a^{-1}\big) + \frac{a\pi}{a^2+1}\\ \end{aligned} so that A = 2 A=2 , B = 2 B=2 , C = 2 C=2 , D = 1 D=1 , E = 1 E=1 , and hence A + B + C + D + E = 8 A+B+C+D+E = \boxed{8} .

2 Helpful 2 Interesting 6 Brilliant 0 Confused

@Mark Hennings , I want to know how did you make a 2 + 1 x ( a − x ) ( 1 − x 2 ) ∗ a 2 − ( ( a 2 + 1 ) ∗ x − a ) 2 \frac{ \sqrt{a^2+1} x (a-x)}{(1-x^2)*\sqrt{a^2-((a^2+1)*x-a)^2}} = a 2 a 2 + 1 ( 1 + s i n θ ) ∗ ( a 2 − s i n θ ) ( a 2 − a + 1 − a ∗ s i n θ ) ( a 2 + a + 1 + a ∗ s i n θ ) \frac{a^2}{\sqrt{a^2+1}}\frac{(1+sin\theta)*(a^2-sin \theta)}{(a^2-a+1-a*sin\theta)(a^2+a+1+a*sin\theta)} by plugging in the L.H.S of the equation ( a 2 + 1 ) ∗ x − a = a ∗ s i n θ (a^2+1)*x-a=a*sin\theta . Would you explain it ?

Winod DHAMNEKAR - 1 year, 8 months ago

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Use the substitution ( a 1 + 1 ) x − a = a sin ⁡ θ (a^1+1)x - a = a\sin\theta carefully, and check what happens to all the terms - the square root in the denominator is a multiple of cos ⁡ θ \cos\theta , so cancels out...

Mark Hennings - 1 year, 8 months ago

@Mark Hennings , ( 1 + s i n θ ) = ( a 2 + 1 ) ∗ x a (1+sin\theta)=\frac{(a^2+1)*x}{a} and s i n θ = ( a 2 + 1 ) ∗ x − a a sin \theta=\frac{(a^2+1)*x-a}{a} Now plugging these values in the equation a 2 + 1 ∗ x ( a − x ) ( 1 − x 2 ) ∗ a 2 − ( ( a 2 + 1 ) ∗ x − a ) 2 = a 2 a 2 + 1 ( 1 + s i n θ ) ∗ ( a 2 − s i n θ ) ( a 2 − a + 1 − a ∗ s i n θ ) ∗ ( a 2 + a + 1 + a ∗ s i n θ ) … ( 1 ) \frac{\sqrt{a^2+1}*x(a-x)}{(1-x^2)*\sqrt{a^2-((a^2+1)*x-a)^2}}=\frac{a^2}{\sqrt{a^2+1}}\frac{(1+sin\theta)*(a^2-sin\theta)}{(a^2-a+1-a*sin\theta)*(a^2+a+1+a*sin\theta)}…(1) , I get on the R.H.S.,

x ∗ ( a 3 − ( a 2 + 1 ) ∗ x + a ) a 2 + 1 ∗ ( a 2 + 1 ) ∗ ( 1 − x 2 ) \frac{x*(a^3-(a^2+1)*x+a)}{\sqrt{a^2+1}*(a^2+1)*(1-x^2)} . Now how to make this term equal to the L.H.S. of the equation(1)?

Winod DHAMNEKAR - 1 year, 8 months ago

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Put the following identities together: x = a a 2 + 1 ( 1 + sin ⁡ θ ) a − x = a a 2 + 1 ( a 2 − sin ⁡ θ ) 1 − x = 1 a 2 + 1 ( a 2 − a + 1 − a sin ⁡ θ ) 1 + x = 1 a 2 + 1 ( a 2 + a + 1 + a sin ⁡ θ ) a 2 − ( ( a 2 + 1 ) x − a ) 2 = a cos ⁡ θ d x = cos ⁡ θ d θ \begin{aligned} x & = \; \tfrac{a}{a^2+1}(1+\sin\theta) \\ a-x & = \; \tfrac{a}{a^2+1}(a^2 - \sin\theta) \\ 1-x & = \; \tfrac{1}{a^2+1}(a^2-a+1-a\sin\theta) \\ 1+x & = \; \tfrac{1}{a^2+1}(a^2+a+1+a\sin\theta) \\ \sqrt{a^2 - ((a^2+1)x-a)^2} & = \; a\cos\theta \\ dx & = \; \cos\theta\,d\theta \end{aligned} Work the rest out.

Mark Hennings - 1 year, 8 months ago

@Mark Hennings,How did you compute this term 1 a 2 + ( a 2 + 1 ) ∗ ( a − 1 ) 2 ∗ a 2 ∗ ( a 2 − a + 1 − a ∗ s i n θ ) + ( a 2 + 1 ) ∗ ( a + 1 ) 2 ∗ a 2 ∗ ( a 2 + a + 1 + a ∗ s i n θ ) ? \frac{1}{a^2}+\frac{(a^2+1)*(a-1)}{2*a^2*(a^2-a+1-a*sin\theta)}+\frac{(a^2+1)*(a+1)}{2*a^2*(a^2+a+1+a*sin\theta)}? You used t = t a n θ 2 t=tan\frac{\theta}{2} , so, d t = 1 2 ( 1 + t a n 2 ( θ 2 ) ) d θ dt=\frac12(1+tan^2(\frac{\theta}{2}))d\theta How did you use that here?

Winod DHAMNEKAR - 1 year, 8 months ago

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Expand by partial fractions (in sin ⁡ θ \sin\theta ). That step is just algebra.

Mark Hennings - 1 year, 8 months ago

@Mark Hennings , Okay, You performed partial fraction. But it should be multiplied by 2 s e c 2 ( θ 2 ) \frac{2}{sec^2(\frac{\theta}{2})} because d θ = 2 ∗ d t s e c 2 ( θ 2 ) d\theta=\frac{2*dt}{sec^2(\frac{\theta}{2})}

Winod DHAMNEKAR - 1 year, 8 months ago

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This is getting ridiculous. The term 1 + t 2 = sec ⁡ 2 1 2 θ 1+t^2=\sec^2\tfrac12\theta cancels out the 1 + t 2 1+t^2 in the denominator of sin ⁡ θ \sin\theta . Please try to solve these for yourself. I am not going to explain every line.

Mark Hennings - 1 year, 8 months ago

@Mark Hennings , Thanks for your math help in understanding the answer to this difficult question.

Winod DHAMNEKAR - 1 year, 8 months ago

@Mark Hennings ,Any particular reason for the condition a ≥ 3 a \geq 3 ? A more natural range for this integral would be a > 1 a>1 .

Winod DHAMNEKAR - 1 year, 8 months ago

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The condition is stronger than it needs to be, but is chosen to ensure that the various angles (defined as inverse tangents) are of the right size so that the formulae I am using to simplify expressions of the form tan ⁡ − 1 A + tan ⁡ − 1 B \tan^{-1}A + \tan^{-1}B are valid... I seem to recall that a > 1 a > 1 is too weak.

Mark Hennings - 1 year, 8 months ago

@Mark Hennings , Though you find it ridiculous, my curiosity is not stopping. I understood whole of your answer except these two steps in your answer. How did you make 2 tan ⁡ − 1 ( ( a 2 + 1 ) ∗ a 2 − 1 ) + 2 tan ⁡ − 1 ( a 2 + 1 a 2 ) = 2 tan ⁡ − 1 ( 1 a 2 − 1 ) 2\tan^{-1} \left((a^2+1)* \sqrt{a^2-1}\right) + 2\tan^{-1} \left(\frac{\sqrt{a^2+1}}{a^2}\right) = 2\tan^{-1}\left(\frac{1}{\sqrt{a^2-1}}\right)

My second question is how did you make 2 tan ⁡ − 1 ( 1 a 2 − 1 ) = 2 cos ⁡ − 1 ( a − 1 ) 2\tan^{-1}\left(\frac{1}{\sqrt{a^2-1}}\right)= 2\cos^{-1}\left(a^{-1}\right) What is the logic behind these two steps?

For your further information, no online and offline integral calculator was able to step by step answer this question. But you have given a step by step answer with the help of classical calculus and trigonometry. I appreciate your step by step answer to this question.

Winod DHAMNEKAR - 1 year, 8 months ago

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This time I have spotted a couple of small typos. In the formula for F ( a − 1 ) F(a^{-1}) , some of the square roots have been changed from a 2 − 1 \sqrt{a^2-1} to a 2 + 1 \sqrt{a^2+1} , and in the final calculation tan ⁡ − 1 1 a 2 − 1 \tan^{-1}\frac{1}{\sqrt{a^2-1}} has become tan ⁡ − 1 a 2 − 1 \tan^{-1}\sqrt{a^2-1} . You should find that all calculations work now.

Mark Hennings - 1 year, 8 months ago

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