A complicated integral

Calculus Level 5

Find the value of ∫ 0 5 − 1 2 arctan ⁡ ( x + x 1 − x 2 1 − x 2 − x 2 ) d x \displaystyle\int_{0}^{\sqrt{\frac{\sqrt{5}-1}{2}}} \arctan \left(\frac{x+x\sqrt{1-x^2}}{\sqrt{1-x^2} -x^2}\right) dx .


The answer is 0.612.

Amal Hari by Amal Hari , 22, Canada

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3 solutions

I = ∫ 0 5 − 1 2 tan ⁡ − 1 ( x + x 1 − x 2 1 − x 2 − x 2 ) d x = ∫ 0 5 − 1 2 tan ⁡ − 1 ( x 1 − x 2 + x 1 − x 2 1 − x 2 ) d x = ∫ 0 5 − 1 2 ( tan ⁡ − 1 ( x 1 − x 2 ) + tan ⁡ − 1 x ) d x = ∫ 0 5 − 1 2 ( sin ⁡ − 1 x + tan ⁡ − 1 x ) d x By integration by parts = x sin ⁡ − 1 x − ∫ x 1 − x 2 d x + x tan ⁡ − 1 x − ∫ x 1 + x 2 d x ∣ 0 5 − 1 2 = x sin ⁡ − 1 x + 1 − x 2 + x tan ⁡ − 1 x − ln ⁡ ( 1 + x 2 ) 2 ∣ 0 5 − 1 2 ≈ 0.612 \begin{aligned} I & = \int_0^{\sqrt{\frac {\sqrt 5-1}2}} \tan^{-1} \left(\frac {x+x\sqrt{1-x^2}}{\sqrt{1-x^2}-x^2} \right) dx \\ & = \int_0^{\sqrt{\frac {\sqrt 5-1}2}} \tan^{-1} \left(\frac {\frac x{\sqrt{1-x^2}}+x}{1-\frac {x^2}{\sqrt{1-x^2}}} \right) dx \\ & = \int_0^{\sqrt{\frac {\sqrt 5-1}2}} \left(\tan^{-1} \left(\frac x{\sqrt{1-x^2}} \right) + \tan^{-1} x \right) dx \\ & = \int_0^{\sqrt{\frac {\sqrt 5-1}2}} \left(\sin^{-1} x + \tan^{-1} x \right) dx & \small \blue{\text{By integration by parts}} \\ & = x \sin^{-1} x - \int \frac x{\sqrt{1-x^2}} dx + x\tan^{-1}x - \int \frac x{1+x^2} dx \bigg|_0^{\sqrt{\frac {\sqrt 5-1}2}} \\ & = x \sin^{-1} x + \sqrt{1-x^2} + x\tan^{-1}x - \frac {\ln(1+x^2)}2 \bigg|_0^{\sqrt{\frac {\sqrt 5-1}2}} \\ & \approx \boxed{0.612} \end{aligned}

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Mark Hennings
Dec 3, 2019

We calculate that d d x tan ⁡ − 1 ( x + x 1 − x 2 1 − x 2 − x 2 ) = 1 1 − x 2 + 1 1 + x 2 \frac{d}{dx}\,\tan^{-1}\left(\frac{x + x\sqrt{1-x^2}}{\sqrt{1-x^2} -x^2}\right) \; = \; \frac{1}{\sqrt{1-x^2}} + \frac{1}{1+x^2} and hence, integrating by parts, ∫ tan ⁡ − 1 ( x + x 1 − x 2 1 − x 2 − x 2 ) d x = x tan ⁡ − 1 ( x + x 1 − x 2 1 − x 2 − x 2 ) + 1 − x 2 + 1 2 ln ⁡ ( 1 + x 2 ) + c \int \tan^{-1}\left(\frac{x + x\sqrt{1-x^2}}{\sqrt{1-x^2} -x^2}\right)\,dx \; = \; x\tan^{-1}\left(\frac{x + x\sqrt{1-x^2}}{\sqrt{1-x^2} -x^2}\right) + \sqrt{1-x^2} + \tfrac12\ln(1 + x^2) + c which makes the definite integral equal to 1 2 π 5 − 1 2 + 1 2 ( 5 − 3 ) − 1 2 ln ⁡ ( 5 + 1 2 ) = 0.612312 \tfrac12\pi\sqrt{\frac{\sqrt{5}-1}{2}} + \tfrac12(\sqrt{5}-3) - \tfrac12\ln\left(\frac{\sqrt{5}+1}{2}\right) \; = \; \boxed{ 0.612312}

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we can also do it by tan-1(a) + tan-1(b) = tan-1((a+b)/(1-ab)), here a = x/√(1-x²) and b = x

Kukkadapu Sudhamsh - 1 year, 6 months ago

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That is Amal Hari's approach, since tan ⁡ − 1 x 1 − x 2 = sin ⁡ − 1 x \tan^{-1} \tfrac{x}{\sqrt{1-x^2}} = \sin^{-1}x .

Mark Hennings - 1 year, 6 months ago
Amal Hari
Dec 3, 2019

arcsin ⁡ x + arctan ⁡ x = arctan ⁡ ( x + x 1 − x 2 1 − x 2 − x 2 ) \arcsin x +\arctan x = \displaystyle\arctan \left(\frac{x+x\sqrt{1-x^2}}{\sqrt{1-x^2} -x^2}\right) for the valid domain

This can be proved as follows:

Suppose arcsin ⁡ x + arctan ⁡ y = z \arcsin x +\arctan y =z Let

sin ⁡ a = x \sin a =x and tan ⁡ b = y \tan b =y

we have a + b = z a+b =z

tan ⁡ ( a + b ) = tan ⁡ a + tan ⁡ b 1 − tan ⁡ a tan ⁡ b \tan\left( a+b\right) =\displaystyle\frac{\tan a +\tan b}{1-\tan a \tan b}

From sin ⁡ a = x \sin a=x we can get

tan ⁡ a = x 1 − x 2 \tan a =\displaystyle\frac{x}{\sqrt{1-x^{2}}}

and tan ⁡ b = y \tan b =y

tan ⁡ ( a + b ) = x 1 − x 2 + y 1 − x × y 1 − x 2 \tan\left( a+b\right) =\displaystyle\frac{\frac{x}{\sqrt{1-x^{2}}} +y}{1-\frac{x\times y}{\sqrt{1-x^{2}}}}

( a + b ) = arctan ⁡ ( x 1 − x 2 + y 1 − x × y 1 − x 2 ) \left( a+b\right) =\displaystyle\arctan\left(\frac{\frac{x}{\sqrt{1-x^{2}}} +y}{1-\frac{x\times y}{\sqrt{1-x^{2}}}}\right)

( a + b ) = z = arcsin ⁡ x + arctan ⁡ y = arctan ⁡ ( x 1 − x 2 + y 1 − x × y 1 − x 2 ) \left( a+b\right)=z=\arcsin x +\arctan y=\displaystyle\arctan\left(\frac{\frac{x}{\sqrt{1-x^{2}}} +y}{1-\frac{x\times y}{\sqrt{1-x^{2}}}}\right)

now make y = x y=x

arcsin ⁡ x + arctan ⁡ x = arctan ⁡ ( x 1 − x 2 + x 1 − x 2 1 − x 2 ) \arcsin x +\arctan x=\displaystyle\arctan\left(\frac{\frac{x}{\sqrt{1-x^{2}}} +x}{1-\frac{x^{2}}{\sqrt{1-x^{2}}}}\right)

factoring out 1 − x 2 \sqrt{1-x^{2}} we have

arcsin ⁡ x + arctan ⁡ x = arctan ⁡ ( x + x 1 − x 2 1 − x 2 − x 2 ) \arcsin x +\arctan x = \displaystyle\arctan \left(\frac{x+x\sqrt{1-x^2}}{\sqrt{1-x^2} -x^2}\right)

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