Trig: Hyperbolic Integral?

Question

The integral from 4 to 6((1/(t^2-9)^1/2)dt)

I need some help on getting to the end. It looks like a cosh^-1(x) type.

Answer 1

You can use hyperbolic substitution here [ with t=3cosh(u) ], but you do not have to. You may use secant substition.

∫ 1/√( t² - 9 ) dt

= ∫ 1/√( t² - 3² ) dt

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Use secant substition:

Let 3·sec(u) = t

Then 3·sec(u)·tan(u) du = dt

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→ ∫ 1/√( [ 3·sec(u) ]² - 3² ) 3·sec(u)·tan(u) du

= ∫ 1/√( 3²·sec²(u) - 3² ) 3·sec(u)·tan(u) du

Factor out 1/√(3²) as 1/3 and cancel it with 3:

= ∫ 1/√[ 3² · (sec²(u) - 1) ] 3·sec(u)·tan(u) du

= ∫ 1/[ 3·√( sec²(u) - 1 ) 3·sec(u)·tan(u) du

= ∫ (1/3)·1/√( sec²(u) - 1 ) 3·sec(u)·tan(u) du

= ∫ 1/√( sec²(u) - 1 ) sec(u)·tan(u) du

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With the Pythagorean trigonometric identity, it can be shown that:

sin²θ + cos²θ = 1

sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ

tan²θ + 1 = sec²θ

tan²θ = sec²θ - 1

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Replace secant squared minus one with tangent squared:

→ ∫ 1/√( tan²(u) ) sec(u)·tan(u) du

The square root of a squared value gives absolute value:

= ∫ 1/| tan(u) | sec(u)·tan(u) du

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Since 3·sec(u) = t

Then u = arcsec(t/3)

Then tan(u) = tan[ arcsec(t/3) ]

Over the interval 4 to 6, this arcsec is not negative. Thus you can remove the absolute value bars. Generally at this stage of calculus, you can just get rid of the absolute value bars and hope that you were justified in doing so (as I will do soon).

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→ ∫ 1/tan(u) sec(u)·tan(u) du

Cancel tangent:

= ∫ sec(u) du

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The integral of secant is a little tricky to derive, but it is possible to memorize its result instead...

Multiply by 1/1 in the form of [ sec(u) + tan(u) ]/[ sec(u) + tan(u) ]

= ∫ sec(u) · [ sec(u) + tan(u) ]/[ sec(u) + tan(u) ] du

= ∫ [ sec²(u) + sec(u)·tan(u) ]/[ sec(u) + tan(u) ] du

Let q = sec(u) + tan(u)

Then dq = [ sec(u)·tan(u) + sec²(u) ] du

→ ∫ dq/q

= ln| q | + C,

Reverse substitution for q:

= ln| sec(u) + tan(u) | + C,

This is the integral of secant. If it is never negative:

= ln( sec(u) + tan(u) ) + C,

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Reverse substitution for u, where u = arcsec(t/3):

= ln{ sec[ arcsec(t/3) ] + tan[ arcsec(t/3) ] } + C,

Simplify these trig compositions to:

= ln[ t/3 + √(t²-9)/3 ] + C,

Factor the 1/3:

= ln{ 1/3· [ t + √(t²-9) ] } + C,

Use the properties of logs to write this product as a sum of logs:

= ln(1/3) + ln[ t + √(t²-9) ] + C,

Since ln(1/3) is a constant, adding this to C, produces some other constant:

= ln[ t + √(t²-9) ] + C

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Now, evaluate the definite integral:

6

∫ 1/√( t² - 9 ) dt

4

= { ln[ t + √(t²-9) ] } as t goes from 4 to 6

= { ln[ 6 + √(6²-9) ] } - { ln[ 4 + √(4²-9) ] }

= ln[ 6 + √(36-9) ] - ln[ 4 + √(16-9) ]

= ln[ 6 + √(27) ] - ln[ 4 + √(7) ]

If you factor out of the square root and combine them using the difference property of logarithms, you get:

= ln[ 6 + √(3²·3) ] - ln[ 4 + √(7) ]

= ln[ 6 + 3·√(3) ] - ln[ 4 + √(7) ]

= ln{ [ 6 + 3·√(3) ] / [ 4 + √(7) ] }

Approximating this gives:

≈ 0.5215924357

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Note that if you only need an approximate answer, you could stop at:

ln{ sec[ arcsec(t/3) ] + tan[ arcsec(t/3) ] } + C,

Just plug in the upper and lower bounds of t in a calculator.

(If you change the bounds for the substitution variables, you could approximate it earlier.)

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Note:

ln[ t/3 + √(t²-9)/3 ] + C,

= arcosh(t/3) + C,

Answer 2

This integral can be easily solved by knowing the basic integral identity as follows :

∫ 1 / √(x^2 - a^2) dx = log ( x + √(x^2 + a^2) ) + C

Now, integral from 4 to 6 (1 / √(t^2 - 3^2) ) dt   [write 9 as 3^2]

= log ( t + √(t^2 - 9) ) + C from 4 to 6

Now put the limits to get the final result as follows,

log ( 6 + 3√3) - log (4 + √7)

= log [ (6 + 3√3) / ( 4 + √7)  ]

= 0.521592435700911078

Solved.