Integrate: 1/sqrt(25+x^2) from 13 to 0?

Question

Nice! good work. but the -sign wasn't supposed to be in the final answer so i must have asked the question wrong.

Answer 1

RESPONSE to update

Yes, I figured that it was probably supposed to be 0 to 13.

It's always lower to upper bound. I've changed the end.

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Use a trig substitution:

∫ 1/√( 25 + x² ) dx

= ∫ 1/√( 5² + x² ) dx

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Use tangent substition:

Let 5·tan(u) = x

Then 5·sec²(u) du = dx

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→ ∫ 1/√( 5² + [ 5·tan(u) ]² ) 5·sec²(u) du

= ∫ 1/√( 5² + 5²·tan²(u) ) 5·sec²(u) du

Factor out 1/√(5²) as 1/5 and cancel it with 5:

= ∫ 1/√[ 5² · ( 1 + tan²(u) ) ] 5·sec²(u) du

= ∫ 1/[ 5·√( 1 + tan²(u) ) ] 5·sec²(u) du

= ∫ (1/5) · 1/√( 1 + tan²(u) ) 5·sec²(u) du

= ∫ 1/√( 1 + tan²(u) ) sec²(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

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Replace tangent squared plus one with secant squared:

→ ∫ 1/√( sec²(u) ) sec²(u) du

The square root of a squared value gives absolute value:

= ∫ 1/| sec(u) | sec²(u) du

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Since 5·tan(u) = x

Then u = arctan(x/5)

Then sec(u) = sec[ arctan(x/5) ]

Over the interval 0 to 13, this value 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/sec(u) sec²(u) du

Cancel secant:

= ∫ 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 = arctan(t/5):

= ln{ sec[ arctan(x/5) ] + tan[ arctan(x/5) ] } + C,

Simplify these trig compositions to:

= ln[ √(x² + 25)/5 + x/5 ] + C,

Factor the 1/5:

= ln{ 1/5 · [ √(x² + 25) + x ] } + C,

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

= ln(1/5) + ln[ √(x² + 25) + x ] + C,

Since ln(1/5) is a constant, adding this to C, produces some other constant:

= ln[ √(x² + 25) + x ] + C

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Now, evaluate the definite integral (bounds corrected):

13

∫ 1/√( 25 + x² ) dx

0

= { ln[ √(x² + 25) + x ] ] } as x goes from 0 to 13

= { ln[ √(13² + 25) + 13 ] } - { ln[ √(0² + 25) + 0 ] }

= { ln[ √(194) + 13 ] } - { ln[ √(25) ] }

= ln[ √(194) + 13 ] - ln(5)

Using the properties of logs, this can be rewritten as:

= ln{ [√(194) + 13] / 5 }

Approximating this gives:

≈ 1.683743144

Answer 2

∫ [1 / √(25 + x²)] dx =

multiply and divide the integrand by [√(25 + x²) + x] :

∫ {[√(25 + x²) + x] /√(25 + x²)} {1/ [√(25 + x²) + x]} dx =

distribute the first factor as:

∫ {[√(25 + x²)/√(25 + x²)] + [x /√(25 + x²)]} {1/ [√(25 + x²) + x]} dx =

∫ {1 + [x /√(25 + x²)]} {1/ [√(25 + x²) + x]} dx =

rearrange it as:

∫ { {1 + [x /√(25 + x²)]} / [√(25 + x²) + x] } dx =

let [√(25 + x²) + x] = u

differentiate both sides:

d[√(25 + x²) + x] = du →

{ {(1/2)(2x)(25 + x²)^[(1/2) -1]} + 1 } dx = du →

{[x (25 + x²)^(-1/2)] + 1} dx = du →

{[x /√(25 + x²)] + 1} dx = du →

thus, substituting, you get:

∫ { {1 + [x /√(25 + x²)]} / [√(25 + x²) + x] } dx = ∫ {[x /√(25 + x²)] + 1} dx / [√(25 + x²) + x] =

∫ du / u = ln | u | + C

that is, substituting back u = [√(25 + x²) + x]:

ln [√(25 + x²) + x] + C

(abs value is no longer needed, since ln argument is positive)

finally, evaluate the definite integral from 13 to 0, that is:

ln [√(25 + 0²) + 0] - ln [√(25 + 13²) + 13] =

ln (√25) - ln [√(25 + 169) + 13] =

ln (5) - ln (√194 + 13) = (approx.) - 1.68374

thus:

0

∫ [1 / √(25 + x²)] dx = - 1.68374

13

if your correct result is supposed to be positive, maybe the integration limits have been switched, being 13 the upper limit so that:

13

∫ [1 / √(25 + x²)] dx = 1.68374

0

I hope it helps...

Bye!