Integral of (dx/x*sqrt(36+x^2)) using hyperbolic functions

Question

The final answer should not include hyperbolic functions

Answer 1

Let x = 6tanθ. Then dx = 6sec²θdθ

Substitute to get ∫[6sec²θ / (6tanθ√(36 + 36tan²θ))]dθ

= (1/6)∫(secθ / tanθ)dθ = (1/6)∫cscθdθ = -(1/6)ln l cscθ + cotθ l + C

= -(1/6)ln l √(36+x²) / x + 6 / x l + C

Answer 2

Given:

I = ∫ dx / ( x * √( 36 + x² )

Find:

I(x) = ?

Solution:

x = 6 sinh u

dx = 6 cosh u

x² = 36 sinh² u

36 + x² = 36 ( 1 + sinh² u ) = 36 cosh² u

√( 36 + x² ) = 6 cosh u

I = ∫ 6 cosh u du / ( 6 sinh u * 6 cosh u)

I = 1/6 ∫ csch u du

I = 1/6 ln| tanh( u / 2 ) | + C

From searching online:

tanh( x / 2 ) = ( cosh x - 1 ) / sinh x

∴ tanh( u / 2 ) = ( √( 36 + x² ) / 6 - 1 ) / ( x / 6 )

tanh( u / 2 ) = ( √( 36 + x² ) - 6 ) / x )

∴ I = 1/6 ln| ( [ √( 36 + x² ) - 6 ) ] / x ) | + C

∴ I = 1/6 * [ ln| [ √( 36 + x² ) - 6 ] - ln| x | ] + C = 1/6 * ln| ( [ √( 36 + x² ) - 6 ) ] / x ) | + C

I hope this helps! Message me in the comments if you have any questions, comments, or concerns on the solution I did above!

Answer 3

Your question is a little confusing with respect to hyperbolic functions.

The substitution x=6 tan θ will convert the integral to a well-known trigonometric integral.

You should be able to do the manipulations and then look up the answer in a table of integrals or actually perform the integration yourself.