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# Integration question involving cosh^-1???

Evaluate the integral (1/sqrt(x^2 - 5))dx on limits of 3 to 4, using the formula: integral of (1/sqrt(x^2 - a^2))dx.
Recall that this formula was found by using the substitution x = a cosh t, and observing that cosh^-1 (x/a) and ln(x + sqrt(x^2 − a^2) differ by a constant.

### 1 Answer by Expert Tutors

Dom V. | Cornell Engineering grad specializing in advanced math subjectsCornell Engineering grad specializing in...
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cosh-1(x) has for its derivative 1/√[x2-1], so if we get something that looks similar in form (by a u-substitution, for example), we can integrate.

1/√[x2-5] is almost there, but we need to do something about the 5 (based on the derivative, we need a 1 there). If we factor out the 5 from the quadratic, we can recover the 1:

1/√[5(x2/5-1)]

And in fact, we can take √5 out of the whole denominator now. Because it's just a constant, we can take it out of the entire integral:

(1/√5)*Int(1/√[x2/5-1] dx)

Next we want to pull the x2/5 together into a single term squared:

x2/5 = (x/√5)2

This is important because now we can make a simple u-sub and get rid of it entirely:

u=x/√5
√5 du=dx

The integral is transformed into

dx/√[ (x/√5)2-1 ]  ==>  (√5 du)/√[u2-1]

The √5 we pulled out earlier (in the denominator) and the √5 we created with the u-substitution (in the numerator) cancel each other out. The integral we're left with will exactly give us cosh-1(u)=cosh-1(x/√5).

When you evaluate cosh-1 at the two limits of integration, you can substitute in the ln(...) definition for evaluation instead (since they differ by a constant, the constant will disappear when you subtract). Just remember to carry over the x/√5 when you go from cosh-1 to ln