Dom V. answered • 05/03/16

Cornell Engineering grad specializing in advanced math subjects

cosh

^{-1}(x) has for its derivative 1/√[x^{2}-1], so if we get something that looks similar in form (by a u-substitution, for example), we can integrate.1/√[x

^{2}-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(x

^{2}/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/√[x

^{2}/5-1] dx)Next we want to pull the x

^{2}/5 together into a single term squared:x

^{2}/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)/√[u^{2}-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