How do you integrate #int 1/sqrt(9x^2-36x+37)# using trig substitutions?

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How do you integrate #int 1/sqrt(9x^2-36x+37)# using trig substitutions?

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Answer 1 · 2016-12-16T01:15:16

By substituting #x=u/3+2# the square root becomes #sqrt(u^2+1)# which is standard form or can be further substituted with #u=sinhv# and then use #cosh^2 v - sinh^2v=1#, #d/dx(sinh v)=cosh(v)# etc.

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How do you integrate #int 1/sqrt(9x^2-36x+37)# using trig substitutions?

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