How do you integrate #int 1/sqrt(9x^2-36x+37)# using trig substitutions? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Roy E. Dec 16, 2016 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. Answer link Related questions How do you find the integral #int1/(x^2*sqrt(x^2-9))dx# ? How do you find the integral #intx^3/(sqrt(x^2+9))dx# ? How do you find the integral #intx^3*sqrt(9-x^2)dx# ? How do you find the integral #intx^3/(sqrt(16-x^2))dx# ? How do you find the integral #intsqrt(x^2-1)/xdx# ? How do you find the integral #intsqrt(x^2-9)/x^3dx# ? How do you find the integral #intx/(sqrt(x^2+x+1))dx# ? How do you find the integral #intdt/(sqrt(t^2-6t+13))# ? How do you find the integral #intx*sqrt(1-x^4)dx# ? How do you prove the integral formula #intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C# ? See all questions in Integration by Trigonometric Substitution Impact of this question 3680 views around the world You can reuse this answer Creative Commons License