How do you integrate #int 1/(x^2sqrt(16x^2-9))# by trigonometric substitution? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Ratnaker Mehta Mar 3, 2018 # sqrt(16x^2-9)/(9x)+C#. Explanation: Suppose that, #I=int1/(x^2sqrt(16x^2-9))dx=int1/(x^2sqrt{(4x)^2-3^2})dx#. We subst. #4x=3secy," so that, "4dx=3secytanydy#. #:. I=int1/{(3/4*secy)^2sqrt(9sec^2y-9)}*(3/4*secytany)dy#, #=16/9*1/4int1/(sec^2ytany)*secytanydy#, #=4/9int1/secydy#, #=4/9intcosydy#, #=4/9siny#, #=4/9sqrt(1-cos^2y)#, #=4/9sqrt(1-1/sec^2y)#, #=4/9sqrt{1-1/(4/3*x)^2}#, #=4/9sqrt{1-9/(16x^2)}#. # rArr I=sqrt(16x^2-9)/(9x)+C#. 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 4321 views around the world You can reuse this answer Creative Commons License