How do you integrate #int x/sqrt(144-x^2)dx# using trigonometric substitution? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Bio Dec 14, 2015 Sub #x = 12sinu#. Restrict #-frac{pi}{2}<\u<=frac{pi}{2}#. Note that #cosu>=0#. #frac{dx}{du} = 12cosu# #int frac{x}{sqrt(144-x^2)} dx = int frac{12sinu}{sqrt(144-(12sinu)^2)} * frac{dx}{du} du# #= int frac{12sinu}{sqrt{144(1-sin^2u)}} * (12cosu) du# #= int frac{12sinucosu}{sqrt{cos^2u}} du# #= 12 int sinu du# #= -12 cosu + C #, where #C# is the constant of integration. #= -sqrt{144-(12sinu)^2} + C # #= -sqrt{144-x^2} + 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 2268 views around the world You can reuse this answer Creative Commons License