What is the integral of #tan^3xsecxdx#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Bill K. Mar 17, 2015 The answer can be written as: #\frac{1}{3}\sec^{3}(x)-\sec(x)+C#. Use the fact that #\tan^{2}(x)=\sec^{2}(x)-1# to write #\tan^{3}(x)\sec(x)# as #(sec^{2}(x)-1)\tan(x)\sec(x)#. Note now that if #u=\sec(x)# then #du=\sec(x)\tan(x)dx#. Hence # \int\tan^{3}(x)\sec(x)dx=\int(u^{2}-1)du=u^{3}/3-u+C=\frac{1}{3}\sec^{3}(x)-\sec(x)+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 2762 views around the world You can reuse this answer Creative Commons License