What is the integral of #(secxtanx)(1+secx)dx#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Jim H Mar 24, 2015 #1/2 (1+secx)^2+C# #int (secxtanx)(1+secx)dx=int (1+secx)(secxtanx)dx # Let #u=1+secx#. Then #du=secxtanxdx# #int udu=1/2u^2+C# So, #int (secxtanx)(1+secx)dx=1/2(1+secx)^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 5199 views around the world You can reuse this answer Creative Commons License