How do you integrate #int sqrttanxsec^2xdx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Douglas K. Nov 29, 2016 Let #u = tan(x)#, then #du = sec^2(x)dx#. Please see the explanation. Explanation: Let #u = tan(x)#, then #du = sec^2(x)dx# #intsqrt(tan(t))sec^2dx = int(u^(1/2))du# Integrate: #intsqrt(tan(t))sec^2dx = 2/3u^(3/2) + C# Reverse the substitution: #intsqrt(tan(t))sec^2dx = 2/3(tan(x))^(3/2) + C# Answer link Related questions How do I evaluate the indefinite integral #intsin^3(x)*cos^2(x)dx# ? How do I evaluate the indefinite integral #intsin^6(x)*cos^3(x)dx# ? How do I evaluate the indefinite integral #intcos^5(x)dx# ? How do I evaluate the indefinite integral #intsin^2(2t)dt# ? How do I evaluate the indefinite integral #int(1+cos(x))^2dx# ? How do I evaluate the indefinite integral #intsec^2(x)*tan(x)dx# ? How do I evaluate the indefinite integral #intcot^5(x)*sin^4(x)dx# ? How do I evaluate the indefinite integral #inttan^2(x)dx# ? How do I evaluate the indefinite integral #int(tan^2(x)+tan^4(x))^2dx# ? How do I evaluate the indefinite integral #intx*sin(x)*tan(x)dx# ? See all questions in Integrals of Trigonometric Functions Impact of this question 1743 views around the world You can reuse this answer Creative Commons License