How do you integrate #{[(secx)^3] tanx} dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Ratnaker Mehta Aug 2, 2016 #1/3sec^3x+C#. Explanation: Let #I=intsec^3xtanxdx#, rewriting it as, #I=intsec^2x(secxtanx)dx#. Let us use subst. #secx=t#, so that, #secxtanxdx=dt#. Hence, #I=intt^2dt=t^3/3=1/3sec^3x+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 3182 views around the world You can reuse this answer Creative Commons License