What's the integral of #int (tanx)^5*(secx)^4dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer maganbhai P. Mar 16, 2018 #I=tan^6x/6+tan^8x/8+c# Explanation: We know that, #color(red)(int[f(x)]^n*f^'(x)dx=[f(x)]^(n+1)/(n+1)+c)# #f(x)=tanx=>f^'(x)=sec^2x# So, #I=int(tanx)^5*(secx)^4dx=int(tanx)^5(sec^2x)(secx)^2dx# #I=int(tanx)^5(1+tan^2x)(sec^2x)dx# #I=int(tanx)^5sec^2xdx+int(tanx)^7sec^2xdx# #I=int(tanx)^5d/(dx)(tanx)dx+int(tanx)^7d/(dx)(tanx)dx# #I=(tanx)^(5+1)/(5+1)+(tanx)^(7+1)/(7+1)+c# #I=tan^6x/6+tan^8x/8+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 3566 views around the world You can reuse this answer Creative Commons License