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 3765 views around the world You can reuse this answer Creative Commons License