How do you integrate #tan^3xsec^4xdx#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Guillaume L. Jun 28, 2018 #inttan³xsec⁴xdx=1/(6cos^6(x))-1/(4cos^4(x))+C#, #C in RR# Explanation: #inttan³xsec⁴xdx# #=int(sin³(x))/(cos^7(x))dx# #=-int(-sin(x)(1-cos²(x)))/(cos^7(x))dx# Let #u=cos(x)# #du=-sin(x)dx# #<=> -int(1-u²)/(u^7)du# #=-int1/u^7du+int1/u^5du# #=1/(6u^6)-1/(4u⁴)# #=1/(6cos^6(x))-1/(4cos^4(x))+C#, #C in RR# \0/ here's our answer ! 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 6080 views around the world You can reuse this answer Creative Commons License