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 6435 views around the world You can reuse this answer Creative Commons License