How do you integrate tan^3xsec^4xdx?

1 Answer
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 !