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

Question

6117 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-28T21:37:25

#inttan³xsec⁴xdx=1/(6cos^6(x))-1/(4cos^4(x))+C#, #C in RR#

#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 !