Integration of : #intdx/(3sinx-4cosx)^2# ?

Question

#intdx/(3sinx-4cosx)^2#

6275 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-14T13:52:22

#int dx/(3sinx-4cosx)^2=-cosx/(9sinx-12cosx)+C#

#int dx/(3sinx-4cosx)^2#

=#int ((secx)^2*dx)/(3tanx-4)^2#

After using #y=tanx# and #dy=(secx)^2*dx# transforms, this integral became

#int dy/(3y-4)^2#

=#-1/3*(3y-4)^(-1)+C#

=#-1/3*(3tanx-4)^(-1)+C#

=#-1/(9tanx-12)+C#

=#-cosx/(9sinx-12cosx)+C#