Integral 3x (sinx/cos^4x) dx

413 · Oct 15, 2006

can anybody help me with this problem
Evaluate :
integral 3x (sinx/cos^4x) dx

courtrigrad · Oct 15, 2006

Is it [tex] \int \frac{3x\sin x}{\cos^{4} x} [/tex]?

Rewrite it as [tex] 3\int x\tan x\ sec^{3} x [/tex] and use integration by parts

413 · Oct 16, 2006

i know the equation for intergration by parts is
intergral u dv = uv -intergral v du

can u tell me which variable is which?...u, du, v, dv=?...there seems to have 3 different variable.

courtrigrad · Oct 17, 2006

Let [tex] u = x [/tex] and [tex] dv = \ tan x \sec^{3} x [/tex]. You will then need to use integration by parts on [tex] dv [/tex] to get [tex] v [/tex].

HallsofIvy · Oct 17, 2006

I'm sure courtrigrad meant [itex]u= x[/itex] and [itex]dv= tan x sec^3 xdx[/itex] (with out the "x" in dv).

courtrigrad · Oct 17, 2006

yep, thanks catching that