can anybody help me with this problem Evaluate : integral 3x (sinx/cos^4x) dx
Oct 15, 2006 #1 413 41 0 can anybody help me with this problem Evaluate : integral 3x (sinx/cos^4x) dx
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Oct 15, 2006 #2 courtrigrad 1,235 1 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 Last edited: 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
Oct 16, 2006 #3 413 41 0 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.
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.
Oct 17, 2006 #4 courtrigrad 1,235 1 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]. Last edited: 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].
Oct 17, 2006 #5 HallsofIvy Science Advisor Homework Helper 41,833 961 I'm sure courtrigrad meant [itex]u= x[/itex] and [itex]dv= tan x sec^3 xdx[/itex] (with out the "x" in dv).
I'm sure courtrigrad meant [itex]u= x[/itex] and [itex]dv= tan x sec^3 xdx[/itex] (with out the "x" in dv).