- #1

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can anybody help me with this problem

Evaluate :

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

Evaluate :

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

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- #1

- 41

- 0

can anybody help me with this problem

Evaluate :

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

Evaluate :

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

- #2

- 1,235

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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

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

Last edited:

- #3

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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.

- #4

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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:

- #5

HallsofIvy

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- #6

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yep, thanks catching that

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