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Katen [24]
3 years ago
14

Which applies the power of a product rule simplify

Mathematics
1 answer:
liberstina [14]3 years ago
5 0

<em>so</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>of</em><em> </em><em>option</em><em> </em><em>C</em><em>.</em>

<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> </em><em>⤴</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>

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Solve the following integral.<br><br> <img src="https://tex.z-dn.net/?f=%5Cint4x%5Ccos%282-3x%29dx" id="TexFormula1" title="\int
bagirrra123 [75]

Hi there!

\boxed{-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C}

To find the indefinite integral, we must integrate by parts.

Let "u" be the expression most easily differentiated, and "dv" the remaining expression. Take the derivative of "u" and the integral of "dv":

u = 4x

du = 4

dv = cos(2 - 3x)

v = 1/3sin(2 - 3x)

Write into the format:

∫udv = uv - ∫vdu

Thus, utilize the solved for expressions above:

4x · (-1/3sin(2 - 3x)) -∫ 4(1/3sin(2 - 3x))dx

Simplify:

-4x/3 sin(2 - 3x) - ∫ 4/3sin(2 - 3x)dx

Integrate the integral:

∫4/3(sin(2 - 3x)dx

u = 2 - 3x

du = -3dx ⇒ -1/3du = dx

-1/3∫ 4/3(sin(2 - 3x)dx ⇒ -4/9cos(2 - 3x) + C

Combine:

-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C

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Let say x = -(-2) where x is the distance of number on number line from 0.
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Step-by-step explanation:

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