This is a hard problem to price in this format. It took me a page on paper. Right here are the vital points:

1) Pull out the 20π. The is simply a constant.

You are watching: Integral of sin^3x cos^2x

2) usage a trig substitution: sin(3x) = 3sin(x) - 4 sin3(x) and also cos(2x) = 1 - 2sin2(x)

3) Multiply the end the resulting 2 binomials to acquire 20π∫8sin5(x) -10sin3(x) + 3sin(x) dx

4) rest this into 3 separate integrals. Because that the first 160π∫sin5(x), rest it into 160π∫sin4(x)sin(x) and then substitute in the trig identification <1-cos2(x)>2 because that sin4(x) then usage a u substitution u = cos(x). The an outcome after you space all completed is 160π<-cos(x) +2/3cos3(x) - 1/5cos5(x)>

5) for the second integral -200π∫sin3(x)dx, repeat the procedure above and also you eventually get: -200π<-cos(x) +1/3cos3(x)

6) The third integral conveniently solves to -60πcos(x)

7) placed them all together and also combine like terms and you get: -20πcos(x) +40πcos3(x) - 160/5πcos5(x) + C

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Note sin x cos y = 1/2

sin(3x) cos(2x) = 1/2

**∫π20sin(3x)cos(2x)dx**

**= 20 π ∫(1/2)**dx

**= 10π ∫**dx

**= 10 π <(-1/5) cos(5x) + (-1)cos(x)> + C**

**=-2 π** + C

**where C is any kind of constant**

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QZ P.

you re welcome remember to add the + C because that the indefinite integralReport

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