# Calculus

Integral of sin2xcos3x using Integration by Parts

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1. so, one way is to let
u = sin2x, du = 2 cos2x dx
dv = cos3x, v = 1/3 sin3x
Now we have
I = ∫sin2x cos3x dx = ∫ u dv
= uv - ∫ v du
= 1/3 sin2x sin3x - 2/3 ∫cos2x sin3x dx
For the second integral, let
u = cos2x, du = -2sin2x dx
dv = sin3x dx, v = -1/3 cos3x
∫cos2x sin3x dx = -1/3 cos2x cos3x - 2/3 ∫ sin2x cos3x dx
Putting all that together, we end up with
I = 1/3 sin2x sin3x - 2/3 (-1/3 cos2x cos3x - 2/3 ∫ sin2x cos3x dx)
I = 1/3 sin2x sin3x + 2/9 cos2x cos3x + 4/9 I
5/9 I = 1/3 sin2x sin3x + 2/9 cos2x cos3x
I = 3/5 sin2x sin3x + 2/5 cos2x cos3x
That can be massaged in various ways, one of which is
1/2 cosx - 1/10 cos5x

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oobleck
2. I really really like the algorithm for integration by parts shown in

where "blackpenredpen" shows the DI method and goes through the 3-Stop method.

-----D ------- I , D for differentiate, I for integrate
+| sin2x .... cos3x
-| 2cos2x .. (1/3)sin3x
+| (-4sin2x) .. (-1/9)cos3x <--- This product is a multiple of the original, so we stop
-| ....

∫sin2x cos3x dx = (1/3)sin2xsin3x + (2/9)cos2xcos3x + (4/9)∫sin2x cos3x dx
(5/9)∫sin2x cos3x dx = (1/3)sin2xsin3x + (2/9)cos2xcos3x

∫sin2x cos3x dx = (3/5)sin2xcos3x + (2/5)cos2xcos3x <------ oobleck's answer

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Reiny

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