Integral of sin2xcos3x using Integration by Parts
I really really like the algorithm for integration by parts shown in
https://www.youtube.com/watch?v=2I-_SV8cwsw
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
To solve the integral of sin(2x)cos(3x) using integration by parts, we will follow these steps:
Step 1: Identify the functions in the integrand.
Here, we have sin(2x) and cos(3x).
Step 2: Determine which function you will differentiate and which function you will integrate.
To choose which function to differentiate (u) and which function to integrate (dv), we can use the acronym "LIATE":
L – Logarithmic function
I – Inverse trigonometric function
A – Algebraic function
T – Trigonometric function (other than the highest power)
E – Exponential function
In this case, we will choose u = sin(2x) and dv = cos(3x).
Step 3: Calculate du and v.
Differentiating u = sin(2x), we get du = 2cos(2x)dx.
Integrating dv = cos(3x), we get v = (1/3)sin(3x).
Step 4: Apply the integration by parts formula.
The integration by parts formula is:
∫u dv = uv - ∫v du
Using this formula, we can solve the integral:
∫sin(2x)cos(3x) dx = (sin(2x))(1/3)sin(3x) - ∫(1/3)sin(3x)(2cos(2x)) dx
= (1/3)sin(2x)sin(3x) - (2/3)∫sin(3x)cos(2x) dx
Step 5: Repeat the process for the new integral on the right.
Now, we have a new integral to solve: ∫sin(3x)cos(2x) dx.
Step 6: Choose new functions to differentiate and integrate.
For this new integral, we will let u = sin(3x) and dv = cos(2x).
Step 7: Calculate du and v.
Differentiating u = sin(3x), we get du = 3cos(3x)dx.
Integrating dv = cos(2x), we get v = (1/2)sin(2x).
Step 8: Apply the integration by parts formula to the new integral.
∫sin(3x)cos(2x) dx = (sin(3x))(1/2)sin(2x) - ∫(1/2)sin(2x)(3cos(3x)) dx
= (1/2)sin(3x)sin(2x) - (3/2)∫sin(2x)cos(3x) dx
Step 9: Rearrange the equation and solve for the unknown integral.
Now, we will rearrange the equation containing the unknown integral and solve for it:
(3/2)∫sin(2x)cos(3x) dx = (1/2)sin(3x)sin(2x) - (1/2)sin(2x)sin(3x)
Step 10: Solve for the unknown integral.
(3/2)∫sin(2x)cos(3x) dx = 0
Dividing both sides by (3/2):
∫sin(2x)cos(3x) dx = 0
Therefore, the integral ∫sin(2x)cos(3x) dx equals zero.
To find the integral of sin(2x)cos(3x) using integration by parts, we'll use the formula:
∫u dv = uv - ∫v du
Let's assign the following:
u = sin(2x), dv = cos(3x) dx
Now, let's find du and v:
First, find du by differentiating u with respect to x:
du/dx = d/dx (sin(2x))
= 2cos(2x)
Now, find v by integrating dv with respect to x:
∫ cos(3x) dx = (1/3) sin(3x)
Using the formula for integration by parts, we have:
∫ sin(2x) cos(3x) dx = uv - ∫v du
= sin(2x) * (1/3) sin(3x) - ∫ (1/3) sin(3x) * 2cos(2x) dx
= (1/3) sin(2x) sin(3x) - (2/3) ∫ sin(3x) cos(2x) dx
Now, we can apply integration by parts again to solve the new integral, ∫ sin(3x) cos(2x) dx.
Let's assign the new values:
u = sin(3x), dv = cos(2x) dx
Find du by differentiating u with respect to x:
du/dx = d/dx (sin(3x))
= 3cos(3x)
Find v by integrating dv with respect to x:
∫ cos(2x) dx = (1/2) sin(2x)
Using integration by parts again, we have:
∫ sin(3x) cos(2x) dx = uv - ∫v du
= sin(3x) * (1/2) sin(2x) - ∫ (1/2) sin(2x) * 3cos(3x) dx
= (1/2) sin(3x) sin(2x) - (3/2) ∫ sin(2x) cos(3x) dx
Notice that we obtained another integral of sin(2x) cos(3x), which is the original integral we started with. Let's call it I:
I = ∫ sin(2x) cos(3x) dx
Now, substituting this back into our equation:
I = (1/2) sin(3x) sin(2x) - (3/2) * I
Simplifying the equation, we get:
4I = (1/2) sin(3x) sin(2x)
Dividing both sides of the equation by 4, we have:
I = (1/8) sin(3x) sin(2x) + C
Therefore, the integral of sin(2x) cos(3x) using integration by parts is:
∫ sin(2x) cos(3x) dx = (1/8) sin(3x) sin(2x) + C
where C is the constant of integration.
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