∫ sin(x)sin(3x)dx
find the integral
one way:
sin(3x) = sinx cos2x + cosx sin2x
= sinx(2cos^2(x)-1) + 2cosx sinx cosx
= cos^2(x) sinx - sinx + 2cos^2(x) sinx
= 3cos^2(x)sinx - sinx
∫ = -cos^3(x) + cos(x)
another way:
sinx sin3x = 1/2 cos(2x) - 1/2 cos(4x)
∫ = 1/4 sin2x - 1/8 sin4x
a little manipulation of that also yields
sin^3(x) - cos(x)
the various expressions are not identical, but differ only by a constant C
SUIIII
Oh, look, it's the sine of a clown! The integral of sin(x)sin(3x) is like telling a clown joke - it's all about timing. Let's grab our clown shoes and solve this step by step.
We can use a handy trick called the product-to-sum formula for sines. By using sin(a)sin(b) = 0.5(cos(a-b) - cos(a+b)), we can simplify the integral.
∫ sin(x)sin(3x) dx = ∫ 0.5(cos(x - 3x) - cos(x + 3x)) dx
Simplifying further, we get:
∫ 0.5(cos(-2x) - cos(4x)) dx
Now we can integrate term by term:
∫ 0.5cos(-2x) dx - ∫ 0.5cos(4x) dx
The integral of cos(-2x) is sin(-2x)/(-2), which simplifies to -sin(2x)/2.
The integral of cos(4x) is sin(4x)/4.
So, our final answer is:
-0.5(sin(2x)/2) - 0.5(sin(4x)/4) + C
Where C is the clown's constant of hilarity.
There you have it! The integral of sin(x)sin(3x) is -0.5(sin(2x)/2) - 0.5(sin(4x)/4) + C.
To find the integral of sin(x)sin(3x), you can use the product-to-sum identity for sine, which states that:
sin(a)sin(b) = (1/2)[cos(a-b) - cos(a+b)]
Applying this identity, we get:
∫ sin(x)sin(3x)dx = ∫ (1/2)[cos(x-3x) - cos(x+3x)]dx
= (1/2) ∫ [cos(-2x) - cos(4x)]dx
Next, we can integrate each term separately:
∫ cos(-2x)dx = (1/2)sin(-2x)/(-2) + C = (-1/4)sin(2x) + C1
∫ cos(4x)dx = (1/4)sin(4x)/4 + C = (1/16)sin(4x) + C2
Combining the two results, we have:
∫ sin(x)sin(3x)dx = (-1/4)sin(2x) + (1/16)sin(4x) + C,
where C is the constant of integration.
To find the integral of sin(x)sin(3x)dx, we will use the integration by parts method. Integration by parts is a technique for finding the integral of a product of two functions.
The formula for integration by parts is ∫ u dv = uv - ∫ v du, where u and v are functions of x.
Let's assign u = sin(x) and dv = sin(3x)dx.
To find du, we differentiate u with respect to x.
du/dx = cos(x)
To find v, we integrate dv with respect to x.
∫ sin(3x)dx = -1/3 cos(3x)
Now, we have u, du, v, and dv, so we can plug them into the integration by parts formula:
∫ sin(x)sin(3x)dx = sin(x) (-1/3 cos(3x)) - ∫ (-1/3 cos(3x)) cos(x) dx
Simplifying this expression gives:
= -1/3 sin(x)cos(3x) + 1/3 ∫ cos(x)cos(3x) dx
Now, let's apply integration by parts again to the new integral: ∫ cos(x)cos(3x) dx.
Assign u = cos(x) and dv = cos(3x) dx.
Differentiating u with respect to x gives:
du/dx = -sin(x)
Integrating dv with respect to x gives:
∫ cos(3x) dx = 1/3 sin(3x)
Now we plug these values into the integration by parts formula:
∫ cos(x)cos(3x) dx = cos(x)(1/3 sin(3x)) - ∫ sin(x) (1/3 sin(3x)) dx
Simplifying this expression gives:
= 1/3 cos(x)sin(3x) - 1/3 ∫ sin(x)sin(3x) dx
Notice that we have the original integral on the right side of the equation. So we can solve for it:
∫ sin(x)sin(3x) dx = -1/3 sin(x)cos(3x) + 1/3 cos(x)sin(3x) - 1/3 ∫ sin(x)sin(3x) dx
Rearranging the equation gives:
2/3 ∫ sin(x)sin(3x) dx = -1/3 sin(x)cos(3x) + 1/3 cos(x)sin(3x)
Now, we can solve for the integral by multiplying both sides by 3/2:
∫ sin(x)sin(3x) dx = (3/2) (-1/3 sin(x)cos(3x) + 1/3 cos(x)sin(3x))
Simplifying further, we get:
∫ sin(x)sin(3x) dx = -1/2 sin(x)cos(3x) + 1/2 cos(x)sin(3x)
Therefore, the integral of sin(x)sin(3x)dx is -1/2 sin(x)cos(3x) + 1/2 cos(x)sin(3x) + C, where C is the constant of integration.