The integral of sin(7x)cos(7x)dx
your expression is just 1/2 sin(14x) dx
To find the indefinite integral of sin(7x)cos(7x) dx, you can use the substitution method. Here are the step-by-step instructions:
Step 1: Let u = sin(7x).
Differentiate both sides with respect to x to find du/dx:
du/dx = 7cos(7x).
Step 2: Rearrange the equation from Step 1 to solve for dx:
dx = du / (7cos(7x)).
Step 3: Substitute the value of dx from Step 2 into the original integral:
∫ sin(7x)cos(7x) dx = ∫ u * dx / (7cos(7x)).
= 1/7 ∫ u du.
Step 4: Integrate the expression with respect to u:
= 1/7 ∫ u du = (1/7) * (u^2 / 2) + C,
where C is the constant of integration.
Step 5: Substitute back the value of u from Step 1:
= (1/7) * (sin^2(7x) / 2) + C.
Therefore, the indefinite integral of sin(7x)cos(7x) dx is (1/7) * (sin^2(7x) / 2) + C, where C represents the constant of integration.
To find the integral of sin(7x)cos(7x), you can use the trigonometric identity for the product of sine and cosine:
sin(2θ) = 2sin(θ)cos(θ)
Let's rewrite sin(7x)cos(7x) as sin(2 * 7x):
sin(7x)cos(7x) = 1/2 * sin(2 * 7x)
Now, integrate 1/2 * sin(2 * 7x) with respect to x:
∫ (1/2 * sin(2 * 7x)) dx
To integrate this, you can use a substitution. Let's set u = 2 * 7x, which means that du = 2 * 7 dx:
∫ (1/2 * sin(u)) du / (2 * 7)
Simplifying this expression gives us:
1/28 ∫ sin(u) du
Now, integrate sin(u) with respect to u:
1/28 * [-cos(u)] + C
Finally, substitute back u = 2 * 7x:
= 1/28 * [-cos(2 * 7x)] + C
Therefore, the integral of sin(7x)cos(7x)dx is 1/28 * [-cos(2 * 7x)] + C.