integrate. cos7xcos3xsinxdx.plz help tanks
To solve the integral ∫cos(7x)cos(3x)sin(x) dx, we can use the product-to-sum identity for cosine:
cos(a)cos(b) = (1/2)[cos(a + b) + cos(a - b)].
Applying this identity, we have:
∫cos(7x)cos(3x)sin(x) dx = ∫[cos(7x + 3x) + cos(7x - 3x)]sin(x) dx
= ∫[cos(10x) + cos(4x)]sin(x) dx.
Now, we can expand this integral into two separate integrals:
∫cos(10x)sin(x) dx + ∫cos(4x)sin(x) dx.
To solve these integrals, we can use the product-to-sum identity for sine:
sin(a)sin(b) = (1/2)[cos(a - b) - cos(a + b)].
Applying this identity, we have:
∫cos(10x)sin(x) dx = (1/2) ∫[cos(10x - x) - cos(10x + x)] dx
= (1/2) ∫[cos(9x) - cos(11x)] dx.
∫cos(4x)sin(x) dx = (1/2) ∫[cos(4x - x) - cos(4x + x)] dx
= (1/2) ∫[cos(3x) - cos(5x)] dx.
Now, we can integrate each term individually:
For the first integral:
∫[cos(9x) - cos(11x)] dx = (1/9)sin(9x) - (1/11)sin(11x) + C.
For the second integral:
∫[cos(3x) - cos(5x)] dx = (1/3)sin(3x) - (1/5)sin(5x) + C.
Finally, combining both integrals, we have:
∫cos(7x)cos(3x)sin(x) dx = (1/9)sin(9x) - (1/11)sin(11x) + (1/3)sin(3x) - (1/5)sin(5x) + C,
where C is the constant of integration.