Find the integral given below.
integral cos^2(7x+1)dx
let z = 7x+1
then dz = 7 dx or dx = dz/7
now you have
(1/7 integral cos^2 z dz
remember to turn z back to x when you finish :)
To find the integral of cos^2(7x + 1), we can use the trigonometric identity:
cos^2(x) = (1/2) + (1/2)cos(2x)
Substituting 7x + 1 for x, we have:
cos^2(7x + 1) = (1/2) + (1/2)cos(2(7x + 1))
= (1/2) + (1/2)cos(14x + 2)
Now, let's integrate term by term:
∫ cos^2(7x + 1) dx = ∫ [(1/2) + (1/2)cos(14x + 2)] dx
The integral of (1/2) with respect to x is (1/2)x. To integrate the second term, we use the following substitution:
Let u = 14x + 2,
Then du = 14 dx,
Solving for dx, we have: dx = (1/14) du.
Now, we can rewrite the integral as:
∫ [(1/2) + (1/2)cos(14x + 2)] dx = (1/2) ∫ dx + (1/2) ∫ cos(14x + 2) dx
= (1/2)x + (1/2) ∫ cos(u) (1/14) du
= (1/2)x + (1/28) ∫ cos(u) du
The integral of cos(u) with respect to u is sin(u), so we have:
= (1/2)x + (1/28)sin(u) + C
Finally, substituting back u = 14x + 2, the integral becomes:
= (1/2)x + (1/28)sin(14x + 2) + C
Therefore, the integral of cos^2(7x + 1) is (1/2)x + (1/28)sin(14x + 2) + C, where C is the constant of integration.