How do I find the indefinite integral of h(u)=sin^2(1/5u) this is one fifth u. Does it involve double angle formulas? Thanks.
To find the indefinite integral of h(u) = sin^2(1/5u), we can use a combination of trigonometric identities and basic integration techniques.
First, let's recall the double angle formula for sine:
sin^2(x) = (1/2)(1 - cos(2x))
In this case, our function h(u) = sin^2(1/5u). To apply the double angle formula, we need to rewrite the argument of the sine function as 2x.
Let's substitute 1/5u for 2x:
1/5u = 2x
Solving for x, we find:
x = 1/10u
Now we can rewrite our function h(u):
h(u) = (1/2)(1 - cos(2 * 1/10u))
= (1/2)(1 - cos(1/5u))
Now that we have our function in terms of cosine, we can integrate it.
The indefinite integral of (1 - cos(1/5u)) can be found using basic integration rules.
The integral of 1 with respect to u is simply u:
∫(1 - cos(1/5u)) du = u - ∫cos(1/5u) du
To integrate cos(1/5u), we can apply the chain rule in reverse.
Let w = 1/5u, then dw = (1/5) du. Solving for du, we get du = 5 dw.
Substituting back into the integral, we have:
∫cos(1/5u) du = ∫cos(w) (1/5) du = (1/5) ∫cos(w) dw
The integral of cos(w) with respect to w is sin(w):
(1/5) ∫cos(w) dw = (1/5) sin(w) + C
Going back to our original variable, u, we have:
∫(1 - cos(1/5u)) du = u - (1/5) sin(1/5u) + C
Therefore, the indefinite integral of h(u) = sin^2(1/5u) is:
u - (1/5) sin(1/5u) + C
Note: C represents the constant of integration, which can take any value.