I posted this below but had no response. I would appreciate help with details please.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.
start with the identity
cos 2x = 1 - 2sin^2 x
2sin^2 x = 1 - cos 2x
sin^2 x = 1/2 - (1/2)cos 2x
then sin^2 (u/5) = 1/2 - (1/2)cos (2u/5)
and the integral of that is
(1/2)u - (1/2)sin(2u/5) (5/2)
= (1/2)u - (5/4)sin(2u/5) + c
sin^2(u/5) = (1/2)[1 - cos(2u/5)]
The indefinite integral of that with respect to u is
u/2 - (1/2)*sin(2u/5)*(5/2)
= u/2 - (5/4)*sin(2u/5)
To find the indefinite integral of the function h(u) = sin^2((1/5)u), you can use trigonometric identities and integration techniques. Here's how you can approach it step by step:
1. Start by applying the double-angle formula for sine.
sin^2(x) = (1/2)(1 - cos(2x))
2. Substitute (1/5)u for x in the double-angle formula.
sin^2((1/5)u) = (1/2)(1 - cos(2(1/5)u))
3. Simplify the double angle expression inside the parentheses.
sin^2((1/5)u) = (1/2)(1 - cos(2/5)u)
4. Now, integrate the function. The integral of sin^2((1/5)u) is:
∫ sin^2((1/5)u) du = ∫ (1/2)(1 - cos(2/5)u) du
5. Apply the power rule and the constant rule of integration to solve the integral.
∫ (1/2)(1 - cos(2/5)u) du = (1/2)(u - (1/2)(1/2)(1/2/5)sin(2/5)u) + C
Simplify the expression:
= (1/2)(u - (1/20)sin(2/5)u) + C
So, the indefinite integral of h(u) = sin^2((1/5)u) is:
∫ h(u) du = (1/2)(u - (1/20)sin(2/5)u) + C
Note: C is the constant of integration, and it is added to the result since indefinite integration leaves an undetermined constant.