Calculus/integrals

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Expert: - 1/25/2010


Question
I'm having trouble with integrals, and the problem I'm stuck on is this:

(pi/3
| (3cos(theta)+8sin(theta))d(theta) =
)0

I know cos(x)= -sin(x)
and sin(x)= cos(x)but I'm just not sure what to do here.

Answer
Now are variable in the integral was really theta, but I'll change it to x,
so we're doing ∫3cos(x) + 8sin(x) dx from π/3 downto 0.

Since the integral of cos(x) = sin(x), the integral of 3cos(x) is 3sin(x)
and since the integral of sin(x) = -cos(x), the integral of 8sin(x) is -8cos(x).

Once this has been done, evaluate it.
Note that sin(π/3) = √3/2, cos(π/3) = 1/2, sin(0)= 0, and cos(0) = 1.

So take F(x) = 3sin(x) - 8cos(x) and calculate F(π/3) - F(0).

Calculus

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