Can someone help me with my integration hmwk?

Question

Evaluate the integral from -pi (lower bound) to pi (upper bound)

cos x/(4+3sinx)^1/2 dx

any help would be much appreciated

Answer 1

To do this integral, let u = 4+3sinx, then du = 3cosxdx, so the integral becomes:

Integral[(1/3)u^(-1/2)du] =

(2/3)u^(1/2) =

(2/3)(4 + 3sinx)^1/2

Now you can plug in the limits of integration to get the result is 0.

Answer 2

Integral ( cos(x) / (4 + 3sin(x))^(1/2) dx )

I'm going to rearrange this to make the substitution more obvious.

Integral ( [1 / (4 + 3sin(x))^(1/2)] [cos(x) dx] )

Now, we do the substitution.
Let u = 4 + 3sin(x). Then
du = 3cos(x) dx. Dividing both sides by 3,

(1/3) du = cos(x) dx. {Note: cos(x) dx is the tail end of our integral, and we substitute accordingly.}

Integral ( [1 / (u^(1/2))] [(1/3) du] )

Pulling the constant 1/3 out of the integral,

(1/3) * Integral ( [1 / u^(1/2)] du )

Note that 1/u^(1/2) is the same as u^(-1/2)

(1/3) * Integral ( u^(-1/2) du )

Now, we integrate normally, using the reverse power rule.

(1/3) * [2u^(1/2)], OR

(2/3) u^(1/2)

Substituting back u = 4 + 3sin(x), we have

(2/3) [4 + 3sin(x)]^(1/2)

I purposely didn't add a +C because this is a definite integral evaluated from -pi to pi. We evaluate it as such. If we let
f(x) = (2/3) [4 + 3sin(x)]^(1/2)

By the property of the definite integral, we want to calculate
f(pi) - f(-pi)

[(2/3) [4 + 3sin(pi)]^(1/2)] - [(2/3) [4 + 3sin(-pi)]^(1/2)]

[(2/3) [4 + 0]^(1/2)] - [(2/3) [4 + 0]^(1/2)]

[(2/3) (2)] - [(2/3) (2)] = 0

Answer 3

use w subsitution

first, find the indefinite integral and use w=4+3sin(x) so dw=3cos(x) dx -> dw/3=cos(x) dx
so
(integral) 1/(3*w^(1/2)) dw
2w^(1/2)/3
2*[4+3sin(x)]^(1/2)/3 now integrate that from -pi to pi

I think you can handle it from here