Math / Calculus

QuestionUse the substitution formula $\int_{a}^{b} f(g(x)) g^{\prime}(x) d x=\int_{g(a)}^{g(b)} f(u) d u$ where $g(x)=u$ , to evaluate the following integrals. (a) $\int_{0}^{2 \pi} \frac{\cos z}{\sqrt{6+\sin z}} d z$ (b) $\int_{-\pi}^{\pi} \frac{\cos z}{\sqrt{6+\sin z}} d z$ (a) $\int_{0}^{2 \pi} \frac{\cos z}{\sqrt{6+\sin z}} d z=0$ (Simplify your answer. Type an integer or a fraction.) (b) $\int_{-\pi}^{\pi} \frac{\cos z}{\sqrt{6+\sin z}} d z=$ $\square$ (Simplify your answer. Type an integer or a fraction.)

Studdy Solution

STEP 1

1. We are given integrals that require substitution for evaluation.
2. The substitution formula $\int_{a}^{b} f(g(x)) g^{\prime}(x) d x=\int_{g(a)}^{g(b)} f(u) d u$ will be used.
3. The substitution $g(x) = u$ is applicable, where $g(x)$ is a function whose derivative appears in the integrand.

STEP 2

1. Evaluate integral (a) using substitution.
2. Evaluate integral (b) using substitution.

STEP 3

For integral (a), $\int_{0}^{2 \pi} \frac{\cos z}{\sqrt{6+\sin z}} d z$ , we use the substitution $u = 6 + \sin z$ .
1.1. Differentiate $u = 6 + \sin z$ to find $du$ : $du = \cos z \, dz$
1.2. Change the limits of integration: - When $z = 0$ , $u = 6 + \sin(0) = 6$ . - When $z = 2\pi$ , $u = 6 + \sin(2\pi) = 6$ .
1.3. Substitute into the integral: $\int_{0}^{2\pi} \frac{\cos z}{\sqrt{6+\sin z}} \, dz = \int_{6}^{6} \frac{1}{\sqrt{u}} \, du$
1.4. Evaluate the integral: Since the limits of integration are the same, the integral evaluates to 0.

STEP 4

For integral (b), $\int_{-\pi}^{\pi} \frac{\cos z}{\sqrt{6+\sin z}} d z$ , we use the same substitution $u = 6 + \sin z$ .
2.1. Differentiate $u = 6 + \sin z$ to find $du$ : $du = \cos z \, dz$
2.2. Change the limits of integration: - When $z = -\pi$ , $u = 6 + \sin(-\pi) = 6$ . - When $z = \pi$ , $u = 6 + \sin(\pi) = 6$ .
2.3. Substitute into the integral: $\int_{-\pi}^{\pi} \frac{\cos z}{\sqrt{6+\sin z}} \, dz = \int_{6}^{6} \frac{1}{\sqrt{u}} \, du$
2.4. Evaluate the integral: Since the limits of integration are the same, the integral evaluates to 0.
The value of integral (b) is:
$\boxed{0}$

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