Math  /  Calculus

QuestionUse the substitution formula abf(g(x))g(x)dx=g(a)g(b)f(u)du\int_{a}^{b} f(g(x)) g^{\prime}(x) d x=\int_{g(a)}^{g(b)} f(u) d u where g(x)=ug(x)=u, to evaluate the following integrals. (a) 02πcosz6+sinzdz\int_{0}^{2 \pi} \frac{\cos z}{\sqrt{6+\sin z}} d z (b) ππcosz6+sinzdz\int_{-\pi}^{\pi} \frac{\cos z}{\sqrt{6+\sin z}} d z (a) 02πcosz6+sinzdz=0\int_{0}^{2 \pi} \frac{\cos z}{\sqrt{6+\sin z}} d z=0 (Simplify your answer. Type an integer or a fraction.) (b) ππcosz6+sinzdz=\int_{-\pi}^{\pi} \frac{\cos z}{\sqrt{6+\sin z}} d z= \square (Simplify your answer. Type an integer or a fraction.)

Studdy Solution
For integral (b), ππcosz6+sinzdz\int_{-\pi}^{\pi} \frac{\cos z}{\sqrt{6+\sin z}} d z, we use the same substitution u=6+sinzu = 6 + \sin z.
2.1. Differentiate u=6+sinzu = 6 + \sin z to find dudu: du=coszdz du = \cos z \, dz
2.2. Change the limits of integration: - When z=πz = -\pi, u=6+sin(π)=6u = 6 + \sin(-\pi) = 6. - When z=πz = \pi, u=6+sin(π)=6u = 6 + \sin(\pi) = 6.
2.3. Substitute into the integral: ππcosz6+sinzdz=661udu \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:
0 \boxed{0}

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