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Evaluate a definite integral of cosine using FTC Part 2

Evaluate the definite integral:

∫π2πcos⁡(x) dx\int_{\pi}^{2\pi} \cos(x)\, dx

Use the fact that sin⁡(x)\sin(x) is an antiderivative of cos⁡(x)\cos(x), and apply the Fundamental Theorem of Calculus (Part 2) to find the exact value.

Hint

Recall that if F′(x)=f(x)F'(x) = f(x), then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a). Here, let F(x)=sin⁡(x)F(x) = \sin(x), and evaluate it at the upper and lower limits.

Step-by-step solution

Since ddx[sin⁡(x)]=cos⁡(x)\frac{d}{dx}[\sin(x)] = \cos(x), an antiderivative of cos⁡(x)\cos(x) is F(x)=sin⁡(x)F(x) = \sin(x). By the Fundamental Theorem of Calculus Part 2, ∫π2πcos⁡(x) dx=F(2π)−F(π)=sin⁡(2π)−sin⁡(π)\int_{\pi}^{2\pi} \cos(x)\,dx = F(2\pi) - F(\pi) = \sin(2\pi) - \sin(\pi).

We know sin⁡(2π)=0\sin(2\pi) = 0 and sin⁡(π)=0\sin(\pi) = 0, so the result is 0−0=00 - 0 = 0. Thus ∫π2πcos⁡(x) dx=0\int_{\pi}^{2\pi}\cos(x)\,dx = 0, meaning the net signed area between x=πx=\pi and x=2πx=2\pi under cos⁡(x)\cos(x) is zero.

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