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Chapter 5: Problem 47

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{\pi / 8} \cos 2 x d x$$

Short Answer

Expert verified
Answer: The value of the integral is $$\frac{\sqrt{2}}{4}$$.

Step by step solution

01

Find the Antiderivative of the Given Function

To find the antiderivative of the cosine of 2x, we need to use the reverse chain rule or u-substitution. Let's first set u = 2x. Then, the derivative of u with respect to x is du/dx = 2. We can rewrite the function in terms of u: $$\int\cos 2x\, dx = \int\cos u\,\frac{1}{2}\, du$$ Now, we can find the antiderivative of the cosine function: $$\frac{1}{2}\int\cos u\, du = \frac{1}{2}\sin u + C$$ Now, substitute back the original variable, x, to obtain the antiderivative in terms of x: $$\frac{1}{2}\sin(2x) + C$$
02

Evaluate the Antiderivative at the Given Limits

Now, we can use the Fundamental Theorem of Calculus to evaluate the definite integral: $$\int_{0}^{\pi / 8} \cos 2x\, dx = \left[\frac{1}{2}\sin(2x)\right]_{0}^{\pi / 8}$$ Evaluate the antiderivative at the upper limit, 蟺/8: $$\frac{1}{2}\sin\left(2\left(\frac{\pi}{8}\right)\right) = \frac{1}{2}\sin\left(\frac{\pi}{4}\right) = \frac{1}{2}\left(\frac{\sqrt{2}}{2}\right)$$ Evaluate the antiderivative at the lower limit, 0: $$\frac{1}{2}\sin(2\cdot0) = \frac{1}{2}\sin(0) = 0$$
03

Subtract the Function's Value at the Lower Limit from Its Value at the Upper Limit

Now, subtract the function's value at the lower limit from its value at the upper limit to find the integral's value: $$\int_{0}^{\pi / 8} \cos 2x\, dx = \frac{1}{2}\left(\frac{\sqrt{2}}{2}\right) - 0 = \frac{\sqrt{2}}{4}$$ So, the value of the integral is: $$\int_{0}^{\pi / 8} \cos 2x\, dx = \frac{\sqrt{2}}{4}$$

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