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Chapter 7: Problem 46

Use the approaches discussed in this section to evaluate the following integrals. $$\int_{0}^{\pi / 2} \sqrt{1+\cos 2 x} d x$$

Short Answer

Expert verified
Answer: The value of the definite integral is \(\sqrt{2}\).

Step by step solution

01

Use the double-angle formula for cosine

The first step is to use the identity for the double-angle formula for \(\cos\): $$\cos{2x} = 2\cos^2{x} - 1$$ This will help simplify the function inside the square root. We can rewrite the integrand as: $$\sqrt{1 + \cos{2x}} = \sqrt{1 + (2\cos^2{x} - 1)} = \sqrt{2\cos^2{x}}$$ Now, we can rewrite the given integral as: $$\int_{0}^{\pi / 2} \sqrt{1+\cos 2 x} d x = \int_{0}^{\pi / 2} \sqrt{2\cos^2{x}} d x$$
02

Apply the square root to the integrand

In this step, we will apply the square root to the integrand, which simplifies it further: $$\int_{0}^{\pi / 2} \sqrt{2\cos^2{x}} d x = \int_{0}^{\pi / 2} \sqrt{2}\cos{x} d x$$
03

Integrate the simplified function

Now, we can integrate the simplified integrand with respect to \(x\): $$\int_{0}^{\pi / 2} \sqrt{2}\cos{x} d x = \sqrt{2}\left[\int_{0}^{\pi / 2} \cos{x} d x\right]$$ The definite integral of \(\cos{x}\) with respect to \(x\) on the interval \([0, \frac{\pi}{2}]\) is: $$\int_{0}^{\pi / 2} \cos{x} d x = \sin{x}\Big|_0^{\pi / 2} = \sin{\frac{\pi}{2}} - \sin{0} = 1$$
04

Combine the result

Now that we have found the definite integral of the simplified expression, we can combine the result: $$\sqrt{2}\left[\int_{0}^{\pi / 2} \cos{x} d x\right] = \sqrt{2}(1) = \sqrt{2}$$ Therefore, the value of the definite integral is: $$\int_{0}^{\pi / 2} \sqrt{1+\cos 2 x} d x = \sqrt{2}$$

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