91Ó°ÊÓ

Understanding the double-angle cosine formula is crucial in simplifying trigonometric expressions, especially when dealing with integrals. The double-angle formulas are a subset of trigonometric identities that express trigonometric functions of double angles in terms of single angles. One of these formulas is for the cosine function:

\[ \cos(2x) = \cos^2(x) - \sin^2(x) \]
However, it's worth noting that this particular identity can be expressed in different forms thanks to the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \). Therefore, we can also write the double-angle cosine formula as:

\[ \cos(2x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x) \]
In the context of our integral evaluation problem, the second form is used to transform the integrand from \( \sqrt{1-\cos(2x)} \) to a simpler expression involving \( \sin(x) \) alone. This kind of manipulation is essential as it can significantly simplify the evaluation of integrals involving trigonometric functions.

When evaluating integrals, finding the antiderivative, which is the reverse of taking the derivative, is often necessary. The antiderivative of a given function is another function whose derivative yields the original function. For the sine function, the antiderivative plays a pivotal role in solving a multitude of integral problems.

The antiderivative of \( \sin(x) \) is:

\[ \int \sin(x) dx = -\cos(x) + C \]
Where \( C \) is the constant of integration that appears whenever we take the indefinite integral. In the given exercise, after applying the double-angle cosine formula, we are left with an integral of \( \sqrt{2}\sin(x) \) that needs to be evaluated. Recognizing that the constant \( \sqrt{2} \) can be factored out of the integral, we turn our focus to finding the antiderivative of \( \sin(x) \), which in this case leads us to \( -\sqrt{2}\cos(x) \) once the constant is reintroduced. This step is pivotal to finding the definite integral over the prescribed limits.

The evaluation of definite integrals includes applying the integral limits, which play an essential role in calculating the area under a curve or other related quantities. When we compute a definite integral, we do so over a specific interval, from a lower limit \( a \) to an upper limit \( b \) as follows:

\[ \int_{a}^{b} f(x) dx = F(b) - F(a) \]
Here, \( F(x) \) is the antiderivative of \( f(x) \). Applying the limits means that we evaluate the antiderivative at the upper limit \( b \) and subtract the value of the antiderivative at the lower limit \( a \).

For our exercise, the limits are from 0 to \( \frac{\pi}{2} \). After finding the antiderivative of the simplified integrand, we substitute these limits into the expression, giving us the final value of the integral. It's important to accurately calculate the antiderivative at each limit, which in this case involves evaluating cosines at 0 and \( \frac{\pi}{2} \) to reach the solution.