# Evaluate the integral. {eq}\displaystyle \int_0^{\dfrac \pi 6} \sqrt {1 + \cos 2 x}\ dx {/eq}

## Question:

Evaluate the integral.

{eq}\displaystyle \int_0^{\dfrac \pi 6} \sqrt {1 + \cos 2 x}\ dx {/eq}

## Fundamental Theorem of Calculus:

For evaluating the final result of the definite integral, the theorem that is used is known as the Fundamental theorem of calculus, part 2. This theorem states that to solve a definite integral first we have to find the antiderivative of the function and then evaluate the value of the antiderivative at the boundaries of the given integral and then subtract these two values.

$$\int_{a}^{b}f(x)dx=F(b)-F(a)$$

{eq}F(x) {/eq} is an antiderivative of {eq}f(x) {/eq}

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Given:

$$\displaystyle \int_0^{\dfrac{\pi}{6}} \sqrt{1 + \cos 2 x}\ dx$$

We have to evaluate the integral.

The given definite integral can be...