The value of $\displaystyle\int^{199\pi/2}_{-\pi/2}\sqrt{(1+\cos 2x)}dx$ is?

Let $I=\int_{-\pi / 2}^{199 \pi / 2} \sqrt{(1+\cos 2 x)} d x$

We have to find the value of I.

From trigonometric identity,

$\therefore I=\int_{-\pi / 2}^{1+\cos 2 x=2 \cos ^{2} x} \sqrt{2 \cos ^{2} x} d x=\sqrt{2} \int_{-\pi / 2}^{199 / 2} \sqrt{\cos ^{2} x} d x$

The period of the periodic function $\cos ^{2} x$ is $\pi$.

Using the property of periodicity,

$\int_{a}^{a+n T} f(x) d x=n \int_{0}^{T} f(x) d x$, where $T$ is the period of the function.

$\begin{array}{l} \therefore I=\sqrt{2} \int_{-\pi / 2}^{(-\pi / 2+100 \pi)} \sqrt{\cos ^{2} x} d x \\ \therefore I=100 \sqrt{2} \int_{0}^{\pi} \sqrt{\cos ^{2} x} d x \\ {\left[\sqrt{\cos ^{2} x}=\cos x \quad \forall x \in\left[0, \frac{\pi}{2}\right], \sqrt{\cos ^{2} x}=-\cos x \forall x \in\left[\frac{\pi}{2}, \pi\right]\right]} \\ \therefore I=200 \sqrt{2}\left[\int_{0}^{\pi / 2} \cos x d x+\int_{\pi / 2}^{\pi}(-\cos x) d x\right] \\ \therefore I=100 \sqrt{2}\left[[\sin x]_{0}^{\pi / 2}+[-\sin x]_{\pi / 2}^{\pi}\right] \\ \therefore I=100 \sqrt{2}\left[\left[\sin \frac{\pi}{2}-\sin 0\right]+\left[(-\sin \pi)-\left(-\sin \frac{\pi}{2}\right)\right]\right. \\ \therefore I=100 \sqrt{2}[(1-0)+(0+1)] \\ \therefore I=200 \sqrt{2} \end{array}$