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Evaluate the integral: integral from 0 to pi/2 of sqrt(1 + sin(2x)) dx.

Question:

Evaluate the integral: {eq}\int_{0}^{\frac{\pi}{2}} \sqrt{1 + \sin(2x)} \, \mathrm{d}x {/eq}.

Answer and Explanation:


Given {eq}\begin{align*} & \int_{0}^{\frac{\pi}{2}} \sqrt{1 + \sin(2x)} \ dx \end{align*} {/eq}

Resolving the above terms, we get:

{eq}\begin{align*} & = \int_{0}^{\frac{\pi}{2}} \sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x} \ dx \ \ \ \ \ \ \ \left[ \ Because \ (\sin^2 x + \cos^2 x) = 1, \ \ \ \sin 2x= 2 \sin x \cos x \right] \\ \\ & = \int_{0}^{\frac{\pi}{2}} \sqrt{(\sin x + \cos x)^2} \ dx \\ \\ & = \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x ) \ dx \\ \\ & = \left[ -\cos x + \sin x \right]_{0}^{\frac{\pi}{2}} \\ \\ &= \left[ -\cos \frac{\pi}{2} + \sin \frac{\pi}{2} + \cos 0 - \sin 0 \right] \\ \\ &= \left[ -0 + 1 + 1 - 0 \right] \\ \\ &= 2 \end{align*} {/eq}


Learn more about this topic:

Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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