Give that \( \int \sqrt{1 + \sin 2x} dx = \sqrt{2} \sin(x + a) + b \) and we need to find the constants \( a \) and \( b \).
Solution:
First, simplify the integration:
\( \begin{aligned} &\int \sqrt{1+\sin 2 x} \,d x \\ &=\int \sqrt{1+\cos \left(\frac{\pi}{2}-2 x\right)} \, d x \\ &=\int \sqrt{2 \cos ^{2}\left(\frac{x}{4}-x\right)} \, d x \\ &=\sqrt{2} \cdot \int \cos \left(\frac{\pi}{4}-x\right) \, d x \\ &=\sqrt{2} \cdot \frac{\sin \left(\frac{\pi}{4}-x\right)}{-1}+C, \quad \text{C is a constant} \\ &=\sqrt{2} \cdot \frac{\sin \left(\frac{\pi}{4}-x\right)}{-1}+C \\ &=\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)+C \\ \end{aligned} \)Comparing with \( \sqrt{2} \sin(x + a) + b \), we have: \(a = -\frac{\pi}{4}, \quad b = 0 \)
Thus, the constants are:
\( \boxed{a = -\dfrac{\pi}{4}} \quad \text{and} \quad \boxed{b = 0} \)Please let me know in the comments if you find any errors in this solution.