$\\int_{0}^{\\frac{\\pi^{2}}{4}}\\frac{\\sin\\sqrt{x}}{\\sqrt{x}}dx$

Explanation:

Step1: Use substitution

Let ( t = \sqrt{x}), then (x=t^{2}) and (dx = 2t\ dt). When (x = 0), (t=0); when (x=\frac{\pi^{2}}{4}), (t=\frac{\pi}{2}). The integral becomes (\int_{0}^{\frac{\pi}{2}}\frac{\sin t}{t}\cdot2t\ dt=2\int_{0}^{\frac{\pi}{2}}\sin t\ dt)

Step2: Integrate (\sin t)

We know that (\int\sin t\ dt=-\cos t + C). So (2\int_{0}^{\frac{\pi}{2}}\sin t\ dt=2\left[-\cos t\right]_{0}^{\frac{\pi}{2}})

Step3: Evaluate the definite - integral

(2\left(-\cos\frac{\pi}{2}+\cos0\right)) Since (\cos\frac{\pi}{2} = 0) and (\cos0=1) (2(0 + 1)=2)

Answer:

(2)