Question

\int_{0}^{3\pi/8} \sin\left(2x - \frac{\pi}{4}\
ight) dx

Explanation:

Step1: Use substitution method

Let \( u = 2x - \frac{\pi}{4} \), then \( du = 2dx \), so \( dx=\frac{1}{2}du \).
When \( x = 0 \), \( u=2\times0 - \frac{\pi}{4}=-\frac{\pi}{4} \); when \( x=\frac{3\pi}{8} \), \( u = 2\times\frac{3\pi}{8}-\frac{\pi}{4}=\frac{3\pi}{4}-\frac{\pi}{4}=\frac{\pi}{2} \).
The integral becomes \(\frac{1}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{2}}\sin(u)du\)

Step2: Integrate \(\sin(u)\)

The antiderivative of \(\sin(u)\) is \(-\cos(u)\). So we have \(\frac{1}{2}[-\cos(u)]_{-\frac{\pi}{4}}^{\frac{\pi}{2}}\)

Step3: Evaluate the definite integral

First, substitute the upper limit \(\frac{\pi}{2}\): \(-\cos(\frac{\pi}{2}) = 0\)
Then, substitute the lower limit \(-\frac{\pi}{4}\): \(-\cos(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}\) (since \(\cos(-\theta)=\cos(\theta)\) and \(\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\))
Now, \(\frac{1}{2}[0 - (-\frac{\sqrt{2}}{2})]=\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4}\)

Answer:

\(\frac{\sqrt{2}}{4}\)