Problem 9 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{0}^{3 \pi / 8} \sin \left(2 x-\frac{\pi}{4}\right) d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral \(\int_0^{3\pi/8} \sin(2x - \frac{\pi}{4}) dx\). Answer: The integral evaluates to 0.

Step 1: Substitution

Let us substitute \(u = 2x - \frac{\pi}{4}\). Now, we differentiate with respect to \(x\) to find the derivative: \(\dfrac{du}{dx} = 2\). Then, multiply both sides by \(dx\) to isolate: \(du = 2 dx\). Next, we need to change the bounds of integration. For the lower bound, when \(x = 0\), we get \(u = 2(0) - \frac{\pi}{4} = -\frac{\pi}{4}\). For the upper bound, when \(x = \frac{3\pi}{8}\), we get \(u = 2(\frac{3\pi}{8}) - \frac{\pi}{4} = \frac{3\pi}{4}\). The integral now changes to: $$\frac{1}{2} \int_{-\pi/4}^{3\pi/4} \sin(u) du$$ The extra factor of \(\frac{1}{2}\) is due to the substitution \(du = 2 dx\).

Step 2: Evaluate the Definite Integral

Now we can integrate \(\sin(u)\) with respect to \(u\): $$\frac{1}{2} \left[ -\cos(u) \right]\bigg|_{-\pi/4}^{3\pi/4}$$ Now evaluate the antiderivative at the bounds: $$\frac{1}{2} \left[-\cos\left(\frac{3\pi}{4}\right) - (-\cos\left(-\frac{\pi}{4}\right))\right]$$

Step 3: Return to the Original Variable

Recall our earlier substitution \(u = 2x - \frac{\pi}{4}\). Since we already have our answer in the desired form, we can simply substitute back to \(x\). To find the value of our answer, we evaluate the expression: $$\frac{1}{2} \left[-\cos\left(\frac{3\pi}{4}\right) - (-\cos\left(-\frac{\pi}{4}\right))\right]=\frac{1}{2} \left(\cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{3\pi}{4}\right)\right) = \frac{1}{2} \left(\frac{\sqrt{2}}{2} + \frac{-\sqrt{2}}{2}\right)=0$$ Thus, the value of the integral is 0.