Given that $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} x \cos x \, dx = ax + b$$ find the exact value of $\alpha$ and the exact value of $b$ - AQA - A-Level Maths Pure - Question 8 - 2021 - Paper 3

Continuing from the previous result:

$\int x \cos x \, dx = x \sin x + \cos x$

This leads to a definite integral when evaluated from $\frac{\pi}{4}$ to $\frac{\pi}{3}$.

Now, we evaluate the above expression at the limits:

1. At $x = \frac{\pi}{3}$:

   $\left(\frac{\pi}{3} \sin \left(\frac{\pi}{3}\right)\right) + \cos \left(\frac{\pi}{3}\right)$

   This equals: $\frac{\pi}{3} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\pi \sqrt{3}}{6} + \frac{1}{2}$

2. At $x = \frac{\pi}{4}$:

   $\left(\frac{\pi}{4} \sin \left(\frac{\pi}{4}\right)\right) + \cos \left(\frac{\pi}{4}\right)$

   This equals: $\frac{\pi}{4} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \frac{\pi \sqrt{2}}{8} + \frac{\sqrt{2}}{2}$

Calculating the difference:

$\left(\frac{\pi \sqrt{3}}{6} + \frac{1}{2} \right) - \left(\frac{\pi \sqrt{2}}{8} + \frac{\sqrt{2}}{2}\right)$

After simplifying, we find the exact values for $\alpha$ and $b$.

Finally, by evaluating the difference and reorganizing, we can express the integral result in the desired format:

$ax + b$

Thus, we find:

• $\alpha = \frac{\sqrt{3}}{6} - \frac{\sqrt{2}}{8}$
• $b = \frac{1}{2} - \frac{\sqrt{2}}{2}$.