2023-M2-03 – Solving Master

Ans: (a)

p = 4

q = 1

(b)

π + \frac{3}{2} \ln 2

$\begin{array}{rcl} 11 \sin x + 7 \cos x & \equiv & p (3 \sin x + \cos x) + q (3 \cos x - \sin x) \\ 11 \sin x + 7 \cos x & \equiv & (3 p - 3 q) \sin x + (p + 3 q) \cos x \end{array}$

By comparing the components of both sides, we have

${\begin{cases} 3 p - q = 11 & \dots ① \\ p + 3 q = 7 & \dots ② \end{cases}$

$① - 3 \times ②$ , we have

$\begin{array}{rcl} - 10 q & = & - 10 \\ q & = & 1 \end{array}$

Sub. $q = 1$ into $②$ , we have

$\begin{array}{rcl} p + 3 (1) & = & 7 \\ p & = & 4 \end{array}$
$\begin{array}{cl} \int_{0}^{\frac{π}{4}} \frac{11 \sin x + 7 \cos x}{3 \sin x + \cos x} d x \\ = & \int_{0}^{\frac{π}{4}} \frac{4 (3 \sin x + \cos x) + (3 \cos x - \sin x)}{3 \sin x + \cos x} d x \\ = & \int_{0}^{\frac{π}{4}} 4 d x + \int_{0}^{\frac{π}{4}} \frac{3 \cos x - \sin x}{3 \sin x + \cos x} d x \\ = & 4 \int_{0}^{\frac{π}{4}} d x + \int_{0}^{\frac{π}{4}} \frac{d (3 \sin x + \cos x)}{3 \sin x + \cos x} \\ = & 4 {[x]}_{0}^{\frac{π}{4}} + {[\ln | 3 \sin x + \cos x |]}_{0}^{\frac{π}{4}} \\ = & 4 \times \frac{π}{4} + \ln | 3 \sin \frac{π}{4} + \cos \frac{π}{4} | - \ln | 3 \sin 0 + \cos 0 | \\ = & π + \ln (3 \times \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) - \ln 1 \\ = & π + \ln 2 \sqrt{2} \\ = & π + \ln 2^{\frac{3}{2}} \\ = & π + \frac{3}{2} \ln 2 \end{array}$