Ans: (a) $p=4$, $q=1$ (b) $\pi+\dfrac{3}{2}\ln 2$
-
$\begin{array}{rcl}
11\sin x +7\cos x & \equiv & p(3\sin x +\cos x)+q(3\cos x -\sin x) \\
11 \sin x+7\cos x & \equiv &(3p-3q)\sin x+(p+3q)\cos x
\end{array}$By comparing the components of both sides, we have
$\left\{ \begin{array}{ll}
3p-q=11 & \ldots \unicode{x2460} \\
p+3q=7 & \ldots \unicode{x2461}
\end{array}\right.$$\unicode{x2460}-3\times\unicode{x2461}$, we have
$\begin{array}{rcl}
-10q & = & -10 \\
q & = & 1
\end{array}$Sub. $q=1$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
p+3(1) & = & 7 \\
p & = & 4
\end{array}$ -
$\begin{array}{cl}
& \dint_0^\frac{\pi}{4} \dfrac{11\sin x+7\cos x}{3 \sin x +\cos x} dx \\
= & \dint_0^\frac{\pi}{4}\dfrac{4(3\sin x+\cos x)+(3\cos x-\sin x)}{3\sin x +\cos x}dx \\
= & \dint_0^\frac{\pi}{4} 4dx + \dint_0^\frac{\pi}{4}\dfrac{3\cos x-\sin x}{3\sin x+\cos x} dx \\
= & 4\dint_0^\frac{\pi}{4} dx +\dint_0^\frac{\pi}{4} \dfrac{d(3\sin x+\cos x)}{3\sin x+\cos x} \\
= & 4\left[ x\right]_0^\frac{\pi}{4} +\left[\ln \left| 3\sin x+\cos x\right|\right]_0^\frac{\pi}{4} \\
= & 4\times \dfrac{\pi}{4} + \ln\left| 3\sin \dfrac{\pi}{4}+\cos \dfrac{\pi}{4}\right| -\ln\left| 3\sin 0+\cos 0\right| \\
= & \pi +\ln\left( 3 \times \dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\right) -\ln 1 \\
= & \pi +\ln2\sqrt{2} \\
= & \pi +\ln2^\frac{3}{2} \\
= & \pi +\dfrac{3}{2} \ln 2
\end{array}$
