Let $f ( x ) = 4 \sqrt { 3 } e ^ { - x } \cos x + 6 e ^ { - x }$. (1) Let $a$ and $b$ ($a < b$) be the values of $x$ satisfying $f ( x ) = 0$ on $0 \leqq x < 2 \pi$. Then, $$a = \frac { \mathbf{A} } { \mathbf{B} } \pi , \quad b = \frac { \mathbf{C} } { \mathbf{D} } \pi$$ (2) The values of the constants $p$ and $q$ satisfying $$\frac { d } { d x } \left( p e ^ { - x } \cos x + q e ^ { - x } \sin x \right) = e ^ { - x } \cos x$$ are given by $$p = \frac { \mathbf { E F } } { \mathbf { G } } , \quad q = \frac { \mathbf { H } } { \mathbf { I } } .$$ (3) Using the values of $a$ and $b$ obtained in (1), we set $A = e ^ { - a }$ and $B = e ^ { - b }$. When we calculate the value of $\int _ { a } ^ { b } f ( x ) d x$, we obtain $$\int _ { a } ^ { b } f ( x ) d x = ( \mathbf { J } - \sqrt { \mathbf{J} } \mathbf { K } ) A - ( \mathbf { L } + \sqrt { \mathbf{L} } ) B .$$
Let $f ( x ) = 4 \sqrt { 3 } e ^ { - x } \cos x + 6 e ^ { - x }$.
(1) Let $a$ and $b$ ($a < b$) be the values of $x$ satisfying $f ( x ) = 0$ on $0 \leqq x < 2 \pi$. Then,
$$a = \frac { \mathbf{A} } { \mathbf{B} } \pi , \quad b = \frac { \mathbf{C} } { \mathbf{D} } \pi$$
(2) The values of the constants $p$ and $q$ satisfying
$$\frac { d } { d x } \left( p e ^ { - x } \cos x + q e ^ { - x } \sin x \right) = e ^ { - x } \cos x$$
are given by
$$p = \frac { \mathbf { E F } } { \mathbf { G } } , \quad q = \frac { \mathbf { H } } { \mathbf { I } } .$$
(3) Using the values of $a$ and $b$ obtained in (1), we set $A = e ^ { - a }$ and $B = e ^ { - b }$. When we calculate the value of $\int _ { a } ^ { b } f ( x ) d x$, we obtain
$$\int _ { a } ^ { b } f ( x ) d x = ( \mathbf { J } - \sqrt { \mathbf{J} } \mathbf { K } ) A - ( \mathbf { L } + \sqrt { \mathbf{L} } ) B .$$