Đáp án:
\(S = 12\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad I = \displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{1 + 2(\sin x - \cos x)^2}{1 + 2\cos^2x}dx\\
\to I = \displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{3 - 2\sin2x}{2 + \cos2x}dx\\
\to I =3\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{1}{2+\cos2x}dx -2\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{\sin2x}{2 + \cos2x}dx\\
\to I = I_1+ I_2\\
Với\ \begin{cases}I_1 = 3\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{1}{2+\cos2x}dx\\
I_2=-2\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{\sin2x}{2 + \cos2x}dx\end{cases}\\
+)\quad I_1 = 3\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{1}{2+\cos2x}dx\\
Đặt\,\,u = 2x\\
\to du = 2dx\\
\text{Đổi cận:}\\
x \quad \Big|\quad 0\qquad \dfrac{\pi}{4}\\
\overline{u\quad \Big|\quad 0\qquad \dfrac{\pi}{2}}\\
\text{Ta được:}\\
\quad I_1 = \dfrac32\displaystyle\int\limits_0^{\tfrac{\pi}{2}}\dfrac{1}{2+\cos u}du\\
Đặt\,t = \tan\dfrac u2\\
\to dt = \dfrac{1}{2\cos^2\dfrac u2}du\\
\text{Đổi cận:}\\
u\quad \Big|\quad 0 \qquad \dfrac{\pi}{2}\\
\overline{\ t\quad \Big|\quad 0\qquad 1}\\
\text{Ta được:}\\
\quad I_1 = \dfrac32\displaystyle\int\limits_0^1\dfrac{2}{(t^2 + 1)\left(\dfrac{1-t^2}{t^2 +1} +2\right)}dt\\
\to I_1 = 3\displaystyle\int\limits_0^1\dfrac{1}{t^2 + 3}dt\\
\to I_1 = \sqrt3\cdot \arctan\dfrac{x}{\sqrt3}\Bigg|_0^1\\
\to I_1 = \dfrac{\pi\sqrt3}{6}\\
+)\quad I_2 = -2\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{\sin2x}{2 + \cos2x}dx\\
Đặt\,\,u = 2 + \cos2x\\
\to du = -2\sin2xdx\\
\text{Đổi cận:}\\
x \quad \Big|\quad 0\qquad \dfrac{\pi}{4}\\
\overline{u\quad \Big|\quad 3\qquad 2}\\
\text{Ta được:}\\
\quad I_2 = \displaystyle\int\limits_3^2\dfrac{1}{u}du\\
\to I_2 = \ln|u|\Bigg|_3^2\\
\to I_2 = \ln2 - \ln3\\
\to I_2 = \ln\dfrac23\\
\text{Ta được:}\\
\quad I = \dfrac{\pi\sqrt3}{6} + \ln\dfrac23\\
\Rightarrow \begin{cases}a = 1\\b = 6\\c=2\\d = 3\end{cases}\\
\Rightarrow S = a + b + c + d = 12
\end{array}\)
