Đáp án:
$A.\ I = \dfrac92$
Giải thích các bước giải:
$\quad I = \displaystyle\int\limits_0^{\pi}\sin2xf(\cos x)dx$
$\Leftrightarrow I = 2\displaystyle\int\limits_0^{\pi}\cos xf(\cos x)\sin xdx$
Đặt $t= \cos x$
$\Rightarrow dt = - \sin xdx$
Đổi cận:
$\begin{array}{c|ccc}x&0&&&\pi\\\hline t&1&&&-1\\\end{array}$
Ta được:
$\quad I = - 2\displaystyle\int\limits_1^{-1}tf(t)dt$
$\Leftrightarrow I = 2\displaystyle\int\limits_{-1}^1tf(t)dt$
$\Leftrightarrow I = 2\left(\displaystyle\int\limits_{-1}^0tf(t)dt + \displaystyle\int\limits_{0}^1tf(t)dt\right)$
$\Leftrightarrow I = 2\left[\displaystyle\int\limits_{-1}^0t(2t-2)dt + \displaystyle\int\limits_{0}^1t(t^2 + 4t -2)dt\right]$
$\Leftrightarrow I = 2\left[\displaystyle\int\limits_{-1}^0(2t^2-2t)dt + \displaystyle\int\limits_{0}^1(t^3+ 4t^2 -2t)dt\right]$
$\Leftrightarrow I = 2\left[\left(\dfrac23t^3 - t^2\right)\Bigg|_{-1}^0 + \left(\dfrac{t^4}{4} + \dfrac43t^3 - t^2\right)\Bigg|_0^1\right]$
$\Leftrightarrow I = 2\left(\dfrac53 + \dfrac{7}{12}\right)$
$\Leftrightarrow I = \dfrac92$
