Đáp án:
$\displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{4}}2^x\left(\dfrac{2^{-x}}{\sin^2x} + 3\right)dx=-1+\sqrt3 + \dfrac{3.2^{\tfrac{\pi}{6}}\left(2^{\tfrac{\pi}{12}} -1\right)}{\ln2}$
Giải thích các bước giải:
$\quad I =\displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{4}}2^x\left(\dfrac{2^{-x}}{\sin^2x} + 3\right)dx$
$\Leftrightarrow I = \displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{4}}\left(\dfrac{1}{\sin^2x} + 3.2^x\right)dx$
$\Leftrightarrow I =\left(-\cot x + \dfrac{3.2^x}{\ln2}\right)\Bigg|_{\tfrac{\pi}{6}}^{\tfrac{\pi}{4}}$
$\Leftrightarrow I = -1+\sqrt3 + \dfrac{3.2^{\tfrac{\pi}{6}}\left(2^{\tfrac{\pi}{12}} -1\right)}{\ln2}$
