Đáp án:
$B.\,\dfrac{1}{12}(2\sin2x - \sin4x) + C$
Giải thích các bước giải:
Ta có:
$\quad f'(x)\sin x = f(x)\cos x + 2\sin^2x\cos^3x$
$\Leftrightarrow \dfrac{f'(x)\sin x - f(x)\cos x}{\sin^2x} = 2\cos^3x$
$\Leftrightarrow \left(\dfrac{f(x)}{\sin x}\right)' = 2\cos3x$
$\Leftrightarrow \displaystyle\int\left(\dfrac{f(x)}{\sin x}\right)'dx= \displaystyle\int2\cos3xdx$
$\Leftrightarrow \dfrac{f(x)}{\sin x}=\dfrac23\sin3x + C$
$\Leftrightarrow f(x)= \dfrac23\sin x\sin3x + C\sin x$
Ta lại có:
$\quad f\left(\dfrac{\pi}{4}\right)= \dfrac13$
$\Rightarrow \dfrac23\sin\dfrac{\pi}{4}\cdot\sin\dfrac{3\pi}{4} + C\cdot \sin\dfrac{\pi}{4} = \dfrac13$
$\Rightarrow C = 0$
Do đó: $f(x)= \dfrac23\sin x\sin 3x$
$\Leftrightarrow \displaystyle\int f(x)dx = \displaystyle\int\dfrac23\sin x\sin 3xdx$
$\Leftrightarrow \displaystyle\int f(x)dx =\dfrac13 \displaystyle\int(\cos2x - \cos4x)dx$
$\Leftrightarrow \displaystyle\int f(x)dx = \dfrac16\sin2x - \dfrac{1}{12}\sin4x + C$
$\Leftrightarrow \displaystyle\int f(x)dx = \dfrac{1}{12}(2\sin2x - \sin4x) + C$
