Đáp án: $\dfrac{5}{36}$
Giải thích các bước giải:
Ta có:
$f'(x)=\sin^2x\cdot\cos4x$
$\to f(x)=\int f'(x)dx=\int \sin^2x\cdot\cos4xdx$
$\to f(x)=\int \dfrac12(1-\cos2x)\cdot\cos4xdx$
$\to f(x)=\dfrac12\int \cos4x-\cos2x\cdot\cos4xdx$
$\to f(x)=\dfrac12\int \cos4x-\dfrac12(\cos6x+\cos2x)dx$
$\to f(x)=\dfrac12\int \cos4x-\dfrac12\cos6x-\dfrac12\cos2xdx$
$\to f(x)=\dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4x\right)-\dfrac{1}{12}\sin \left(6x\right)-\dfrac{1}{4}\sin \left(2x\right)+C\right)$
Mà $f(0)=0$
$\to \dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4\cdot 0\right)-\dfrac{1}{12}\sin \left(6\cdot 0\right)-\dfrac{1}{4}\sin \left(2\cdot 0\right)+C\right)=0$
$\to C=0$
$\to f(x)=\dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4x\right)-\dfrac{1}{12}\sin \left(6x\right)-\dfrac{1}{4}\sin \left(2x\right)\right)$
$\to \int^0_{\dfrac{\pi}{2}}f(x)dx=\int^0_{\dfrac{\pi}{2}}\dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4x\right)-\dfrac{1}{12}\sin \left(6x\right)-\dfrac{1}{4}\sin \left(2x\right)\right)dx$
$\to \int^0_{\dfrac{\pi}{2}}f(x)dx=\dfrac{1}{2}\left(-\dfrac{1}{16}\cos \left(4x\right)+\dfrac{1}{72}\cos \left(6x\right)+\dfrac{1}{8}\cos \left(2x\right)\right)\Bigg|^0_{\dfrac{\pi}{2}}$
$\to \int^0_{\dfrac{\pi}{2}}f(x)dx=\dfrac{5}{36}$
