Đáp án:
1) $\displaystyle\int f(x)dx = x\cos x + \sin x + C$
2) $\displaystyle\int\limits_0^5f(x)dx= \dfrac52$
Giải thích các bước giải:
1) Ta có:
$\quad \sin xf'(x) - \cos xf(x)= \sin^2x$
$\Leftrightarrow \dfrac{\sin xf'(x) - \cos xf(x)}{\sin^2x} = 1$
$\Leftrightarrow \left[\dfrac{f(x)}{\sin x}\right]' = 1$
$\Leftrightarrow \dfrac{f(x)}{\sin x} = x + C$
$\Leftrightarrow f(x)= x\sin x + C\sin x$
Ta lại có:
$\quad f\left(\dfrac{\pi}{2}\right) = \dfrac{\pi}{2}$
$\Leftrightarrow \dfrac{\pi}{2}\cdot\sin\dfrac{\pi}{2} + C\cdot \sin\dfrac{\pi}{2} = \dfrac{\pi}{2}$
$\Leftrightarrow C = 0$
Do đó:
$f(x)= x\sin x$
Ta được:
$\quad \displaystyle\int f(x)dx =\displaystyle\int x\sin xdx$
Đặt $\begin{cases}u = x\\dv =\sin xdx\end{cases}\longrightarrow \begin{cases}du = dx\\v = -\cos x\end{cases}$
$\Rightarrow \displaystyle\int f(x)dx = - x\cos x + \displaystyle\int \cos xdx$
$\Rightarrow \displaystyle\int f(x)dx = x\cos x + \sin x + C$
2) $2f^3(x) - 3f^2(x) + 6f(x)= x$
Đặt $u = f(x)$
$\to 2u^3 - 3u^2 + 6u = x$
$\to (6u^2 - 6u + 6)du = dx$
Đổi cận:
$x\quad \Big|\quad 0\qquad 5$
$\overline{u\quad\Big|\quad 0\qquad 1}$
Ta được:
$\quad \displaystyle\int\limits_0^5f(x)dx = \displaystyle\int\limits_0^1u(6u^2 - 6u + 6)du$
$\Leftrightarrow \displaystyle\int\limits_0^5f(x)dx= \displaystyle\int\limits_0^1(6u^3 - 6u^2 + 6u)du$
$\Leftrightarrow \displaystyle\int\limits_0^5f(x)dx= \left(\dfrac{3u^4}{2} - 2u^3 + 3u^2\right)\Bigg|_0^1$
$\Leftrightarrow \displaystyle\int\limits_0^5f(x)dx= \dfrac52$
