利用分部积分法求下列不定积分：∫√xsin√xdx.-中华词典

利用分部积分法求下列不定积分：∫√xsin√xdx.

更新时间:2026-05-31 14:30:56 栏目: 中华词典

利用分部积分法求下列不定积分：∫√xsin√xdx.

好的，我们一步一步来。题目是

\int \sqrt{x} \sin \sqrt{x} \, dx

$sin x$

$d x$

$令$

t = \sqrt{x}, \quad x = t^2, \quad dx = 2t \, dt

$, x = t^{2}, d x = 2 t d t$

$则$

\sqrt{x} = t, \quad \sqrt{x} \sin\sqrt{x} = t \sin t

$= t, x$

$sin x$

$= t sin t$

$原积分变为：$

\int t \sin t \cdot (2t \, dt) = \int 2t^2 \sin t \, dt

$设$

I_1 = \int t^2 \sin t \, dt

$\int u \, dv = uv - \int v \, du$

u = t^2, \quad dv = \sin t \, dt

$则$

du = 2t \, dt, \quad v = -\cos t

$所以$

I_1 = -t^2 \cos t - \int (-\cos t)(2t \, dt)

$再用分部积分：$

u = t, \quad dv = \cos t \, dt

I_1 = -t^2 \cos t + 2\left[ t \sin t + \cos t \right] + C_1

$2 I_1$

\int \sqrt{x} \sin\sqrt{x} \, dx = 2\left[ -t^2 \cos t + 2t \sin t + 2\cos t \right] + C

$sin x$

$= -2t^2 \cos t + 4t \sin t + 4\cos t + C$

= -2x \cos \sqrt{x} + 4\sqrt{x} \sin \sqrt{x} + 4\cos \sqrt{x} + C

$+$

$sin$

$+$

$\cos\sqrt{x}$

$的项：$

\cos\sqrt{x} (4 - 2x) + 4\sqrt{x} \sin\sqrt{x} + C

\boxed{-2x\cos\sqrt{x} + 4\sqrt{x}\sin\sqrt{x} + 4\cos\sqrt{x} + C}