题目 Problem Find (i) ∫sin23x dx\int \sin^2 3x\,dx∫sin23xdx (2) (ii) ∫x(x2+4)32 dx\int x(x^2+4)^{\frac32}\,dx∫x(x2+4)23dx (2) 题目中文翻译 求 (i) ∫sin23x dx\int \sin^2 3x\,dx∫sin23xdx (ii) ∫x(x2+4)32 dx\int x(x^2+4)^{\frac32}\,dx∫x(x2+4)23dx 解答 (i) Use the identity sin23x=1−cos6x2\sin^2 3x=\frac{1-\cos6x}{2}sin23x=21−cos6x Then ∫sin23x dx=∫1−cos6x2 dx=12x−112sin6x+c\begin{aligned} \int \sin^2 3x\,\mathrm{d}x &=\int \frac{1-\cos6x}{2}\,\mathrm{d}x \\ &=\frac12x-\frac{1}{12}\sin6x+c \end{aligned}∫sin23xdx=∫21−cos6xdx=21x−121sin6x+c So ∫sin23x dx=12x−112sin6x+c\boxed{\int \sin^2 3x\,\mathrm{d}x =\frac12x-\frac{1}{12}\sin6x+c}∫sin23xdx=21x−121sin6x+c (ii) Let u=x2+4u=x^2+4u=x2+4 Then dudx=2x\frac{\mathrm{d}u}{\mathrm{d}x}=2xdxdu=2x so x dx=12 dux\,\mathrm{d}x=\frac12\,\mathrm{d}uxdx=21du Therefore ∫x(x2+4)32 dx=12∫u32 du=12⋅25u52+c=15u52+c\begin{aligned} \int x(x^2+4)^{\frac32}\,\mathrm{d}x &=\frac12\int u^{\frac32}\,\mathrm{d}u \\ &=\frac12\cdot\frac{2}{5}u^{\frac52}+c \\ &=\frac15u^{\frac52}+c \end{aligned}∫x(x2+4)23dx=21∫u23du=21⋅52u25+c=51u25+c Substitute back: ∫x(x2+4)32 dx=15(x2+4)52+c\boxed{\int x(x^2+4)^{\frac32}\,\mathrm{d}x =\frac15(x^2+4)^{\frac52}+c}∫x(x2+4)23dx=51(x2+4)25+c
