P. 230 习题 5.3
1. 求下列定积分
(9)
\[
\int_{\frac{1}{4}}^{\frac{3}{4}}\frac{\arcsin\sqrt{x}}{\sqrt{x(1-x)}}\mathrm{d}x
\]
3. 用分部积分法求下列定积分.
(7)
\[
\int_{0}^{\sqrt{3}}\ln(x+\sqrt{1+x^2})\mathrm{d}x
\]
6. 证明下列各等式.
(3) 设 $a > 0$, $f(x)$ 在 $[0,a]\cup[0,a^2]$ 上可积, 则有
\[
\int_{0}^{a}x^3 f(x^2)\mathrm{d}x=\frac{1}{2}\int_{0}^{a^2}xf(x)\mathrm{d}x.
\]
1
1. (9) 令 $t=\sqrt{x}$, 则 $x=t^2$, $\mathrm{d}x=2t\mathrm{d}t$.
\[
\begin{split}
\text{原积分}&=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{\arcsin t}{\sqrt{t^2(1-t^2)}}\cdot 2t\mathrm{d}t\\
&=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{2\arcsin t}{\sqrt{1-t^2}}\mathrm{d}t\\
&=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}2\arcsin t\mathrm{d}\arcsin t\\
&=(\arcsin t)^2\biggr|_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\\
&=(\arcsin\frac{\sqrt{3}}{2})^2-(\arcsin\frac{1}{2})^2\\
&=(\frac{\pi}{3})^2-(\frac{\pi}{6})^2\\
&=\frac{\pi^2}{12}
\end{split}
\]
2
3. (7)
\[
\begin{split}
&\int_{0}^{\sqrt{3}}\ln(x+\sqrt{1+x^2})\mathrm{d}x\\
=&x\ln(x+\sqrt{1+x^2})\biggr|_{0}^{\sqrt{3}}-\int_{0}^{\sqrt{3}}x\mathrm{d}\ln(x+\sqrt{1+x^2})\\
=&\sqrt{3}\ln(\sqrt{3}+2)-\int_{0}^{\sqrt{3}}x\cdot\frac{1}{x+\sqrt{1+x^2}}\cdot(1+\frac{2x}{2\sqrt{1+x^2}})\mathrm{d}x\\
=&\sqrt{3}\ln(\sqrt{3}+2)-\int_{0}^{\sqrt{3}}\frac{x}{\sqrt{1+x^2}}\mathrm{d}x\\
=&\sqrt{3}\ln(\sqrt{3}+2)-(\sqrt{1+x^2})\biggr|_{0}^{\sqrt{3}}\\
=&\sqrt{3}\ln(\sqrt{3}+2)-(2-1)\\
=&\sqrt{3}\ln(2+\sqrt{3})-1.
\end{split}
\]
3
6. (3)
证明:
\[
\begin{split}
\text{LHS}&=\int_{0}^{a}x^2\cdot x\cdot f(x^2)\mathrm{d}x\\
&=\int_{0}^{a}\frac{1}{2}\cdot x^2\cdot f(x^2)\mathrm{d}x^2\\
&=\frac{1}{2}\int_{0}^{a^2}t f(t)\mathrm{d}t\\
&=\text{RHS}
\end{split}
\]