📝 题目
例 2 试计算积分
$$ {\int }_{0}^{\pi /2}{\sin }^{\alpha }x \cdot {\cos }^{\beta }x\mathrm{\;d}x\;\left( {\alpha > - 1,\beta > - 1}\right) . $$
💡 答案与解析
解 做变元替换 $t = {\sin }^{2}x$ 就得到
$$ {\int }_{0}^{\pi /2}{\sin }^{\alpha }x \cdot {\cos }^{\beta }x\mathrm{\;d}x = \frac{1}{2}{\int }_{0}^{1}{t}^{\frac{\alpha - 1}{2}}{\left( 1 - t\right) }^{\frac{\beta - 1}{2}}\mathrm{\;d}t $$
$$ = \frac{1}{2}\mathrm{\;B}\left( {\frac{\alpha + 1}{2},\frac{\beta + 1}{2}}\right) $$
$$ = \frac{1}{2}\frac{\Gamma \left( \frac{\alpha + 1}{2}\right) \Gamma \left( \frac{\beta + 1}{2}\right) }{\Gamma \left( {\frac{\alpha + \beta }{2} + 1}\right) }. $$
对于 $\alpha = 4,\beta = 6$ 的情形,我们得到
$$ {\int }_{0}^{\pi /2}{\sin }^{4}x \cdot {\cos }^{6}x\mathrm{\;d}x = \frac{\Gamma \left( \frac{5}{2}\right) \Gamma \left( \frac{7}{2}\right) }{{2\Gamma }\left( 6\right) } = \frac{3\pi }{512}. $$