根号下x方-4的定积分是什么!
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/25 00:51:12
根号下x方-4的定积分是什么!
根号下(x^2 - 4),
根号下(x^2 - 4),
![根号下x方-4的定积分是什么!](/uploads/image/z/19160236-28-6.jpg?t=%E6%A0%B9%E5%8F%B7%E4%B8%8Bx%E6%96%B9-4%E7%9A%84%E5%AE%9A%E7%A7%AF%E5%88%86%E6%98%AF%E4%BB%80%E4%B9%88%21)
求不定积分∫[√(x²-4)]dx
∫[√(x²-4)]dx=2∫√[(x/2)²-1]dx
令x/2=secu,则dx=2secutanudu,代入上式得:
∫[√(x²-4)]dx=2∫√[(x/2)²-1]dx=4∫secutan²udu=4∫tanud(secu)=4[tanusecu-∫sec³udu]
=4[tanusecu-(1/2)tanusecu-(1/2)ln(secu+tanu)]+C
=2[tanusecu-ln(sec+tanu)]+C
=2{(x/4)√(x²-4)-ln[(x/2)+√(x²-4)]}+C=(x/2)√(x²-4)-2ln[(x/2)+√(x²-4)]}+C
再问: 4∫tanud(secu)=4[tanusecu-∫sec³udu] 这一步是怎么得到的
再答: 套用分部积分公式:∫udv=uv-∫vdu 4∫tanud(secu)=4[tanusecu-∫secudtanu]=4[tanusecu-∫secu(sec²udu)] =4[tanusecu-∫sec³udu]
∫[√(x²-4)]dx=2∫√[(x/2)²-1]dx
令x/2=secu,则dx=2secutanudu,代入上式得:
∫[√(x²-4)]dx=2∫√[(x/2)²-1]dx=4∫secutan²udu=4∫tanud(secu)=4[tanusecu-∫sec³udu]
=4[tanusecu-(1/2)tanusecu-(1/2)ln(secu+tanu)]+C
=2[tanusecu-ln(sec+tanu)]+C
=2{(x/4)√(x²-4)-ln[(x/2)+√(x²-4)]}+C=(x/2)√(x²-4)-2ln[(x/2)+√(x²-4)]}+C
再问: 4∫tanud(secu)=4[tanusecu-∫sec³udu] 这一步是怎么得到的
再答: 套用分部积分公式:∫udv=uv-∫vdu 4∫tanud(secu)=4[tanusecu-∫secudtanu]=4[tanusecu-∫secu(sec²udu)] =4[tanusecu-∫sec³udu]