帮忙算一个定积分 我算来算去是4π/3 但是答案是π 请大家算算多少
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/08 14:34:54
帮忙算一个定积分
![](http://img.wesiedu.com/upload/6/9d/69d8c53702b0fb4c56e916c6d4881290.jpg)
我算来算去是4π/3 但是答案是π 请大家算算多少
![](http://img.wesiedu.com/upload/6/9d/69d8c53702b0fb4c56e916c6d4881290.jpg)
我算来算去是4π/3 但是答案是π 请大家算算多少
![帮忙算一个定积分 我算来算去是4π/3 但是答案是π 请大家算算多少](/uploads/image/z/8278535-47-5.jpg?t=%E5%B8%AE%E5%BF%99%E7%AE%97%E4%B8%80%E4%B8%AA%E5%AE%9A%E7%A7%AF%E5%88%86+%E6%88%91%E7%AE%97%E6%9D%A5%E7%AE%97%E5%8E%BB%E6%98%AF4%CF%80%2F3%26nbsp%3B+%E4%BD%86%E6%98%AF%E7%AD%94%E6%A1%88%E6%98%AF%CF%80%26nbsp%3B+%26nbsp%3B%E8%AF%B7%E5%A4%A7%E5%AE%B6%E7%AE%97%E7%AE%97%E5%A4%9A%E5%B0%91)
方法一:
(16/3)∫(0→π/2) cos⁴θ dθ
= (16/3) * 3!/4!* π/2
= (16/3)(3 * 1)/(4 * 2) * π/2
= π
方法二:
(16/3)∫(0→π/2) cos⁴θ dθ
= (16/3)∫(0→π/2) (cos²θ)² dθ
= (16/3)∫(0→π/2) [(1 + cos2θ)/2]² dθ
= (16/3)(1/4)∫(0→π/2) (1 + 2cos2θ + cos²2θ) dθ
= (4/3)∫(0→π/2) [1 + 2cos2θ + (1 + cos4θ)/2] dθ
= (4/3)∫(0→π/2) [3/2 + 2cos2θ + (1/2)cos4θ] dθ
= (4/3)[3θ/2 + sin2θ + (1/8)sin4θ] |(0→π/2)
= 4/3 * 3/2 * π/2
= π
再问: 帮我看看哪儿错了。
第一个方法我怎么看不懂啊 答案和你第一个写的一样 我不知道怎么得到的公式
再答: cos⁴θ到cos8θ转换那不对吧 cos²θ = (1 + cos2θ)/2 这个公式要二次方才行 4次方的话要先转为2次方再用 cos⁴θ = (cos²θ)² = [(1 + cos2θ)/2]²
(16/3)∫(0→π/2) cos⁴θ dθ
= (16/3) * 3!/4!* π/2
= (16/3)(3 * 1)/(4 * 2) * π/2
= π
方法二:
(16/3)∫(0→π/2) cos⁴θ dθ
= (16/3)∫(0→π/2) (cos²θ)² dθ
= (16/3)∫(0→π/2) [(1 + cos2θ)/2]² dθ
= (16/3)(1/4)∫(0→π/2) (1 + 2cos2θ + cos²2θ) dθ
= (4/3)∫(0→π/2) [1 + 2cos2θ + (1 + cos4θ)/2] dθ
= (4/3)∫(0→π/2) [3/2 + 2cos2θ + (1/2)cos4θ] dθ
= (4/3)[3θ/2 + sin2θ + (1/8)sin4θ] |(0→π/2)
= 4/3 * 3/2 * π/2
= π
再问: 帮我看看哪儿错了。
![](http://img.wesiedu.com/upload/7/9c/79cb12c8aa51127c180dc35062e95ecc.jpg)
再答: cos⁴θ到cos8θ转换那不对吧 cos²θ = (1 + cos2θ)/2 这个公式要二次方才行 4次方的话要先转为2次方再用 cos⁴θ = (cos²θ)² = [(1 + cos2θ)/2]²