使用微积分基本原理 (1)∫(x sin√(x^2+4))/√(x^2+4) dx(2)∫x^2 sin(x^3+5) cos^9 (x^3+5)dx(3)∫_0^(π/2)sinxsin(cosx)dx(4)∫_(-π/2)^(π/2)cosθ cos(π sinθ)dθ
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![使用微积分基本原理 (1)∫(x sin√(x^2+4))/√(x^2+4) dx(2)∫x^2 sin(x^3+5) cos^9 (x^3+5)dx(3)∫_0^(π/2)sinxsin(cosx)dx(4)∫_(-π/2)^(π/2)cosθ cos(π sinθ)dθ](/uploads/image/z/5453483-59-3.jpg?t=%E4%BD%BF%E7%94%A8%E5%BE%AE%E7%A7%AF%E5%88%86%E5%9F%BA%E6%9C%AC%E5%8E%9F%E7%90%86+%EF%BC%881%EF%BC%89%E2%88%AB%28x+sin%E2%88%9A%28x%5E2%2B4%29%29%2F%E2%88%9A%28x%5E2%2B4%29+dx%EF%BC%882%EF%BC%89%E2%88%ABx%5E2+sin%28x%5E3%2B5%29+cos%5E9+%28x%5E3%2B5%29dx%EF%BC%883%EF%BC%89%E2%88%AB_0%5E%28%CF%80%2F2%29sinxsin%28cosx%29dx%EF%BC%884%EF%BC%89%E2%88%AB_%28-%CF%80%2F2%29%5E%28%CF%80%2F2%29cos%CE%B8+cos%28%CF%80+sin%CE%B8%29d%CE%B8)
使用微积分基本原理 (1)∫(x sin√(x^2+4))/√(x^2+4) dx(2)∫x^2 sin(x^3+5) cos^9 (x^3+5)dx(3)∫_0^(π/2)sinxsin(cosx)dx(4)∫_(-π/2)^(π/2)cosθ cos(π sinθ)dθ
使用微积分基本原理
(1)∫(x sin√(x^2+4))/√(x^2+4) dx
(2)∫x^2 sin(x^3+5) cos^9 (x^3+5)dx
(3)∫_0^(π/2)sinxsin(cosx)dx
(4)∫_(-π/2)^(π/2)cosθ cos(π sinθ)dθ
使用微积分基本原理 (1)∫(x sin√(x^2+4))/√(x^2+4) dx(2)∫x^2 sin(x^3+5) cos^9 (x^3+5)dx(3)∫_0^(π/2)sinxsin(cosx)dx(4)∫_(-π/2)^(π/2)cosθ cos(π sinθ)dθ
(1)原式= 1/2 ∫ sin√(x²+4)/(√(x²+4)d(x²+4)= - ∫ sin√(x²+4)d√(x²+4)
=cos√(x²+4)+C
(2)原式= 1/3 ∫ sin(x^3+5) cos^9 (x^3+5)d(x^3+5)= -1/3 ∫ cos^9 (x^3+5)d [cos(x^3+5)]
= -1/30cos^10(x^3+5)+C
(3)不定积分∫ sinxsin(cosx)dx= - ∫ sin(cosx)dcosx=cos(cosx)+C,∴定积分= 1-cos1;
(4)不定积分∫cosθ cos(π sinθ)dθ= 1/π ∫ cos(π sinθ)d(πsinθ)=1/πsin(π sinθ)+C,
∴定积分=0.
PS:第四题中为奇函数,可以不必求得积分即可得到定积分为0.