作业帮∫sin²（根号x）dx
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∫sin²（根号x）dx
楼上都答得好乱,还是我的最清楚~答题步骤清晰是一种美德,也是尊敬考官的行为.
令x=z²,dx=2z dz
∫sin²√x dx
= ∫(sin²z)(2z dz)
= 2∫z*(1-cos2z)/2 dz
= ∫z dz - ∫zcos2z dz
= ∫z dz - (1/2)∫z d(sin2z),分部积分法
= ∫z dz - (1/2)zsin2z + (1/2)∫sin2z dz,分部积分法
= z²/2 - (1/2)zsin2z - (1/4)cos2z + C,回代z=√x
= x/2 - (1/2)(√x)sin(2√x) - (1/4)cos(2√x) + C
令x=z²,dx=2z dz
∫sin²√x dx
= ∫(sin²z)(2z dz)
= 2∫z*(1-cos2z)/2 dz
= ∫z dz - ∫zcos2z dz
= ∫z dz - (1/2)∫z d(sin2z),分部积分法
= ∫z dz - (1/2)zsin2z + (1/2)∫sin2z dz,分部积分法
= z²/2 - (1/2)zsin2z - (1/4)cos2z + C,回代z=√x
= x/2 - (1/2)(√x)sin(2√x) - (1/4)cos(2√x) + C