计算定积分∫(上限1下限-1) (2x^2+x^9*cosx)/(1+√(1-x^2))dx
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计算定积分∫(上限1下限-1) (2x^2+x^9*cosx)/(1+√(1-x^2))dx
![计算定积分∫(上限1下限-1) (2x^2+x^9*cosx)/(1+√(1-x^2))dx](/uploads/image/z/7680983-23-3.jpg?t=%E8%AE%A1%E7%AE%97%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%EF%BC%88%E4%B8%8A%E9%99%901%E4%B8%8B%E9%99%90-1%EF%BC%89+%282x%5E2%2Bx%5E9%2Acosx%29%2F%281%2B%E2%88%9A%281-x%5E2%29%29dx)
因为这一项,x^9*cosx/(1+√(1-x^2)是个奇函数,所以在-1,1上的积分为0
所以元积分=4∫(0到1) x^2/(1+√(1-x^2))dx
令x=sint
元积分=4∫(0到π/2) (sint)^2 cost/(1+cost)dt
=4∫(0到π/2) [1-(cost)^2] cost/(1+cost)dt
=4∫(0到π/2) [1-(cost)] costdt
=4∫(0到π/2) [cost-(cost)^2] dt
=4sint|(0到π/2)-2∫(0到π/2) (cos2t+1) dt
=4-(sin2t+2t)|(0到π/2)
=4-π
再问: x^9*cosx/(1+��(1-x^2)�Ǹ��溯����f(x)������-f(-x)�ɣ�
再答: cosx�Ǹ�ż���� (1+��(1-x^2)Ҳ�Ǹ�ż���� x&9�Ǹ��溯�� ��˵��
所以元积分=4∫(0到1) x^2/(1+√(1-x^2))dx
令x=sint
元积分=4∫(0到π/2) (sint)^2 cost/(1+cost)dt
=4∫(0到π/2) [1-(cost)^2] cost/(1+cost)dt
=4∫(0到π/2) [1-(cost)] costdt
=4∫(0到π/2) [cost-(cost)^2] dt
=4sint|(0到π/2)-2∫(0到π/2) (cos2t+1) dt
=4-(sin2t+2t)|(0到π/2)
=4-π
再问: x^9*cosx/(1+��(1-x^2)�Ǹ��溯����f(x)������-f(-x)�ɣ�
再答: cosx�Ǹ�ż���� (1+��(1-x^2)Ҳ�Ǹ�ż���� x&9�Ǹ��溯�� ��˵��
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