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
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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