高分求解两个关于定积分的题目,求高手详解.

Question

百度大不了公式，word做的，麻烦用word做了截个图给我。

Answer 1

1.令x=a*(sint)^2 dx=2a*sint*costdt t∈(0,π/2)
原式=∫(0,π/2)2*a^4(sint)^8dt
=2a^4*∫(0,π/2)(sint)^8dt
∫(0,π/2)(sint)^ndt=-[(sint)^(n-1)cost]/n+[(n-1)/n]∫(0,π/2)(sint)^(n-2)dt
原式=2a^4*(7/8)*(5/6)*(3/4)*(1/2)∫(0,π/2)dt (由递推公式)
=35*a^4*π/128
2.原式=∫(0,+∞)dx/[(x+2)^2+4]=(1/4)∫(0,+∞)dx/[(x/2+1)^2+1], 令y=x/2+1
=(1/2)∫(1,+∞)dy/[y^2+1]
=(1/2)*arctan[y]|(1,+∞)
=π/8

Answer 2

1.令x=a*(sint)^2 dx=2a*sint*costdt t∈(0,π/2)
原式=∫(0,π/2)2*a^4(sint)^8dt
=2a^4*∫(0,π/2)(sint)^8dt
=2a^4*(7/8)*(5/6)*(3/4)*(1/2)∫(0,π/2)dt (由递推公式)
=35*a^4*π/128
2.原式=∫(0,+)dx/[(x+2)^2+4]
=(1/2)*arctan[(x+2)/2]|(0,+)
=π/8本回答被网友采纳