求2∫（0,π）（6X²＋4－sin²X)dx详细过程

求2∫（0,π）（6X²＋4－sin²X)dx详细过程

2∫(0->π)[6x^2+4－(sinx)^2 ]dx
=∫(0->π)[12x^2+8-2(sinx)^2 ]dx
=∫(0->π)(12x^2+7+cos2x ) dx
= [4x^3+7x+ sin(2x)/2](0->π)
= 4π^3+7π