Question

高等数学 求不定积分

高等数学
求不定积分越详细越好，基础比较差。

Answer 1

I = √2 a^3∫<0, 2π> {2[sin(t/2)]^2}^(5/2) dt
= 8 a^3∫<0, 2π>[sin(t/2)]^5 dt
= -16 a^3∫<0, 2π>[sin(t/2)]^4 dcos(t/2)
= -16 a^3∫<0, 2π>{1-[cos(t/2)]^2}^2 dcos(t/2)
= -16 a^3∫<0, 2π>{1 - 2[cos(t/2)]^2 + [cos(t/2)]^4}dcos(t/2)
= -16 a^3[cos(t/2) - (2/3){cos(t/2)}^3 + (1/5)(cos(t/2)}^5]<0, 2π>
= (256/15) a^3