第一题：=∫下0上pi -(sinx)^2*(cosx)^6 dcosx=∫下0上pi (cosx^2-1)*(cosx)^6dcosx
令cosx=t,则=∫下1上-1 (t^2-1)*t^6dt,答案为4/63
第二题：方法名字忘了,
∫下1上4 xln(√x) dx=1/2∫下1上4 ln(√x) dx^2=1/2*ln(√x) *x^2-1/2∫下1上4 x^2dln(√x)
=1/2*ln(√x) *x^2-1/4∫下1上4 xdx
答案为8ln2-15/8