Question

高等数学积分三角函数

如图，这式子是怎么转化过来的？求解释过程，谢谢啦

Answer 1

根号内：
提取(sinx)^2: =(sinx)^2*[1-(sinx)^2]=(sinx)^2*(cosx)^2
因为x∈[0,π/2],都是正数，开方出来就是sinxcosx

Answer 2

I = (1/4)∫2>(1-cos2t)^2dt - ∫2>[1-(cost)^2]^2dcost = (1/4)∫2>[3/2-2cos2t+(1/2)cos4t]dt - ∫2>[1-2(cost)^2+(cost)^4]dcost = (1/4)[3t/2-sin2t+(1/8)sin2t])2> - [cost-(2/3)(cost)^3+(1/5)(cost)^5]2> = 3π/16 +(1-2/3+1/5) = 3π/16 + 8/15 .