Question

请问这道高数题目，这两步怎么化的，求大佬解答，谢谢啦

Answer 1

=π/2∫（0，π/2）（sin2x/2）dx
-π/2∫（π/2，π）（sin2x/2）dx
=-π/2×1/4cos2x丨（0，π/2）
+π/2×1/4cos2x丨(π/2，π）
=π/8+π/8+π/8+π/8
=π/2
解法分析：利用解法的性质，进行积分，就可以很快得出结果。

Answer 2

(π/2)∫(0->π) |cosx| sinx dx
=(π/2)∫(0->π/2) cosx. sinx dx -(π/2)∫(π/2->π) cosx. sinx dx
=(π/2)∫(0->π/2) sinx dsinx -(π/2)∫(π/2->π) sinx dsinx
=(π/4) [ (sinx)^2]|(0->π/2) -(π/4) [ (sinx)^2]|(π/2->π)
=(π/4)(1-0) -(π/4)(0-1)
=π/2