回答:
#int cos(x)/(sin ^ 2(x)+ sin(x)) "d" x = ln | sin(x)/(sin(x)+1)| + C#
説明:
# int cos(x)/(sin ^ 2(x)+ sin(x)) "d" x#
代替 #u = sin(x)# そして # "d" u = cos(x) "d" x#。これは与える
#= int ( "d" u)/(u ^ 2 + u)#
#= int ( "d" u)/(u(u + 1))#
部分分数に分離する #1 /(u(u + 1))= 1 / u-1 /(u + 1)#:
#= int (1 / u-1 /(u + 1)) "d" u#
#= ln | u | -ln | u + 1 | + C#
#= ln | u /(u + 1)| + C#
代用 #u = sin(x)#:
#= ln | sin(x)/(sin(x)+1)| + C#