Question

积分求解

求下列不定积分：
1、∫（2-sinx）/（2+cosx）dx
2、∫（sinxcosx）/（sinx+cosx）dx
2用万能公式不好解啊

Answer 1

3楼说得对，第二题利用万能公式来计算麻烦。其实这个题简单，想复杂了。我上面做的第一题没问题。第二题：

∫（sinxcosx）/（sinx+cosx）dx

=∫[（sinx+cosx）^2-1]/2（sinx+cosx）dx

=∫[（sinx+cosx）/2]dx-1/2[∫1/（sinx+cosx）dx]

=(1/2)sinx-(1/2)cosx-(√2/2)∫csc(x+pi/4)dx

=(1/2)sinx-(1/2)cosx+(√2/2)ln|csc(x+pi/4)-cot(x+pi/4)|+C

Answer 2

1:2楼方法好~顶~

2：∫（sinxcosx）/（sinx+cosx）dx

=∫（2sinxcosx）/[2√2sin(x+π/4)]dx

=∫sin2x/[2√2sin(x+π/4)]dx

=∫-cos(2x+π/2)/[2√2sin(x+π/4)]dx

=∫[2sin^2(x+π/4)-1]/[2√2sin(x+π/4)]dx

=∫｛√2/2*sin(x+π/4)+√2/4[-1/sin(x+π/4)]｝dx

=∫√2/2*sin(x+π/4)dx+∫√2/4[-1/sin(x+π/4)]dx

=-√2/2*cos(x+π/4)+√2/4*cos(x+π/4)/sin(x+π/4)+C

2用万能公式确实很麻烦~~~~~~~~1主要是因为我想不出......2还是我的方法好~