Question

定积分∫(x/(1+sin(x)))dx 从-π/4积到π/4

Answer 1

简单计算一下即可，答案如图所示

Answer 2

∫(-
π/4→π/4)
x/(1
+
sinx)
dx
=
∫(-
π/4→π/4)
x(1
-
sinx)/cos²x
dx
=
∫(-
π/4→π/4)
(x
-
xsinx)/cos²x
dx
=
∫(-
π/4→π/4)
xsec²x
dx
-
∫(-
π/4→π/4)
xsecxtanx
dx
=
∫(-
π/4→π/4)
x
d(tanx)
-
∫(-
π/4→π/4)
x
d(secx)
=
xtanx
-
∫(-
π/4→π/4)
tanx
dx
-
xsecx
+
∫(-
π/4→π/4)
secx
dx
=
[(π/4)
-
(-
π/4)(-
1)]
+
lncosx
-
[(π/4)√2
-
(-
π/4)√2]
+
ln(secx
+
tanx)
=
ln(√2/2)
-
ln(√2/2)
-
π/√2
+
ln(√2
+
1)
-
ln(√2
-
1)
=
ln[(√2
+
1)/(√2
-
1)]
-
π/√2
=
ln(3
+
2√2)
-
π/√2
过程先上下乘以1
-
sinx
然后分裂为两个积分，之后它们分别运用分部积分法
就这么简单