Question

高数：请问这一步积分是换算的？

Answer 1

利用三角函数周期性，t=x-pi替换旧得到了

Answer 2

let
u=x-π
du=dx
x=π/2, u=-π/2
x=π, u=0
∫(π/2->π) sin2x/[1+(sinx)^4 ] dx
=∫(0->-π/2) sin2u/[1+(sinu)^4 ] (-du)
=∫(-π/2->0) sin2u/[1+(sinu)^4 ] du
=∫(-π/2->0) sin2x/[1+(sinx)^4 ] dx
//
∫(0->π) | sin2x|/[1+(sinx)^4 ] dx
=∫(0->π/2) | sin2x|/[1+(sinx)^4 ] dx +∫(π/2->π) | sin2x|/[1+(sinx)^4 ] dx

=∫(0->π/2) | sin2x|/[1+(sinx)^4 ] dx +∫(-π/2->0) | sin2x|/[1+(sinx)^4 ] dx
=∫(-π/2->π/2) | sin2x|/[1+(sinx)^4 ] dx
=∫(0->π/2) sin2x/[1+(sinx)^4 ] dx +∫(π/2->π) sin2x/[1+(sinx)^4 ] dx

Answer 3

定积分换元法。类似。。

Answer 4

π/4 ∫(0，π) |sin2x|/[1＋(sinx)^4]dx
=π/4 ∫(0，π/2) |sin2x|/[1＋(sinx)^4]dx＋π/4 ∫(π/2，π) |sin2x|/[1＋(sinx)^4]dx
上述式子中的第二部分令x=t＋π，则
=π/4 ∫(0，π/2) |sin2x|/[1＋(sinx)^4]dx＋π/4∫(–π/2，0)
|sin(2t＋2π)|/[1＋(sin(t＋π))^4]d(t＋π)
=π/4 ∫(0，π/2) |sin2x|/[1＋(sinx)^4]dx＋π/4∫(–π/2，0)
|sin2t|/[1＋(–sint)^4]dt
=π/4 ∫(0，π/2) |sin2x|/[1＋(sinx)^4]dx＋π/4∫(–π/2，0)
|sin2t|/[1＋(sint)^4]dt
=π/4 ∫(0，π/2) |sin2x|/[1＋(sinx)^4]dx＋π/4∫(–π/2，0)
|sin2x|/[1＋(sinx)^4]dx
=π/4 ∫(–π/2，π/2) |sin2x|/[1＋(sinx)^4]dx

π/4 ∫(0，π) |sin2x|/[1＋(sinx)^4]dx
=π/4 ∫(0，π/2) sin2x/[1＋(1–cos2x)²/4] dx–π/4 ∫(π/2，π) sin2x/[1＋(1–cos2x)²/4] dx
=π∫(0，π/2) sin2x/[4＋(1–cos2x)²] dx–π ∫(π/2，π) sin2x/[4＋(1–cos2x)²] dx
=π/2 ∫(0，π) sinx/[4＋(1–cosx)²] dx–π/2 ∫(π，2π) sinx/[4＋(1–cosx)²] dx
=π/2 ∫(0，π) 1/[4＋(1–cosx)²] d(1–cosx)–π/2 ∫(π，2π) 1/[4＋(1–cosx)²] d(1–cosx)
=π/4 arctan[(1–cosx)/2]|(0，π)–π/4 arctan[(1–cosx)/2]|(π，2π)
=π/4·π/4–π/4·(–π/4)
=π²/8

π/4 ∫(–π/2，π/2) |sin2x|/[1＋(sinx)^4] dx
=–π/4 ∫(–π/2，0) sin2x/[1＋(1–cos2x)²/4] dx＋π/4 ∫(0，π/2) sin2x/[1＋(1–cos2x)²/4] dx
=–π∫(–π/2，0) sin2x/[4＋(1–cos2x)²] dx＋π ∫(0，π/2) sin2x/[4＋(1–cos2x)²] dx
=–π/2 ∫(–π，0) sinx/[4＋(1–cosx)²] dx＋π/2 ∫(0，π) sinx/[4＋(1–cosx)²] dx
=–π/2 ∫(–π，0) 1/[4＋(1–cosx)²] d(1–cosx)＋π/2 ∫(0，π) 1/[4＋(1–cosx)²] d(1–cosx)
=–π/4 arctan[(1–cosx)/2]|(–π，0)＋π/4 arctan[(1–cosx)/2]|(0，π)
=–π/4·(0–π/4)＋π/4·(π/4–0)
=π²/8

Answer 5

详情如图所示

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