關於微積分三角函數的問題?

Question

請問一下
∫sinx cosx dx這個怎麼去解呢?

∫3sinx dx又是多少呢?

那麼∫[(3-sinx)(-sinx)+cosx cosx+(-cosx)(sinx)] dx

怎麼解出∫cos^2x + sin^2x dx呢?

那-3sinx dx 跟 +cosx sinx dx 怎麼不見呢?

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Answer 1

1. ∫sinx cosx dx 變數代換, 令u=sinx => du=cosx dx
= ∫u du =1/2 u2 +c
= 1/2 * sin2x + c
2. ∫3sinx dx = -3cosx +c
3. ∫[(3-sinx)(-sinx)+cosx cosx - sinx cosx ] dx
=∫(sin2x - 3cosx + cos2x - sinx cosx ) dx
= ∫(1 - 3cosx - sinx cosx )dx (sin2x + cos2x = 1 )
= x - 3sinx - 1/2 sin2x + c

2007-12-21 02:00:07 補充：
更正:第3題:乘錯了!
3. ∫[(3-sinx)(-sinx)+cosx cosx - sinx cosx ] dx
=∫(sin²x - 3sinx + cos²x - sinx cosx ) dx
= ∫(1 - 3sinx - sinx cosx )dx (sin²x + cos²x = 1 )
= x + 3cosx - 1/2 sin²x + c