## Trending News

Promoted

# Calculus - integral(cos(sinx))

integral cosx(cos(sinx))^5 dx=?

請詳述過程!

Update:

=integral cos(y)^4*d(siny)

=integral [1-sin(y)^2]^4*d(siny)

次方數好像怪怪的?

### 4 Answers

Rating

- 金鼎Lv 59 years agoFavorite Answer
大大你好，由我為你解答

∫cosx*(cos(sinx))^5 dx

（這題需要帶換2次，答案才會出來）

Sol: u=sinx

du=cosxdx

=∫cos^5(u) du

=∫cos^4(u)*cos(u) du

v=sinu

dv=cosudu

=∫(1-sin^2)^2 dv

=∫(1-v^2)^2 dv

=∫(1-2v^2+v^4)dv

= v-2/3v^3+1/5v^5+c

(再來把值帶入還原)

＝sinu-2/3sinu^3+1/5sinu^5+c

= sin(sinx)-2/3sin(sinx)^3+1/5sin(sinx)^5+c為正解

Source(s): 我 - 麻辣Lv 79 years ago
a=∫cosx*(cos(sinx))^5 dx=?=∫[cos(sinx)]^5*d(sinx)=∫cos(y)^5*dy..........y=sin(x)=∫cos(y)^4*d(siny)=∫[1-sin(y)^2]^4*d(siny)=∫(1-u^2)^4*du.........u=sin(y)=∫(1-4u^2+6u^4-4u^6+u^8)du=u-4u^3/3+6u^5/5-4u^7/7+u^9/9+c=sin(y)-4/3*sin(y)^3+6/5*sin(y)^5-4/7*sin(y)^7+1/9*sin(y)^9+c=sin(sinx)-4/3*sin(sinx)^3+6/5*sin(sinx)^5-4/7*sin(sinx)^7+1/9*sin(sinx)^9+c

Still have questions? Get your answers by asking now.