风吹动的夜晚
幼苗
共回答了17个问题采纳率:82.4% 举报
问题一:令X=tgt,则dx=sec²tdt
∫x²(1+x²)^½dx=∫tg²·sect·sec²tdt
=∫sin²t/cos^5t·dt
=∫(1-cos²t)/cos^5t·dt
=∫1/cos^5t·dt-∫1/cos³t·dt
=1/(5-1)sint/cos^4t+(5-2)/(5-1)∫1/cos³t·dt-∫1/cos³t·dt
=1/4sint/cos^4t-1/4∫1/cos³t·dt
=1/4sint/cos^4t-1/4{1/2sint/cos²t+1/2∫1/cost·dt}
=1/4sint/cos^4t-1/8sint/cos²t-1/8ln(sect+tgt)+C
由x=tgt,得:sint=x/(1+x)^½ cost=1/(1+x)^½ sect=(1+x)^½
故:∫x²(1+x²)^½dx=1/4sint/cos^4t-1/8sint/cos²t-1/8ln(sect+tgt)+C
=x/8(2x²+1)(x²+1)^½-1/8ln[x+(x²+1)^½]+C
问题二:∫sin²/cos³x·dx=∫(1-cos²x)/cos³x·dx
=∫1/cos³x·dx-∫1/cosx·dx
=1/2sinx/cos²x+1/2∫1/cosx·dx- ∫1/cosx·dx
= 1/2sinx/cos²x-1/2∫1/cosx·dx
= 1/2sinx/cos²x-1/2ln(secx+tgx)+C
注:解决这些问题,我们运用到一个公式,
即:∫1/cos^nx·dx=1/(n-1)sinx/cos^(n-1)x+(n-2)/(n-1)∫1/cos^(n-2)x·dx
这个公式的证明可以运用分部积分法结合回溯求积分法来证明.
证明就略去吧.
还有,楼上的回答算什么呀,莫名其妙.
1年前
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