Integrate y^2/(k^2+y^2)^3/2 with respect to y, where the upper limit and the lower
limit are d/2 and -d/2 respectively. (k and d are constant.)
Definite Integral Question
2013-10-14 6:17 am
回答 (1)
2013-10-14 9:32 am
✔ 最佳答案
Assume k > 0, d > 0by trig. sub. y=k tan(x)
∫ k²tan²(x)/[k³ sec³(x)] k sec²(x) dx
= ∫ sec(x)-cos(x) dx
= ln(secx+tanx)- sinx + C
= ln[ √(k²+y²)+ y ] - y/√(k²+y²) + C
Ans: 2 ln[ √(k²+d²/4)+ d/2 ] - 2 ln(k) - d/√(k²+d²/4)
收錄日期: 2021-04-21 22:45:35
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