请问下图的定积分应该怎么计算？ 请问图中定积分怎么计算
\u8bf7\u95ee\u4e0b\u56fe\u4e2d\u7684\u5b9a\u79ef\u5206\u8981\u600e\u4e48\u8ba1\u7b97\uff1f\u222bdt/ [(sint)^2.cost]
=\u222b [(csct)^2.sect ] dt
=-\u222b sect dcot(t)
=-sect.cot(t) + \u222b cot(t) .sect.tant dt
=-1/sint + \u222b sect dt
=-1/sint + ln|sect+tant| +C
\u222b( arctan(3/4)->arctan(4/3)) dt/ [(sint)^2.cost]
=[ -1/sint + ln|sect+tant| ] | ( arctan(3/4)->arctan(4/3))
=[ -5/4 + ln( 5/3 + 4/3) ] - [ -5/3 +ln(5/4+3/4) ]
=5/12 + ln3 - ln2
=5/12 + ln(3/2)
\u5c31\u8fd9\u6837\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002
其实sec^3的积分是有公式的，先化成含secx的积分的式子，然后就可以再用一次公式求出来了。
∫ (sect)^3 dt
=∫ sect dtant
=sect.tant -∫ sect.(tant)^2 du
=sect.tant -∫ sect.[(sect)^2-1] du
2∫ (sect)^3 dt =sect.tant +∫ sect du
∫ (sect)^3 dt
=(1/2)[ sect.tant +ln|sect+tant| ] +C
∫(0->arctan(4/3)) (sect)^3 dt
=(1/2)[ sect.tant +ln|sect+tant| ]|(0->arctan(4/3))
=(1/2)[ (5/3)(4/3) + ln(5/3 + 4/3) ]
=(1/2)( 20/9 + ln3 )
where
tant =4/3
cost = 3/5
sect =5/3
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绛旓細鍥炵瓟锛氬厛姹傝绉嚱鏁扮殑鍘熷嚱鏁,鍐嶆妸鏁板间唬杩涘幓
