15. 令 x=tant, 则
I = ∫(sect)^3dt/(tant)^2
= ∫sectdt/(sint)^2
= -∫sectdcott
= -[sectcott] + ∫cottsecttantdt
= -[csct] + ∫sectdt
= -[csct] +[ln(sect+tant)]
= √3-√2 + ln(√6+2√3-2√2-2)
定积分题目如图第15题求第15题详细过程
15. 令 x=tant, 则
I = ∫(sect)^3dt/(tant)^2
= ∫sectdt/(sint)^2
= -∫sectdcott
= -[sectcott] + ∫cottsecttantdt
= -[csct] + ∫sectdt
= -[csct] +[ln(sect+tant)]
= √3-√2 + ln(√6+2√3-2√2-2)