tanx^2.secx的积分

网友回答

原式=∫tanx d(secx)
=tanxsecx- ∫(secx)^3 dx
其中∫(sec t)³dt
=∫(1/cos t)³dt
=∫cost/(cos t)^4 dt
=∫1/(1-sin²t)² d sin t
=∫1/(1-sint)²(1+sint)² d sin t
=1/4 ∫[1/(1-sin t) + 1/(1+sint)]² d sin t
=1/4∫[1/(1-sint)²+1/(1+sint)²+2/(1-sint)(1+sint)] d sin t
=1/4[1/(1-sint) - 1/(1+sint) - ln (1-sint) + ln (1+sint)] + C
=2tg t sec t + 1/4 ln (1+sint)/(1-sint) + C
最后原式=tanxsecx+2tg t sec t + 1/4 ln (1+sint)/(1-sint) + C