Question

求定积分∫(1+t^2)*√(1+t^2)的值

Answer 1

∫[0,2π] (1+t^2) *√(1+t^2) dt
=∫[0,2π]√(1+t^2)^3dt
=t√(1+t^2)^3 |[0,2π] - ∫[0,2π] 2*(3/2)*t^2√(1+t^2)dt
=2π√(1+4π^2)^3 - 3∫[0,2π]√(1+t^2)^3dt+3∫[0,2π]√(1+t^2)dt
4∫[0,2π](1+t^2)*√(1+t^2)dt=2π√(1+4π^2)^3+(3/2)t√(1+t^2)|[0,2π]+(3/2)ln|t+√(1+t^2)||[0,2π]
∫[0,2π](1+t^2)*√(1+t^2)dt=(π/2)√(1+4π^2)^3 +(3/4)π√(1+4π^2)+(3/2)ln|2π+√(1+4π^2)|
∫√(1+t^2)dt t=tanu
=∫secu^3du=(1/2)secutanu+(1/2)ln|secu+tanu|=(1/2)t√(1+t^2)+(1/2)ln|t+√(1+t^2)|