Determine the integral ∫π20sec4(t2)dt
Let u=t2, then du=12dt, so dt=2du. When t=0,u=0 and when t=π2,u=π4
∫π20sec4(t2)dt=∫π40sec4u⋅2du∫π20sec4(t2)dt=2∫π40sec4udu∫π20sec4(t2)dt=2∫π40sec2usec2uduApply Pythagorean Identity sec2u=tan2u+1∫π20sec4(t2)dt=2∫π40(tan2u+1)sec2udu
Let v=tanu, then dv=sec2udu. When u=0,v=0 and when u=π4,v=1. Thus,
2∫π40(tan2u+1)sec2udu=2∫21(v2+1)dv2∫π40(tan2u+1)sec2udu=2[v2+12+1+v]102∫π40(tan2u+1)sec2udu=2[v33+v]102∫π40(tan2u+1)sec2udu=2[(1)33+1−(0)33−0]2∫π40(tan2u+1)sec2udu=2(13+1)2∫π40(tan2u+1)sec2udu=2(43)2∫π40(tan2u+1)sec2udu=83
Tuesday, March 8, 2016
Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 22
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