Tuesday, March 8, 2016

Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 22

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=π40sec4u2duπ20sec4(t2)dt=2π40sec4uduπ20sec4(t2)dt=2π40sec2usec2uduApply Pythagorean Identity sec2u=tan2u+1π20sec4(t2)dt=2π40(tan2u+1)sec2udu


Let v=tanu, then dv=sec2udu. When u=0,v=0 and when u=π4,v=1. Thus,


2π40(tan2u+1)sec2udu=221(v2+1)dv2π40(tan2u+1)sec2udu=2[v2+12+1+v]102π40(tan2u+1)sec2udu=2[v33+v]102π40(tan2u+1)sec2udu=2[(1)33+1(0)330]2π40(tan2u+1)sec2udu=2(13+1)2π40(tan2u+1)sec2udu=2(43)2π40(tan2u+1)sec2udu=83

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