Monday, November 16, 2015

Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 30

Determine the integral π30tan5xsec6xdx


π30tan5xsec6xdx=π30tan5xsec4xsec2xdxπ30tan5xsec6xdx=π30tan5x(sec2x)2sec2xdxApply Trigonometric Identity sec2x=tan2x+1π30tan5xsec6xdx=π30tan2x(tan2x+1)2sec2xdxπ30tan5xsec6xdx=π30tan5x(tan4+2tan2x+1)sec2xdxπ30tan5xsec6xdx=π30(tan9x+2tan7x+tan5x)sec2xdx


Let u=tanx, then du=sec2xdx. When x=0,u=0 and when x=π3,u=3. Thus,


π30(tan9x+2tan7x+tan5x)sec2xdx=30(u9+2u7+u5)duπ30(tan9x+2tan7x+tan5x)sec2xdx=[u9+19+1+2(u7+17+1)+u5+15+1]30π30(tan9x+2tan7x+tan5x)sec2xdx=[u1010+2u88+u66]30π30(tan9x+2tan7x+tan5x)sec2xdx=[u1010+u84+u66]30π30(tan9x+2tan7x+tan5x)sec2xdx=(3)1010+(3)84+(3)66(0)1010(0)84(0)66π30(tan9x+2tan7x+tan5x)sec2xdx=[(3)2]510+[(3)2]44+[(3)2]36π30(tan9x+2tan7x+tan5x)sec2xdx=(3)510+(3)44+(3)36π30(tan9x+2tan7x+tan5x)sec2xdx=24310+814+276π30(tan9x+2tan7x+tan5x)sec2xdx=24310+814+92π30(tan9x+2tan7x+tan5x)sec2xdx=486+405+9020π30(tan9x+2tan7x+tan5x)sec2xdx=98120=49.05

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