Determine the integral ∫π30tan5xsec6xdx
∫π30tan5xsec6xdx=∫π30tan5xsec4xsec2xdx∫π30tan5xsec6xdx=∫π30tan5x(sec2x)2sec2xdxApply Trigonometric Identity sec2x=tan2x+1∫π30tan5xsec6xdx=∫π30tan2x(tan2x+1)2sec2xdx∫π30tan5xsec6xdx=∫π30tan5x(tan4+2tan2x+1)sec2xdx∫π30tan5xsec6xdx=∫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)du∫π30(tan9x+2tan7x+tan5x)sec2xdx=[u9+19+1+2(u7+17+1)+u5+15+1]√30∫π30(tan9x+2tan7x+tan5x)sec2xdx=[u1010+2u88+u66]√30∫π30(tan9x+2tan7x+tan5x)sec2xdx=[u1010+u84+u66]√30∫π30(tan9x+2tan7x+tan5x)sec2xdx=(√3)1010+(√3)84+(√3)66−(0)1010−(0)84−(0)66∫π30(tan9x+2tan7x+tan5x)sec2xdx=[(√3)2]510+[(√3)2]44+[(√3)2]36∫π30(tan9x+2tan7x+tan5x)sec2xdx=(3)510+(3)44+(3)36∫π30(tan9x+2tan7x+tan5x)sec2xdx=24310+814+276∫π30(tan9x+2tan7x+tan5x)sec2xdx=24310+814+92∫π30(tan9x+2tan7x+tan5x)sec2xdx=486+405+9020∫π30(tan9x+2tan7x+tan5x)sec2xdx=98120=49.05
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