What is #int_0^(pi/4) 7secthetatanthetad theta# ?

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Answer 1 · 2016-11-22T14:28:28

#7[sqrt2-1]#

#d/(dx)(secx)=secxtanx#

so#""int_0^(pi/4)7secthetatanthetad theta#

#=7[sectheta]_0^(pi/4)#

#=7[sec(pi/4)-sec0]#

#7[1/cos(pi/4)-1/cos0]#

#7[1/(1/sqrt2)-1/1]#

#7[sqrt2-1]#