What is #f(x) = int sec^2x- cos^2x dx# if #f((5pi)/4) = 0 #?

1 Answer
Monzur R.
Feb 26, 2017

#f(x)=1/8(2sin2x-4+5pi-10)+tanx#

Explanation:

#f(x)=intsec^2x-cos^2x# #dx, f(5/4pi)=0#

#f(x)=intsec^2x-cos^2x# #dx=intsec^2x# #dx-intcos^2x# #dx=#
#tanx+1/4(sin2x-2x)+"C"#

#f(5/4pi)=0=tan(5/4pi)+1/4(sin(5/2pi)-5/2pi)+"C"=#
#1+1/4-5/8pi+"C"=0#

#"C"=(5pi-10)/8#

#f(x)=tanx+1/4(sin2x-2x)+(5pi-10)/8=#
#1/8(2sin2x-4+5pi-10)+tanx#

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