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

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Answer 1 · 2018-04-08T14:35:54

#f(x)=tan(x)-ln|tan(x/2)|+ln|-sqrt2-1|-1#

We got: #f(x)=intsec^2x-cscx \ dx#

#=intsec^2x \ dx-intcscx \ dx#

#=tan(x)-ln|tan(x/2)|+C#

Also, we know that #f((5pi)/4)=0#, and we have to find #C#.

#f((5pi)/4)=0#

#=>tan((5pi)/4)-ln|tan(((5pi)/4)/2)|+C=0#

#=>tan((5pi)/4)-ln|tan((5pi)/8)|+C=0#

#=>1-ln|-sqrt2-1|+C=0#

#:.C=ln|-sqrt2-1|-1#

#:.f(x)=tan(x)-ln|tan(x/2)|+ln|-sqrt2-1|-1#