What is #f(x) = int x-cotx dx# if #f((5pi)/4) = 0 #?

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Answer 1 · 2017-03-03T04:18:46

#f(x)=1/2x^2-lnabssinx-(25pi^2+16ln2)/32#

#f(x)=intx-cotxdx#

#f(x)=intxdx-intcosx/sinxdx#

The first can be integrated using #intx^ndx=x^(n+1)/(n+1)#:

#f(x)=1/2x^2-intcosx/sinxdx#

For the remaining integral let #u=sinx#. This implies that #du=cosxdx#. Then:

#f(x)=1/2x^2-int(du)/u#

Which is the natural logarithm integral:

#f(x)=1/2x^2-lnabsu#

#f(x)=1/2x^2-lnabssinx+C#

Use the initial condition #f((5pi)/4)=0#:

#0=1/2((5pi)/4)^2-lnabssin((5pi)/4)+C#

#0=(25pi^2)/32-ln(1/sqrt2)+C#

Rewriting with log rules:

#0=(25pi^2)/32+1/2ln2+C#

#C=-(25pi^2+16ln2)/32#

So:

#f(x)=1/2x^2-lnabssinx-(25pi^2+16ln2)/32#