What is f(x) = int tanx dx if f(pi/8) = 0 ?
2 Answers
Explanation:
Use the definition of
f(x)=inttanxdx=intsinx/cosxdx
We can put this in the form
f(x)=-int(-sinx)/cosxdx=-int(du)/u=-lnabsu+C=-lnabscosx+C
We can use the initial condition
0=-lnabscos(pi/8)+C
C=ln(cos(pi/8))
We can find this using a form of the cosine double angle formula:
cos(pi/8)=sqrt(1/2(1+cos(pi/4)))=sqrt(1/2(1+sqrt2/2))
=sqrt(1/2((2+sqrt2)/2))=sqrt(2+sqrt2)/2
Thus,
C=ln(sqrt(2+sqrt2)/2)
and
f(x)=-lnabscosx+ln(sqrt(2+sqrt2)/2)
Explanation:
We shall evaluate
Let
Then