What is #f(x) = int -cos^3x +tanx dx# if #f(pi)=-2 #?

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Answer 1 · 2016-03-20T20:43:03

#f(x)=-sinx+sin^3x/3-lnabscosx-2#

This can be split into two integrals:

#f(x)=-intcos^3xdx+inttanxdx#

The second integral #inttanxdx=-lnabscosx+C#, this is a common integral.

To find the first, do what follows:

#-intcos^3xdx=-intcos^2xcosxdx#

#=-int(1-sin^2x)cosxdx#

Here, use substitution: let #u=sinx# and #du=cosxdx#. This gives:

#=-int1-u^2du#

Which we can integrate using #intu^ndu=u^(n+1)/(n+1)+C#, going term by term:

#=-u+u^3/3+C=-sinx+sin^3x/3+C#

Thus, the whole expression equals

#f(x)=-sinx+sin^3x/3-lnabscosx+C#

Since #f(pi)=-2#, we see that

#-2=-sinpi+sin^3pi/3-lnabscospi+C#

#-2=-0+0^3/3-lnabs(-1)+C#

#-2=0+0+ln1+C#

#-2=C#

Thus, we obtain

#f(x)=-sinx+sin^3x/3-lnabscosx-2#