What is f(x) = int -cos^3x +tanx dxf(x)=cos3x+tanxdx if f(pi)=-2 f(π)=2?

1 Answer
Mar 20, 2016

f(x)=-sinx+sin^3x/3-lnabscosx-2f(x)=sinx+sin3x3ln|cosx|2

Explanation:

This can be split into two integrals:

f(x)=-intcos^3xdx+inttanxdxf(x)=cos3xdx+tanxdx

The second integral inttanxdx=-lnabscosx+Ctanxdx=ln|cosx|+C, this is a common integral.

To find the first, do what follows:

-intcos^3xdx=-intcos^2xcosxdxcos3xdx=cos2xcosxdx

=-int(1-sin^2x)cosxdx=(1sin2x)cosxdx

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

=-int1-u^2du=1u2du

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

=-u+u^3/3+C=-sinx+sin^3x/3+C=u+u33+C=sinx+sin3x3+C

Thus, the whole expression equals

f(x)=-sinx+sin^3x/3-lnabscosx+Cf(x)=sinx+sin3x3ln|cosx|+C

Since f(pi)=-2f(π)=2, we see that

-2=-sinpi+sin^3pi/3-lnabscospi+C2=sinπ+sin3π3ln|cosπ|+C

-2=-0+0^3/3-lnabs(-1)+C2=0+033ln|1|+C

-2=0+0+ln1+C2=0+0+ln1+C

-2=C2=C

Thus, we obtain

f(x)=-sinx+sin^3x/3-lnabscosx-2f(x)=sinx+sin3x3ln|cosx|2