What is f(x) = int -cos^3x +tanx dxf(x)=∫−cos3x+tanxdx if f(pi)=-2 f(π)=−2?
1 Answer
Explanation:
This can be split into two integrals:
f(x)=-intcos^3xdx+inttanxdxf(x)=−∫cos3xdx+∫tanxdx
The second integral
To find the first, do what follows:
-intcos^3xdx=-intcos^2xcosxdx−∫cos3xdx=−∫cos2xcosxdx
=-int(1-sin^2x)cosxdx=−∫(1−sin2x)cosxdx
Here, use substitution: let
=-int1-u^2du=−∫1−u2du
Which we can integrate using
=-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+sin3x3−ln|cosx|+C
Since
-2=-sinpi+sin^3pi/3-lnabscospi+C−2=−sinπ+sin3π3−ln|cosπ|+C
-2=-0+0^3/3-lnabs(-1)+C−2=−0+033−ln|−1|+C
-2=0+0+ln1+C−2=0+0+ln1+C
-2=C−2=C
Thus, we obtain
f(x)=-sinx+sin^3x/3-lnabscosx-2f(x)=−sinx+sin3x3−ln|cosx|−2