Answer
\[ = \frac{a}{{{a^2} + {b^2}}}\,\,and\,\frac{b}{{{a^2} + {b^2}}}\]
Work Step by Step
\[\begin{gathered}
part\,a \hfill \\
\hfill \\
we\,\,\,know\,\,the\,\,formula \hfill \\
\hfill \\
\int_{}^{} {{e^{ - ax}}\,\cos \,\,bx\,dx = \frac{{{e^{ - ax}}}}{{{a^2} + {b^2}}}} \,\,\left[ { - a\cos bx + b\sin bx} \right] \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_0^\infty {{e^{ - ax}}\,\cos \,\,bx\,dx = \mathop {\lim }\limits_{c \to \infty } \frac{{{e^{ - ax}}}}{{{a^2} + {b^2}}}} \,\,\left[ { - a\cos bx + b\sin bx} \right]_0^c \hfill \\
\hfill \\
use\,\,the\,\,ftc \hfill \\
\hfill \\
= \mathop {\lim }\limits_{c \to \infty } \frac{{{e^{ - ac}}}}{{{a^2} + {b^2}}}\,\,\left[ { - a\cos bc + b\sin bc} \right] + \frac{a}{{{a^2} + {b^2}}} \hfill \\
\hfill \\
{\text{evaluate}}\,\,{\text{the}}\,\,{\text{limit}} \hfill \\
\hfill \\
= \frac{a}{{{a^2} + {b^2}}} \hfill \\
\hfill \\
part\,\,b \hfill \\
\hfill \\
we\,\,\,know\,\,the\,\,formula \hfill \\
\hfill \\
\int_{}^{} {{e^{ - ax}}\,\sin \,bx\,dx = } \frac{{{e^{ - ax}}}}{{{a^2} + {b^2}}}\,\,\left[ { - a\sin bx - b\cos bx} \right] \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \mathop {\lim }\limits_{c \to \infty } \frac{{{e^{ - ax}}}}{{{a^2} + {b^2}}}\,\,\left[ { - a\sin bx - b\cos bx} \right]_0^c \hfill \\
\hfill \\
use\,\,the\,\,ftc \hfill \\
\hfill \\
= \mathop {\lim }\limits_{c \to \infty } \frac{{{e^{ - ac}}}}{{{a^2} + {b^2}}}\,\,\left[ { - a\sin bx - b\cos bx} \right] + \frac{b}{{{a^2} + {b^2}}} \hfill \\
\hfill \\
{\text{evaluate}}\,\,{\text{the}}\,\,{\text{limit}} \hfill \\
\hfill \\
= \frac{b}{{{a^2} + {b^2}}} \hfill \\
\end{gathered} \]
