# Find the Laplace transform for f(t)=2u(t-u)cos^2 2t-6e^(2t+7) delta(t+3)

Find the Laplace transform for
$f\left(t\right)=2u\left(t-3\right){\mathrm{cos}}^{2}2t-6{e}^{2t+7}\delta \left(t+3\right)$

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Benedict
Step 1
Calculate the Laplace transform:
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left(2u\left(t-3\right){\mathrm{cos}}^{2}2t-6{e}^{2t+7}\delta \left(t+3\right)\right){e}^{-st}dt$
$=2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}u\left(t-3\right){\mathrm{cos}}^{2}2t\cdot {e}^{-st}dt-6{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2t+7}\delta \left(t+3\right){e}^{-st}dt$
$=2{\int }_{3}^{\mathrm{\infty }}{e}^{-st}{\mathrm{cos}}^{2}2tdt-6{e}^{3s+1}$
$={\int }_{3}^{\mathrm{\infty }}{e}^{-st}\left(\mathrm{cos}4t+1\right)dt-6{e}^{3s+1}$
$={\int }_{3}^{\mathrm{\infty }}{e}^{-st}\mathrm{cos}4tdt+{\int }_{3}^{\mathrm{\infty }}{e}^{-st}dt-6{e}^{3s+1}$
$={\int }_{3}^{\mathrm{\infty }}{e}^{-st}\mathrm{cos}4tdt+\frac{1-{e}^{-3s}}{s}-6{e}^{3s+1}$
$={\int }_{3}^{\mathrm{\infty }}{e}^{-st}\frac{{e}^{4jt}+{e}^{-4jt}}{2}dt+\frac{1-{e}^{-3s}}{s}-6{e}^{3s+1}$
$={\int }_{3}^{\mathrm{\infty }}\frac{{e}^{-st+4jt}+{e}^{-st-4jt}}{2}dt+\frac{1-{e}^{-3s}}{s}-6{e}^{3s+1}$