Question

# Find the laplace transform of L\left\{f(x)\right\} being f(x)=\int_0^3 \cos (3t)dt

Laplace transform

Find the laplace transform of $$L\left\{f(x)\right\}$$ being $$\displaystyle{f{{\left({x}\right)}}}={\int_{{0}}^{{3}}}{\cos{{\left({3}{t}\right)}}}{\left.{d}{t}\right.}$$

2021-09-11

Find $$L\left\{f(x)\right\}$$ where $$\displaystyle{f{{\left({x}\right)}}}={\int_{{0}}^{{3}}}{\cos{{\left({3}{t}\right)}}}{\left.{d}{t}\right.}$$
Given that $$f(x)=\int_0^3 \cos(3t)dt$$ , so we first calculate this integral.
$$\displaystyle{f{{\left({x}\right)}}}={\int_{{0}}^{{3}}}{\cos{{\left({3}{t}\right)}}}{\left.{d}{t}\right.}={{\left[{\frac{{{\sin{{\left({3}{t}\right)}}}}}{{{3}}}}\right]}_{{0}}^{{3}}}={\frac{{{1}}}{{{3}}}}{\left[{\sin{{9}}}-{\sin{{0}}}\right]}$$
$$\displaystyle\Rightarrow{f{{\left({x}\right)}}}={\frac{{{\sin{{9}}}}}{{{3}}}}$$
$$\displaystyle\Rightarrow{f{{\left({x}\right)}}}$$ is a constant function
Now $$F(s)=L\left\{f(x)\right\}=\int_0^{\infty}f(x)e^{-sx}dx$$
$$\displaystyle={\int_{{0}}^{{\infty}}}{\frac{{{\sin{{9}}}}}{{{3}}}}{e}^{{-{s}{x}}}{\left.{d}{x}\right.}={\frac{{{\sin{{9}}}}}{{{3}}}}{\int_{{0}}^{{\infty}}}{e}^{{-{s}{x}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{\sin{{9}}}}}{{{3}}}}{{\left[{\frac{{{e}^{{-{s}{x}}}}}{{-{s}}}}\right]}_{{0}}^{{\infty}}}={\frac{{-{\sin{{9}}}}}{{{3}{s}}}}{\left[{e}^{{-\infty}}-{e}^{{0}}\right]}$$
$$\Rightarrow F(s)=L\left\{f(x)\right\}=\frac{1}{3s}\sin 9$$