# Use integration by parts to find the Laplace transform of the given function f(t)=4tcos h(at)

Question
Laplace transform
Use integration by parts to find the Laplace transform of the given function
$$f(t)=4t\cos h(at)$$

2020-11-01
Step 1
Given function is
$$f(t)=4t\cos h(at)$$
We have to find Laplace transformation by integration by part.
Step 2
So Laplace Transformation is
$$Lf(t)=4\int_0^\infty t\cos h(at)e^{-st}dt$$
$$\Rightarrow\ Lf(t)=4\int_0^\infty t\bigg(\frac{e^{at}+e^{-at}}{2}\bigg)e^{-st}dt$$
$$\Rightarrow\ Lf(t)=2\int_0^\infty t(e^{(a-s)t}+e^{-(a+s)t})dt$$
$$\Rightarrow\ Lf(t)=2\int_0^\infty te^{(a-s)t}dt+2\int_0^\infty te^{-(a+s)t}dt$$
$$\Rightarrow\ Lf(t)=2\left[-\frac{te^{(a-s)t}}{s-a}-\frac{e^{(a-s)t}}{(s-a)^{2}}\right]_0^\infty+2\left[-\frac{te^{-(a+s)t}}{s+a}-\frac{e^{-(a+s)t}}{(s+a)^{2}}\right]_0^\infty$$
$$\Rightarrow\ Lf(t)=-2\left[0-\frac{1}{(s-a)^{2}}\right]-2\left[0-\frac{1}{(s+a)^{2}}\right]$$
$$\Rightarrow\ Lf(t)=\frac{2}{(s-a)^{2}}+\frac{2}{(s+a)^{2}}$$
$$\Rightarrow\ Lf(t)=\frac{2(2s^{2}+2a^{2}+2as-2as)}{(s-a)^{2}(s+a)^{2}}$$
$$\Rightarrow\ Lf(t)=\frac{4(s^{2}+a^{2})}{(s^{2}-a^{2})^{2}}$$

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