Find $\mathcal{L}\left\{t e^{2t}\cos\left(5t\right)\right\}$

Question

This is what I have so far:

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\begin{align} \mathcal{L}\left\{t e^{2t}\cos\left(5t\right)\right\}=\int_0^\infty e^{-st}t e^{2t}\cos\left(5t\right)\:dt,\tag{1} \end{align} but notice that if \begin{align} F'\left(s\right)=\int e^{-st}t e^{2t}\cos\left(5t\right)\:dt,\tag{2} \end{align} then \begin{align} F\left(s\right)=\int -e^{-st} e^{2t}\cos\left(5t\right)\:dt = \left(\frac{1}{s}\right)\int\left[e^{2t}\cos\left(5t\right)\right]\:d\left(e^{-st}\right),\tag{3} \end{align} due to Leibniz's Rule. Therefore, an integration by parts reveals that \begin{align} \int\left[e^{2t}\cos\left(5t\right)\right]\:d\left(e^{-st}\right) & = e^{-st} e^{2t}\cos\left(5t\right)-\int e^{-st}\:d\left(e^{2t}\cos\left(5t\right)\right) \\[3ex] & = e^{-st} e^{2t}\cos\left(5t\right)-\int e^{-st}\frac{d}{dt}\left\{e^{2t}\cos\left(5t\right)\right\}\:dt\tag{4} \end{align} which is \begin{align} =e^{-st} e^{2t}\cos\left(5t\right)+5\int e^{-st}e^{2t}\sin\left(5t\right)\:dt -2\int e^{-st}e^{2t}\cos\left(5t\right)\:dt.\tag{5} \end{align} Now \begin{align} \int -e^{-st}e^{2t}\sin\left(5t\right)\:dt & =\left(\frac{1}{s}\right)\int \left[e^{2t}\sin\left(5t\right)\right]\:d\left(e^{-st}\right),\tag{6} \end{align} which results in \begin{align} \int \left[e^{2t}\sin\left(5t\right)\right]\:d\left(e^{-st}\right) & = e^{2t}\sin\left(5t\right)e^{-st}-\int e^{-st}\:d\left(e^{2t}\sin\left(5t\right)\right) \\[3ex] & = e^{2t}\sin\left(5t\right)e^{-st} - \int e^{-st}\frac{d}{dt}\left\{e^{2t}\sin\left(5t\right)\right\}\:dt\tag{7} \end{align} which gives us something similar to (5): \begin{align} = e^{2t}\sin\left(5t\right)e^{-st}-5\int e^{-st}e^{2t}\cos\left(5t\right)\:dt-2\int e^{-st}e^{2t}\sin\left(5t\right)\:dt. \end{align}

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Where do I go from here? Or perhaps there is a better way. But if what I have is correct, is it possible to solve it from here?

Answer 1

If you try and derive $$\int e^{ax} \sin(bx) \ dx \ \ \text{ or } \int e^{ax} \cos(bx) \ dx$$ you end up going around in circles like this. The trick is the recognize you have got the original integrals back but with different constants. Then you can add/subtract them to arrive at the standard results

$$\int e^{ax} \sin(bx) \ dx = \frac{1}{a^2 + b^2} e^{ax} \left( a\sin(bx) - b\cos(bx) \right) + C$$

and similarly for $\cos$.