Question

Evaluate the Integral\(\int {{e^t}\sin \left( {\alpha t - 3} \right)dt} \)

Short answer

The value of the integral is\(\int {{e^t}\sin \left( {\alpha t - 3} \right)} dt = \frac{{{e^t}}}{{\left( {1 + {\alpha ^2}} \right)}}(\sin (\alpha - 3) - \alpha \cos \left( {\alpha t - 3} \right)) + C\)

Step-by-step solution

Step(1):- Using substitution we substitute \((\alpha t - 3) = x\) and differentiate to get the value of dt.

Then using the formula \(\int {{e^{ax}}\sin du} \) we get our desired results.

\(I = \int {{e^t}\sin \left( {\alpha t - 3} \right)dt} \)

Let, \(\left( {\alpha t - 3} \right) = x\) and \(t = \frac{{x + 3}}{\alpha }\)

\(\begin{aligned}{l} \Rightarrow \alpha dt = dx\\ \Rightarrow dt = \frac{{dx}}{\alpha }\end{aligned}\)

Step(2):- we substitute  \((\alpha t - 3) = x\) in integral I, also \(t = \frac{{x + 3}}{\alpha }\).

\(\begin{aligned}{l}I = \int {\frac{{{e^t}\sin xdx}}{\alpha } = \int {\frac{{{e^{\left( {\frac{{x - 3}}{\alpha }} \right)}}\sin xdx}}{\alpha }} } \\I = \frac{1}{\alpha }\int {{e^{x/\alpha }}{e^{3/\alpha }}\sin xdx} \\I = \frac{{{e^{3/\alpha }}}}{\alpha }\int {{e^{x/\alpha }}\sin xdx} \end{aligned}\)

Step(3):- Now we use the formula:-

\(\int {{e^{au}}\sin budu = \frac{{{e^{au}}}}{{{a^2} + {b^2}}}\left( {a\sin bu + b\cos bu} \right)} + C\)

Integral is:

\(\begin{aligned}{l}I = \frac{{{e^{3/\alpha }}}}{\alpha }(\frac{{{e^{3/\alpha }}}}{{{{\left( {\frac{1}{\alpha }} \right)}^2} + 1}}\left( {\frac{1}{\alpha }\sin x - \cos x} \right)) + C\\I = \frac{{{\alpha ^2}{e^{\left( {3 + x} \right)/\alpha }}}}{{{{\left( {1 + \alpha } \right)}^2}}}\left( {\sin x - \alpha \cos x} \right) + C\end{aligned}\)

Now, substitute all values back in terms of t.

Hence, the integration of \(I = \int {{e^t}\sin \left( {\alpha t - 3} \right)dt} \)

\(I = \frac{{{e^t}}}{{\left( {1 + {\alpha ^2}} \right)}}\left( {\sin \left( {\alpha t - 3} \right) - \alpha \cos \left( {\alpha t - 3} \right)} \right) + C\)