Question

How do you evaluate \[\int{{{e}^{6x}}\sin \left( {{e}^{2x}} \right)dx}\] using substitution and tabular integration?

Answer 1

Hint: In this problem, we have to evaluate the given integral using the substitution and the integration integral. Here we can assume \[t={{e}^{2x}}\] and substitute it in the given integral. We can then use the integration by parts formula and integrate in steps, we can then substitute the t value in the final answer to get the result.

Complete step by step solution:
We know that the given integral expression is,
\[\int{{{e}^{6x}}\sin \left( {{e}^{2x}} \right)dx}\]
We can now assume,
 Let \[t={{e}^{2x}}\]
\[\begin{align}
  & \Rightarrow dt=2{{e}^{2x}}dx \\
 & \Rightarrow dx=\dfrac{dt}{2{{e}^{2x}}} \\
\end{align}\]
We can substitute the above value in the given integral, we get
\[\Rightarrow \int{{{e}^{6x}}\sin \left( {{e}^{2x}} \right)dx}=\int{\dfrac{1}{2}{{t}^{2}}\sin \left( t \right)dt}\]
We can now integrate using integration by parts.
We know that the integration by parts formula is,
\[\Rightarrow \int{udv=uv-\int{vdu}}\] ……… (2)
We can now use the ILATE rule, where we have a product of two functions.
We can now take the above indefinite integral (1).
We can now use the ILATE method to find which functions to be used in integration by parts.
We know that ILATE stands for Inverse Logarithmic Algebraic Trigonometric Exponential.
We can choose the functions according to the order of letters in ILATE.
We have to take integration for \[{{t}^{2}}\sin \left( t \right)\], where \[{{t}^{2}}\] is the algebraic function and \[\sin \left( t \right)\] is trigonometric function.
Then taking by parts we choose \[{{t}^{2}}\] for A (of ILATE) as first function, u and \[\sin \left( t \right)\] for T (of ILATE) as second function, \[dv\]
 Such that \[u={{t}^{2}}\] and \[dv=\sin tdt\]
Then, \[du=2tdt\] and \[v=-\cos t\] .
We can now substitute in the formula, we get
\[\Rightarrow \dfrac{1}{2}\left( \left[ -{{t}^{2}}\cos \left( t \right) \right]+2\int{t\cos \left( t \right)dt} \right)\]
We can again integrate using the u-v method, we get
Such that \[u=t\] and \[dv=\cos tdt\]
Then, \[du=dt\] and \[v=\sin t\] .
We can now substitute in the formula, we get
\[\Rightarrow \dfrac{1}{2}\left[ -{{t}^{2}}\cos \left( t \right) \right]+2\left( t\sin t-\int{\sin tdt} \right)\]
We can now simplify the above step, we get
\[\Rightarrow t\sin \left( t \right)+\cos \left( t \right)-\dfrac{{{t}^{2}}}{2}\cos \left( t \right)+C\]
We can now substitute \[t={{e}^{2x}}\], we get
\[\Rightarrow {{e}^{2x}}\sin \left( {{e}^{2x}} \right)+\cos \left( {{e}^{2x}} \right)-\dfrac{{{e}^{4x}}}{2}\cos \left( {{e}^{2x}} \right)+C\]

Therefore, the answer is \[{{e}^{2x}}\sin \left( {{e}^{2x}} \right)+\cos \left( {{e}^{2x}} \right)-\dfrac{{{e}^{4x}}}{2}\cos \left( {{e}^{2x}} \right)+C\].

Note: We should also remember that \[du\] is obtained by differentiating u and v is obtained by integrating \[dv\], which we use in the integration by parts formula. We know that ILATE stands for Inverse Logarithmic Algebraic Trigonometric Exponential. We can choose the functions according to the order of letters in ILATE.