Question

Write the value of $\int {{e^x}(\sin x + \cos x)dx} $

Answer 1

Hint: Here we have to use a standard property formula.
Also, we can derive that the formula was correct.
Finally, we use that formula to get the required value of the given integral.

Formula used: Formula for by parts rule of integration: $\int {u\dfrac{{dv}}{{dx}}dx} = uv - \int {v\dfrac{{du}}{{dx}}dx} $
Basic formulas, \[\int {{e^x}dx = } {e^x}\]and\[\dfrac{d}{{dx}}(f(x) = f'(x)\]
Differentiation of \[\sin x = \cos x\]
Differentiation of \[\cos x = - \sin x\]

Complete step-by-step answer:
It is given that the integral be ${\text{I}}$.
Let ${\text{I}} = \int {{e^x}(\sin x + \cos x)dx} ...\left( 1 \right)$
We have to find the value of equation (1),
Here we use that the standard formula $\int {{e^x}\{ f(x) + f'(x)\} dx} $
Then, we have to derive that
First we have to separate the integral
$\int {{e^x}f(x)dx + \int {{e^x}f'(x)dx} } ....\left( 2 \right)$
 Formula for by parts rule of integration: $\int {u\dfrac{{dv}}{{dx}}dx} = uv - \int {v\dfrac{{du}}{{dx}}dx} $
Applying by parts rule of integration on the standard property $\int {{e^x}} f(x)dx + \int {{e^x}f'(x)dx} ...\left( 3 \right)$
First we need to isolate the functions to apply by parts rule of integration.
We see in the first integral expression, that is $\int {{e^x}} f(x)dx....\left( 4 \right)$
Here we consider $u = f(x)$ and $v = \int {{e^x}dx} $,
Now we put the above function in the formula we get,
We can write that, \[\int {{e^x}} f(x)dx = f(x)\int {{e^x}dx - \int {\left[ {\dfrac{d}{{dx}}(f(x))\int {{e^x}dx} } \right]} } dx + {\text{C}}\]
We use these formulas, \[\int {{e^x}dx = } {e^x}\]and\[\dfrac{d}{{dx}}(f(x) = f'(x)\], we get the next step
$\int {{e^x}} f(x)dx = f(x){e^x} - \int {f'(x){e^x}dx} ....\left( 5 \right)$
Substitute $\left( 5 \right)$ in $\left( 3 \right)$ we get,
$\int {{e^x}} f(x)dx + \int {{e^x}f'(x)dx} = f(x){e^x} - \int {f'(x){e^x}dx} + \int {{e^x}f'(x)dx + c} $
In the above step now we can cancel the opposite expressions in the same step
We get,$\int {{e^x}\{ f(x) + f'(x)\} dx} $$ = {e^x}f(x) + {\text{C}}$, where ${\text{C}}$ is the constant term when we do integrate a constant term just added.
Now we have to find the value of given integral, by using the above standard formula,
Given that${\text{I}} = \int {{e^x}(\sin x + \cos x)dx} $
 We can equate to this $\left[ {\int {{e^x}\left\{ {f(x) + f'(x)} \right\}dx} } \right]$
Since, it is a standard property.
Now, function of x, \[f(x) = \sin x\] and differentiation of$f(x)$, $f'(x) = \cos x$
Therefore $I = {e^x}\sin x + c$$\left[ {\because \int {{e^x}\{ f(x) + f'(x)\} dx = {e^x}f(x) + c} } \right]$
Hence we got the required result.

Note: \[{\text{ILATE}}\] rule in integration is as follows:
In the integration by parts, we have two functions in the product, we need to give them priority. This rule helps to decide which term should you differentiate first term and second term to integrate first.
\[\begin{array}{*{20}{l}}
  {{\text{I = Inverse function}}} \\
  {{\text{L = Logarithmic function}}} \\
  {{\text{A = Algebraic function}}} \\
  {{\text{T = trigonometric function}}} \\
  {{\text{E = Exponential function}}}
\end{array}\]
The term is closer to \[{\text{I}}\] is differentiated first and the term which is closer to \[{\text{E}}\] is integrated first.