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The integral of $\int {{e^x}\left( {\sin x + \cos x} \right)dx} $ is:
A.${e^x}\cos x + c$
B.${e^x}\sin x + c$
C.${e^x}\sec x + c$
D.None of this

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Last updated date: 13th Jun 2024
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Answer
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Hint: First of all we will take the given equation and break it in two parts by opening the bracket with integration symbols in each. Let these two terms as${I_1}\& {I_2}$. After that solve any one term and integrate with respect to $x$, then put this value in the original equation, thus we will get the answer, and add an integrating constant c with the answer.

Complete step-by-step answer:
We have to integrate the given term i.e.:
$\int {{e^x}\left( {\sin x + \cos x} \right)dx} $
Now we can write the given term as:
By separate both functions:
$ \Rightarrow \int {{e^x}\sin xdx + } \int {{e^x}\cos xdx} .........\left( 1 \right)$
Let the first term in equation $\left( 1 \right)$ is ${I_1}$ and second term is ${I_2}$
Where:
${I_1} = $ \[\int {{e^x}\sin xdx} \]\[\]
\[
  {I_2} = \int {{e^x}\cos xdx} \\
    \\
 \]
Now differentiate the term ${I_2}$ with respect to $x$
By using the $uv$ rule of integration:
$dx = d\left( {uv} \right)dx = udvdx + vdudx$
We will apply this rule and integrate it:
\[
  {I_2} = \int {{e^x}\cos xdx} \\
   \Rightarrow {I_2} = {e^x}\int {\cos xdx - \int {\left( {\dfrac{d}{{dx}}\left( {{e^x}} \right).\int {\cos x} } \right)} } dx \\
   \Rightarrow {I_2} = {e^x}\sin x - \int {{e^x}} \sin xdx \\
    \\
 \]
Put the value of ${I_2}$in term${I_1}$
Thus we get:
\[
  {I_1} = \int {{e^x}\sin xdx + {e^x}\sin x - \int {{e^x}\sin xdx} } \\
   \Rightarrow {e^x}\sin x + c \\
    \\
 \]
Where c is integrating constant.
Hence the correct answer is option B.


Note: For the given question we have to remember that to integrate the given equation we have to remember that to break the equation in two integrations like $\int {{e^x}\sin xdx + } \int {{e^x}\cos xdx} $
called as ${I_1}\& {I_2}$ then solve ${I_2}$ part of the equation and put this value in equation $1$ and add an integrating constant c with this and this is our answer.