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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

Last updated date: 13th Jun 2024
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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.

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.