Question

How do you integrate $\int{{{e}^{2x}}.\sin x}$ by parts?

Answer 1

Hint: In this question, we have to find the value of integration. Thus, we will use the by-parts method to solve the same, because the terms given in the question are multiplied by each other. So, first, we will find the value of u and v, and then put it in the by-parts formula \[\int{u.vdx=u\int{v.dx-\int{(u{)}'\left( \int{vdx} \right)}}}dx\] . Also, we will put the integration and differentiation of sinx, cosx, and ${{e}^{2x}}$ in the steps. After further solving the equation, we will again put the by-parts formula in the solution and make the necessary calculations to get the required value of the problem.

Complete step-by-step answer:
According to the question, we have to find the value of the integration.
Thus, we will use the by-parts method, which is
If $I=\int{u.vdx}$ ------- (1)
Then, \[\int{u.vdx=u\int{v.dx-\int{(u{)}'\left( \int{vdx} \right)}}}dx\] --------- (2)
The equation given to us is $I=\int{{{e}^{2x}}.\sin x}dx$ --------- (3)
So, on comparing equation (1) and (3), we get
$u={{e}^{2x}}$ and $v=\sin x$
Therefore, now we will put the above values in equation (2), we get
\[\Rightarrow I={{e}^{2x}}\int{\sin x}dx-\int{({{e}^{2x}}{)}'}\left( \int{\sin xdx} \right)dx\]
Now, we know that the integration of $\int{\sin xdx=-\cos x}$ , therefore we get
\[\Rightarrow I={{e}^{2x}}(-\cos x)-\int{({{e}^{2x}}{)}'}.(-\cos x)dx\]
Also, we know that the differentiation of ${{\left( {{e}^{2x}} \right)}^{\prime }}=2{{e}^{2x}}$ , thus, we put the value in the above equation, we get
\[\Rightarrow I={{e}^{2x}}(-\cos x)-\int{2{{e}^{2x}}}.(-\cos x)dx\]
On further solving, we get
\[\Rightarrow I=-{{e}^{2x}}\cos x+2\int{{{e}^{2x}}}.\cos xdx\] --------- (4)
So, now we will solve ${{I}_{1}}={{\int{e}}^{2x}}.\cos x.dx$ and then put the value in equation (4), so we will again use by-parts to solve this equation, we get
$\Rightarrow {{I}_{1}}={{e}^{2x}}{{\int{\cos x.dx-\int{\left( {{e}^{2x}} \right)}}}^{\prime }}\left( \int{\cos x.dx} \right)dx$
Now, we know that the integration of $\int{\cos xdx=\sin x}$ and the differentiation of ${{\left( {{e}^{2x}} \right)}^{\prime }}=2{{e}^{2x}}$ , therefore we get
$\Rightarrow {{I}_{1}}={{e}^{2x}}.\sin x-\left( \int{2{{e}^{2x.}}\cos x.dx} \right)$
On further solving, we get
$\begin{align}
  & \Rightarrow {{I}_{1}}={{e}^{2x}}.\sin x-2\left( \int{{{e}^{2x.}}\cos x.dx} \right) \\
 & \Rightarrow {{I}_{1}}={{e}^{2x}}.\sin x-2{{I}_{1}} \\
\end{align}$
Now add $2{{I}_{1}}$ on both sides of the equation, we get
$\Rightarrow {{I}_{1}}+2{{I}_{1}}={{e}^{2x}}.\sin x-2{{I}_{1}}+2{{I}_{1}}$
As we know, the same terms with opposite signs cancel out each other, therefore we get
$\Rightarrow 3{{I}_{1}}={{e}^{2x}}.\sin x$
Now, we will divide 3 on both sides of the equation, we get
$\Rightarrow \dfrac{3}{3}{{I}_{1}}=\dfrac{{{e}^{2x}}.\sin x}{3}$
Thus, on further solving, we get
$\Rightarrow {{I}_{1}}=\dfrac{{{e}^{2x}}.\sin x}{3}$
Now, we will put the above value in equation (4), we get
\[\Rightarrow I=-{{e}^{2x}}\cos x+2.\left( \dfrac{{{e}^{2x}}.\sin x}{3} \right)\]
Thus, on further simplification, we get
\[\Rightarrow I=-{{e}^{2x}}\cos x+\dfrac{2}{3}{{e}^{2x}}\sin x\]
Therefore, for the problem $\int{{{e}^{2x}}.\sin x}$ , its value is equal to \[-{{e}^{2x}}\cos x+\dfrac{2}{3}{{e}^{2x}}\sin x\] , which is our required answer.

Note: While solving this problem, do mention all the steps and the formula you are using, while solving this problem. Do remember the integration of sinx is negative of cosx and not the addition of cosx.