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# Find the value of the integral $\int {e_{}^{2x}.\sin (3x + 1)dx}$

Last updated date: 13th Sep 2024
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Hint: In this type of question we have to be solved by applying the formula.
Here we apply integration by parts formula and do some simple simplification.
Finally we get the required answer.

Formula used: $\left( 1 \right)$ Integration by parts,$\int {uvdx = u\int {vdx - \int {(\dfrac{{du}}{{dx}}} } } - vdx)dx$
$\left( 2 \right)\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sinx = cosx}}$

It is given that the integer and let us consider as $I = \int {e_{}^{2x}.\sin (3x + 1)dx} ....\left( 1 \right)$
Now by applying the formula integration by parts that is $\int {uvdx = u\int {vdx - \int {(\dfrac{{du}}{{dx}}} } } - vdx)dx$
Here, in the given integral we have to consider as, $u = $$\sin (3x + 1)dx be the first term and v =$$e_{}^{2x}$be the second one.
Now, substitute the value in the formula we get,
Therefore, we can write it as,
$\Rightarrow$$I = \sin (3x + 1)\int {e_{}^{2x}} dx - \int {\{ \dfrac{d}{{dx}}} \sin (3x + 1)\int {e_{}^{2x}} dx\} dx....\left( 2 \right) Here we have to split the term and using some formula Since \dfrac{d}{{dx}}\sin x = \cos x Therefore, in the terms has to be change that \dfrac{d}{{dx}}\sin (3x + 1) = 3\cos (3x + 1)....\left( 3 \right) Also, since \int {e_{}^x} dx = {e^x} \Rightarrow$$\int {e_{}^{2x}} dx = \dfrac{{e_{}^{2x}}}{2}....\left( 4 \right)$
Substitute these values $\left( 3 \right)$and $\left( 4 \right)$ in $\left( 2 \right)$we can write it as,
$\Rightarrow$$I = \dfrac{{\sin (3x + 1)e_{}^{2x}}}{2} - \int {3\cos (3x + 1).\dfrac{{e_{}^{2x}}}{2}} dx Taking common \dfrac{3}{2} we get- \Rightarrow$$I = \dfrac{{e_{}^{2x}\sin (3x + 1)}}{2} - \dfrac{3}{2}\int {e_{}^{2x}} \cos (3x + 1)dx$
Again we have to apply integration by parts for $\int {e_{}^{2x}} \cos (3x + 1)dx$,
In this case, we have to take the parts like, $u = $$\cos (3x + 1)dx and v = e_{}^{2x} we get- \Rightarrow$$I = \dfrac{{e_{}^{2x}\sin (3x + 1)}}{2} - \dfrac{3}{2}[\cos (3x + 1)\int {e_{}^{2x}} dx - \int {\{ \dfrac{d}{{dx}}} \cos (3x + 1)\int {e_{}^{2x}} dx\} dx$
On doing some simplification and using the formula we get,
$\Rightarrow$$= \dfrac{{e_{}^{2x}\sin (3x + 1)}}{2} - \dfrac{3}{2}[\cos (3x + 1).\dfrac{{e_{}^{2x}}}{2} - \int { - 3\sin (3x + 1)\dfrac{{e_{}^{2x}}}{2}} dx] + {C_1} On multiplying \dfrac{3}{2} in second term we get, \Rightarrow$$I = \dfrac{{e_{}^{2x}\sin (3x + 1)}}{2} - \dfrac{3}{4}e_{}^{2x}\cos (3x + 1) - \dfrac{9}{4}\int {e_{}^{2x}} \sin (3x + 1)dx + {C_1}$
From equation $\left( 1 \right)$ we can write,
$\Rightarrow$$I = \dfrac{{e_{}^{2x}\sin (3x + 1)}}{2} - \dfrac{3}{4}e_{}^{2x}\cos (3x + 1) - \dfrac{9}{4}I + {C_1} We can take {\text{I}} as RHS, and we get \Rightarrow$$I + \dfrac{9}{4}I = \dfrac{{e_{}^{2x}\sin (3x + 1)}}{2} - \dfrac{3}{4}e_{}^{2x}\cos (3x + 1) + {C_1}$
We can take the LCM for RHS and we can write it as,
$\Rightarrow$$\dfrac{{4I + 9I}}{4} = \dfrac{{e_{}^{2x}\sin (3x + 1)}}{2} - \dfrac{3}{4}e_{}^{2x}\cos (3x + 1) + {C_1} We can add the numerator in the RHS we get, \Rightarrow$$\dfrac{{13}}{4}I = \dfrac{{e_{}^{2x}\sin (3x + 1)}}{2} - \dfrac{{3e_{}^{2x}\cos (3x + 1)}}{4} + {C_1}$
On dividing $\dfrac{4}{{13}}$ on both side we get,
$\Rightarrow$$I = \dfrac{{2e_{}^{2x}\sin (3x + 1)}}{{13}} - \dfrac{{3e_{}^{2x}\cos (3x + 1)}}{{13}} + \dfrac{{4{C_1}}}{{13}}$
Therefore we can write that

$I = \dfrac{{2e_{}^{2x}\sin (3x + 1)}}{{13}} - \dfrac{{3e_{}^{2x}\cos (3x + 1)}}{{13}} + C$, where $C = \dfrac{{4{C_1}}}{{13}}$

Note: Here while applying the integration by parts formula consider the term which can be easily integrated as ‘v’ and the term which is hard to integrate as ‘u’, so that it will make the process easier. It is a question of indefinite integration as no limit is given but in case of definite integration we will have the limit of minimum and maximum range.