Question

How do you integrate $\int {{e^{2x}}\sin (3x)}$ by integrating by parts method?

Answer 1

According to the question we have to integrate $\int {{e^{2x}}\sin (3x)}$ by integrating by parts method. So, first of all to integrate $\int {{e^{2x}}\sin (3x)}$ by integrating by parts method we have to understand about integrate by part rule which is as explained below:
Integrate by parts method: According to this method if we have to integrate more than function we have to use this rule in which first of all we have to choose our first and the second terms in which one of the function is to be integrated and other to be differentiated with the help of the formulas to integrate and differentiate and formula for the by parts method is as mentioned below:
$\Rightarrow \int {uvdx = } u\int {vdx - \left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} dx...........(A)$
Where, u is the first term and v is the second term and to choose the term we have to follow the process as explained below:
Ilate Rule: To choose the first and the second term we have to follow the ilate rule which is as explained below:
I – Inverse,
L – Logarithm,
A – Algebraic,
T – Trigonometric and,
E – Exponential
Now, we have to use the formula (A) and substitute all the values in the formula (A) which and then we have to integrate and differentiate the terms.
Now to integrate the exponential terms we have to use the formula which is as mentioned below:
$\Rightarrow \int {{e^{ax}} = \dfrac{1}{a}{e^{ax}} + c.............(B)}$
Where, c is the constant term.
Now, to find the differentiation of the functions in the expression (3) as obtained in the solution step 4 we have to use the formula (C) as mentioned below:
$\Rightarrow \dfrac{d}{{dx}}\sin ax = a\cos ax...............(C)$

Complete step-by-step solution:
Step 1: First of all we have to let that our given integration is I which is as mentioned below:
$\Rightarrow I = \int {{e^{2x}}\sin (3x)}$………………….(A)
Step 2: Now, we have to use the ilate rule to determine the first and the second terms which is already explained in the solution hint. Hence,
$\Rightarrow$First term $u = \sin 3x$
$\Rightarrow$Second term $v = {e^{2x}}$
Step 3: Now, to find the integration of both of the terms as obtained in the solution step 2 we have to use the formula (A) to find the integration with the help of integration by parts method. Hence, on substituting all the values in the formula (A),
$\Rightarrow I = \sin 3x\int {{e^{2x}}} dx - \int {\left( {\dfrac{d}{{dx}}\sin x\int {{e^{2x}}dx} } \right)} dx$…………………(2)
Step 4: Now, to find the integration of the expression (2) we have to use the formula (B) which is as explained in the solution hint. Hence,
$\Rightarrow I = \sin 3x\left( {\dfrac{{{e^{2x}}}}{2}} \right) - \int {\left( {\dfrac{d}{{dx}}\sin x\left( {\dfrac{{{e^{2x}}}}{2}} \right)} \right)} dx$……………(3)
Step 5: Now, to find the differentiation of the functions in the expression (3) as obtained in the solution step 4 we have to use the formula (C) as mentioned in the solution hint. Hence,
$\Rightarrow I = \sin 3x\left( {\dfrac{{{e^{2x}}}}{2}} \right) - \dfrac{3}{2}\int {\left( {\cos 3x{e^{2x}}} \right)} dx$
Step 6: Now, we have to solve the integration of $\int {\left( {\cos 3x{e^{2x}}} \right)} dx$ using the integration by parts rule. Hence,
$\Rightarrow I = \sin 3x\left( {\dfrac{{{e^{2x}}}}{2}} \right) - \dfrac{3}{2}\left[ {\cos 3x\int {{e^{2x}} - \int {\left( {\dfrac{d}{{dx}}\cos 3x\int {{e^{2x}}dx} } \right)} } } \right]dx \\\ \Rightarrow I = \sin 3x\left( {\dfrac{{{e^{2x}}}}{2}} \right) - \dfrac{3}{2}\left[ {\cos 3x\dfrac{{{e^{2x}}}}{2} - \int {\left( { - \sin 3x\dfrac{{{e^{2x}}}}{2}} \right)} dx} \right] \\\ \Rightarrow I = \sin 3x\left( {\dfrac{{{e^{2x}}}}{2}} \right) - \dfrac{3}{4}\cos 3x{e^{2x}} - \dfrac{3}{4}\int {\sin 3x{e^{2x}}dx} ..........(4) \\\$
Step 7: Now, as from the step 4 we can see that after integration we have obtained $\int {{e^{2x}}\sin (3x)}$which we already let as I in the solution step 1 hence,
$\Rightarrow I = \sin 3x\left( {\dfrac{{{e^{2x}}}}{2}} \right) - \dfrac{3}{4}\cos 3x{e^{2x}} - \dfrac{3}{4}I$
Now on taking the terms I in the same side of the expression as obtained just above,
$\Rightarrow I + \dfrac{3}{4}I = \sin 3x\left( {\dfrac{{{e^{2x}}}}{2}} \right) - \dfrac{3}{4}\cos 3x{e^{2x}} \\\ \Rightarrow \dfrac{{7I}}{4} = \dfrac{{\sin 3x{e^{2x}}}}{2} - \dfrac{3}{4}\cos 3x{e^{2x}} \\\$
Now, on finding the L.C.M of the expression as obtained just above, Hence,
$\Rightarrow \dfrac{{7I}}{4} = \dfrac{{\sin 3x{e^{2x}} - 3\cos 3x{e^{2x}}}}{4} \\\ \Rightarrow I = \dfrac{{{e^{2x}}}}{7}(2\sin 3x - 3\cos 3x)$

Hence, with the help of the formula (A), (B), and (C) we have determined the value of the given integration $\int {{e^{2x}}\sin (3x)}$ which is $= \dfrac{{{e^{2x}}}}{7}(2\sin 3x - 3\cos 3x)$.

Note: To find the integration by parts rule it is necessary that we have to determine the first and the second terms with the help of the ilate rule according to which our first term should be Inverse, Logarithm, Algebraic, Trigonometric and, Exponential.
To find the integration by integration by part rule first of all we have to integrate the first term and then find the integration of the differentiation of the first term and integration of the second term.