Question

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

Answer 1

Hint: Here in this question we have to find the derivatives of the given function. We have to find the derivative of the function in two parts first is ${e^{2x}} $ and the second part is $\sin (3x + 1) $. Solve the question using the integrations by parts method of derivative.

Complete step-by-step answer:
We have
= $\int {{e^{2x}}.\sin (3x + 1)dx.} $
Now derive the function by using integration by parts
The formula of integration by parts is
$\int {u.vdx = u\int {vdx - \int {{u^,}(\int {vdx)dx} } } } $
Here $u = {e^{2x}},v = \sin (3x + 1)$
Do the derivative using the above formula
$ = \sin (3x + 1)\int {{e^{2x}}dx - \int {\{ \dfrac{d}{{dx}}\sin (3x + 1)\int {{e^{2x}}dx\} dx} } } $
Do the derivative of the function
$ = \dfrac{{\sin (3x + 1).{e^{2x}}}}{2} - \int {3\cos (3x + 1).\dfrac{{{e^{2x}}}}{2}dx} $
Separate the numeral part
$ = \dfrac{{{e^{2x}}\sin (3x + 1)}}{2} - \dfrac{3}{2}\int {{e^{2x}}\cos (3x + 1)dx} $
We have again two parts in the second parts which is
1-${e^{2x}} $
2-$\cos (3x + 1) $
Do the derivative of second part using the integration by parts
$ = \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]} } } $
Do the derivative of only second part write the first part same as written above
\[ = \dfrac{{{e^{2x}}\sin (3x + 1)}}{2} - \dfrac{3}{2}[\cos (3x + 1)\dfrac{{{e^{2x}}}}{2} - \int { - 3\sin (3x + 1).\dfrac{{{e^{2x}}}}{2}dx} ] + C\]
\[ = \dfrac{{{e^{2x}}\sin (3x + 1)}}{2} - \dfrac{3}{4}{e^{2x}}\cos (3x + 1) - \dfrac{9}{4}\int {{e^{2x}}\sin (3x + 1)dx + C} \]
$I = \dfrac{{{e^{2x}}\sin (3x + 1)}}{2} - \dfrac{3}{4}{e^{2x}}\cos (3x + 1) - \dfrac{9}{4}I + C$
In the above equation we put
$\int {{e^{2x}}.\sin (3x + 1)dx.}$ $= I$
Now equalizing the equation, we get
$\dfrac{{13}}{4}I = \dfrac{{{e^{2x}}\sin (3x + 1)}}{2} - \dfrac{{3{e^{2x}}\cos (3x + 1)}}{4} + C$
$I = \dfrac{{2{e^{2x}}\sin (3x + 1)}}{{13}} - \dfrac{{3{e^{2x}}\cos (3x + 1)}}{{13}} + C$
Hence, we have the derivative of our function given in the question
$\int {{e^{2x}}.\sin (3x + 1)dx.} $$ = \dfrac{2}{{13}}{e^{2x}}\sin (3x + 1) - \dfrac{3}{{13}}{e^{2x}}\cos (3x + 1)$

Note: Here in this question use the integration by parts method. Solve the function step by step. Students mostly make mistakes while doing the derivative of ${e^{2x}} $. When there is more than one function in the question, first separate the parts of the functions and then do the derivative of the function. Do not get confused between the power and the real numbers. Always write down the C at the end of the function when you did the derivative where c means a consent part of the integration.at the end separate the fractional part and the non-fractional part.