Question

How do you integrate $[({e^{2x}})\sin x]dx$ ?

Answer 1

Hint : In this question, we see that two functions are in multiplication with each other, to find the integration of the product of two functions, we use integration by parts. According to the integration by parts method, the integration of the product of two functions is equal to the difference of the product of the first function and the integration of the second function and integration of the product of the derivative of the first function and the integration of the second function, that is, $\int {u(x)v(x) = u(x)\int {v(x)} } - \int {u'(x)(\int {v(x} )}$ . So, we solve the given question by integration by parts using the above-mentioned definition

** Complete step-by-step answer** :
$[({e^{2x}})\sin x]dx$ is the product of two functions that are ${e^{2x}}$ and $\sin x$ , so we integrate using integration by parts as follows –
$\int {\sin x.{e^{2x}}dx} = \sin x\int {{e^{2x}}dx} - \int {\dfrac{{d\sin x}}{{dx}}(\int {{e^{2x}}dx} )dx} \\\ \Rightarrow \int {\sin x.{e^{2x}}dx} = \dfrac{{{e^{2x}}}}{{\dfrac{{d(2x)}}{{dx}}}}\sin x - \int {\cos x\dfrac{{{e^{2x}}}}{{\dfrac{{d(2x)}}{{dx}}}}} dx + c \\\ \Rightarrow \int {\sin x.{e^{2x}}dx} = \dfrac{{{e^{2x}}}}{2}\sin x - \int {\cos x\dfrac{{{e^{2x}}}}{2}} dx + c \\\ \Rightarrow \int {\sin x.{e^{2x}}dx} = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{1}{2}\int {\cos x{e^{2x}}dx} + c \;$
Now we see do the integration of $\cos x{e^{2x}}$ by using integration by parts –
$\int {\cos x{e^{2x}}dx} = \cos x\int {{e^{2x}}dx} - \int {\dfrac{{d\cos x}}{{dx}}(\int {{e^{2x}}dx} )dx} \\\ \Rightarrow \int {\cos x{e^{2x}}dx} = \dfrac{{{e^{2x}}\cos x}}{2} - \int { - \dfrac{{{e^{2x}}\sin x}}{2}dx} \\\ \Rightarrow \int {\cos x{e^{2x}}dx} = \dfrac{{{e^{2x}}\cos x}}{2} + \dfrac{1}{2}\int {\sin x{e^{2x}}dx} \;$
Using the above result in the obtained expression, we get –
$\int {\sin x.{e^{2x}}dx} = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{1}{2}[\dfrac{{{e^{2x}}\cos x}}{2} + \dfrac{1}{2}\int {\sin x{e^{2x}}dx} ] \\\ \Rightarrow \int {\sin x.{e^{2x}}dx} = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4} - \dfrac{1}{4}\int {\sin x{e^{2x}}dx} \\\ \Rightarrow \int {\sin x.{e^{2x}}dx} + \dfrac{1}{4}\int {\sin x{e^{2x}}dx} = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4} \\\ \Rightarrow \dfrac{5}{4}\int {\sin x.{e^{2x}}dx} = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4} \\\ \Rightarrow \int {\sin x.{e^{2x}}dx} = \dfrac{4}{5}(\dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4}) + c \;$
Hence $\int {\sin x.{e^{2x}}dx}$ is equal to $\dfrac{4}{5}(\dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4}) + c$ .
So, the correct answer is “ $\dfrac{4}{5}(\dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4}) + c$ .”.

Note : For solving such type of questions, we must know the integration of basic functions like trigonometric functions, exponential functions or $\int {\sin x} ,\int {\cos x} ,\int {{e^x},\,etc.}$ Now we see that while applying the integration by parts method, we are not given in the question that which function we have to take as the first function and which function as the second, to resolve this confusion, we have a rule called ILATE (I: Inverse trigonometric function, L: Logarithm function, A: Algebraic function, T: Trigonometric function, E: Exponential function). This rule tells us the order of preference for different types of functions.