Integration by parts is a method to calculate indefinite integrals by using the differential of the product of two functions.
If we have two functions, u and v, the differential of their product will be:
\[d(u \cdot v)=u \cdot dv + v \cdot du \tag{1}\]If we integrate both sides of the mathematical expression, we’ll get:
\[\int d(u \cdot v)=\int \left (u \cdot dv + v \cdot du \right ) \tag{2}\]We know that the integral of the derivative of a function gives the function itself. So:
\[\int d(u \cdot v)= u \cdot v \tag{3}\]Also, the integral of the sum of two functions is the sum of the integral of each function:
\[\int \left (u \cdot dv + v \cdot du \right ) = \int u \cdot dv + \int v \cdot du \tag{4}\]Replacing equations (3) and (4) in equation (2) we get:
\[ u \cdot v =\int u \cdot dv + \int v \cdot du \tag{5} \]Rearranging equation (5) gives the mathematical expression for integration by parts:
\[\bbox[#FFFF9D]{ \int u \cdot dv =u \cdot v -\int v \cdot du} \tag{6} \]Example 1. Solve the following integral using integration by parts:
\[\int x \cdot cos(x) dx\]As we can see, the integral contains the product of two functions: x and cos(x). We are going to replace these with u and v.
Step 1. Replace the functions with u and v
First we replace:
\[u = x \tag{7}\]Applying differentiation to equation (7) we get:
\[du = dx \tag{8}\]Second we replace:
\[dv = cos(x)dx \tag{9}\]Applying integration to equation (9) we get:
\[\int dv = \int cos(x)dx \tag{10}\] \[v = sin(x) \tag{11}\]Step 2. Use general equation (6) and replace u and v with calculated functions:
\[\int x \cdot cos(x) dx = x \cdot sin(x) – \int sin(x)dx \tag{12}\]Step 3. Solve the right side of the equation
We know that:
\[\int sin(x) dx = – cos(x) + C \tag{13}\]Replacing the result of the integral (13) in equation (12) we get the result of our integration by parts:
\[\int x \cdot cos(x) dx = x \cdot sin(x) + cos(x) + C\]where C is a constant of integration.
Example 2. Solve the following integral using integration by parts:
\[\int x \cdot e^x dx\]As we can see, the integral contains the product of two functions: x and ex. We are going to replace these with u and v.
Step 1. Replace the functions with u and v
First we replace:
\[u = x \tag{14}\]Applying differentiation to equation (14) we get:
\[du = dx \tag{15}\]Second we replace:
\[dv = e^x dx \tag{16}\]Applying integration to equation (16) we get:
\[\int dv = \int e^x dx \tag{17}\] \[v = e^x \tag{18}\]Step 2. Use general equation (6) and replace u and v with calculated functions:
\[\int x \cdot e^x dx = x \cdot e^x – \int e^x dx \tag{19}\]Step 3. Solve the right side of the equation
We know that:
\[\int e^x dx = e^x + C \tag{20}\]Replacing the result of the integral (20) in equation (19) we get the result of our integration by parts:
\[\int x \cdot e^x dx = x \cdot e^x – e^x + C\]which in simplified form is:
\[\int x \cdot e^x dx = e^x \cdot (x-1) + C\]where C is a constant of integration.
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