Problems based on Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and is formally stated by the integration by parts formula. This method is particularly useful when the integrand is a product of an algebraic expression and a transcendental function.

Integration by Parts Formula

The formula for integration by parts is derived from the product rule for differentiation and is given by:

[ \int u \, dv = uv - \int v \, du ]

where:

• ( u ) and ( v ) are functions of ( x ),
• ( du ) is the derivative of ( u ) with respect to ( x ),
• ( dv ) is the derivative of ( v ) with respect to ( x ).

Choosing ( u ) and ( dv )

The choice of ( u ) and ( dv ) is crucial for simplifying the integration process. A common mnemonic to decide which function to choose for ( u ) is "LIATE":

Priority	Function Type	Description
1	Logarithmic	Functions like ( \ln(x) )
2	Inverse trig	Functions like ( \arctan(x) ), ( \arcsin(x) )
3	Algebraic	Polynomials, ( \sqrt{x} ), etc.
4	Trigonometric	Functions like ( \sin(x) ), ( \cos(x) )
5	Exponential	Functions like ( e^x ), ( a^x )

The function that comes first in the LIATE list should be chosen as ( u ), and the remaining function as ( dv ).

Examples

Let's go through some examples to illustrate the use of integration by parts.

Example 1: Algebraic and Exponential Functions

Integrate ( \int x e^x \, dx ).

1. Choose ( u = x ) (Algebraic) and ( dv = e^x \, dx ) (Exponential).
2. Compute ( du = dx ) and ( v = \int e^x \, dx = e^x ).
3. Apply the integration by parts formula:

[ \int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C ]

where ( C ) is the constant of integration.

Example 2: Trigonometric and Algebraic Functions

Integrate ( \int x^2 \sin(x) \, dx ).

1. Choose ( u = x^2 ) (Algebraic) and ( dv = \sin(x) \, dx ) (Trigonometric).
2. Compute ( du = 2x \, dx ) and ( v = -\cos(x) ).
3. Apply the integration by parts formula:

[ \int x^2 \sin(x) \, dx = -x^2 \cos(x) - \int -\cos(x) \cdot 2x \, dx ]

Now, you need to integrate by parts again for the remaining integral:

[ \int -\cos(x) \cdot 2x \, dx ]

Choose ( u = 2x ) and ( dv = -\cos(x) \, dx ), then:

[ \int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2x \sin(x) - 2 \int \sin(x) \, dx ]

Finally:

[ \int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C ]

Example 3: Logarithmic and Algebraic Functions

Integrate ( \int \ln(x) \, dx ).

1. Choose ( u = \ln(x) ) (Logarithmic) and ( dv = dx ) (Algebraic).
2. Compute ( du = \frac{1}{x} \, dx ) and ( v = x ).
3. Apply the integration by parts formula:

[ \int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx = x \ln(x) - \int 1 \, dx ]

Thus:

[ \int \ln(x) \, dx = x \ln(x) - x + C ]

Important Points to Remember

• Integration by parts is not always straightforward; sometimes it requires multiple applications or clever manipulation.
• The choice of ( u ) and ( dv ) can greatly simplify the problem, so it's important to choose wisely.
• Sometimes integration by parts will lead to an integral that includes the original integral. In such cases, you can solve for the original integral algebraically.
• Always remember to add the constant of integration ( C ) at the end of an indefinite integral.

Integration by parts is a powerful tool in calculus, and mastering it requires practice. By working through various problems and becoming familiar with different types of functions, you can develop an intuition for the best way to approach each integral.