Integration by parts is a formula used to integrate the product of two functions.
If and , the following equation holds:
where and .
The formula for integration by parts requires one of the functions appearing in the integrand to be differentiated, and the other to be integrated. To apply this formula to a product of two functions, choose one of the functions to be the function that is differentiated and label it . The other function is to be integrated, and is therefore labelled .
Note: On some occasions the integrand appearing in the term may still be a product of two functions of . In this case the integration by parts formula would need to be applied again to compute .
Some care is required when making the choices for and . The function to be differentiated, , should be a function that will be reduced to a simpler function after differentiating. Polynomial functions are a good choice for , since repeated differentiation of a polynomial will always produce a constant. Trigonometric functions such as are not good choices for as they retain their trigonometric form after differentiation, and are therefore better choices for .
The formula can also be applied to definite integration between limits and :
Suppose there are two functions and . Then by the product rule, the derivative of their product is
Integrate both sides with respect to :
By the fundamental theorem of calculus,
hence:
Rearranging this gives the formula for integration by parts:
Use integration by parts to find .
Recall the formula for integration by parts:
The first step is to make suitable choices for and . The two functions appearing in the integrand are and . The choice for should be a function that will become simpler when differentiated; here would be a bad choice, as the derivative of is just another trigonometric function, . The derivative of , however, is a constant, so we choose .
The choices for and are and .
It can be convenient to lay out the expressions for , and their derivatives in the following way:
The gaps for and can be filled in after performing the relevant calculations.
Differentiating with respect to gives :
\[u' = \frac{\mathrm{d </div>{\mathrm{d} x}\bigl[1+x\bigr] = 1,\]
and integrating with respect to gives :
Note: It is not necessary to include the constant of integration when computing .
The expressions for , , and are:
Substituting these into the formula for integration by parts gives:
First evaluate
Substitute this back in the formula to obtain:
The value of the integral is .
Note: Suppose different choices were made for and . Let and . Then and . Substituting these expressions into the formula for integration by parts yields: It is immediately obvious that the integral on the right hand side is more complicated than the original problem, thus emphasising the importance of making appropriate choices for and .
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Prof. Robin Johnson uses integration by parts to find .
Prof. Robin Johnson uses integration by parts to find .
Prof. Robin Johnson uses integration by parts to find .
Prof. Robin Johnson uses integration by parts to find .
Prof. Robin Johnson uses integration by parts to find .
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.
Test yourself: Numbas test on integration by parts