AP Calculus AB/BC · Unit 6 · Topic 6.11 · BC Only
Integrating Using Integration by Parts
Reverse the product rule and choose u to simplify after differentiation.
1. Topic Focus
Interpret definite integrals as accumulated change, connect sums to integrals, apply both Fundamental Theorems, and select antiderivative techniques.
This topic: Reverse the product rule and choose u to simplify after differentiation.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
For ∫x e^x dx, choose u=x and dv=e^x dx to get xe^x−e^x+C.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
Integration by parts is an AP Calculus BC technique
This method evaluates products and single functions that can be rewritten as products. Its purpose is to exchange the original integral for one that is easier to evaluate.
The formula comes from the product rule
Integrating both sides and solving for the first integral gives the integration-by-parts identity.
Know the formula and its sign
The subtraction is inherited from rearranging the integrated product rule; it is not optional.
The new integral should be simpler
A successful choice makes \(\int v\,du\) more recognizable than the original integral. Before committing, check whether both \(du\) and \(v\) can actually be found.
Choose u to simplify under differentiation
Polynomials lose degree, logarithms lose their logarithm, and inverse-trigonometric functions become algebraic rational expressions when differentiated.
Choose dv so it can be integrated
The remaining factor, including \(dx\), becomes \(dv\). If finding \(v=\int dv\) is harder than the original problem or has no elementary form, reconsider the choice.
LIATE is a guide, not a theorem
Logarithmic, inverse trigonometric, algebraic, trigonometric, and exponential is a useful preference order for \(u\). The decisive test is still whether the resulting integral becomes simpler.
Record all four parts before substituting
Write \(u\), \(du\), \(dv\), and \(v\) explicitly. This small table reduces coefficient and sign errors, especially when \(dv=e^{kx}dx\) or \(dv=\sin(kx)dx\).
Polynomial times exponential or trigonometric is a common signal
Choose the polynomial as \(u\) so repeated differentiation lowers its degree, while the exponential or trigonometric factor remains integrable.
A single logarithm can be multiplied by one
Choose \(u=\ln x\) and \(dv=dx\). The same idea works for inverse-trigonometric functions.
Some products require repeated application
For \(x^n e^{ax}\), \(x^n\sin bx\), or \(x^n\cos bx\), each use lowers the polynomial degree. Continue until the remaining integral is basic.
Tabular organization is shorthand for repeated parts
A derivative column for the polynomial and an antiderivative column for the other factor can organize alternating signs. It is bookkeeping for the same formula, not a different theorem.
Cyclic integrals are solved algebraically
For products such as \(e^{ax}\sin bx\), applying parts twice can reproduce the original integral. Move that repeated integral to the other side and solve for it.
Simplify or substitute before using parts when helpful
An integral may need substitution first to expose a polynomial-times-exponential form. Method selection should follow the structure rather than a fixed sequence of techniques.
Definite integration by parts keeps the original bounds
When no variable change occurs, the remaining integral retains the same bounds.
Evaluate the entire boundary product
The notation \([uv]_a^b\) means \(u(b)v(b)-u(a)v(a)\). Apply both endpoints to the complete product before subtracting the remaining definite integral.
Differentiate to verify
Use the product rule on every \(uv\)-type term in the proposed antiderivative. Cancellation should leave exactly the original integrand.
A reliable workflow and common errors
Inspect simpler methods first, choose and record \(u,du,dv,v\), apply the formula with its minus sign, repeat or solve algebraically if needed, add \(+C\) only for indefinite integrals, and verify. Common errors include choosing a nonintegrable \(dv\), losing coefficients, omitting the boundary term, and stopping while the new integral is no simpler.
6. Detailed Worked Example and Error Check
Example 1: Polynomial times exponential. Choose \(u=x\), \(dv=e^{2x}dx\), so \(du=dx\) and \(v=e^{2x}/2\):
Example 2: Apply parts twice. Starting with \(u=x^2\) and \(dv=\cos xdx\),
Differentiation makes the extra product-rule terms cancel and leaves \(x^2\cos x\).
Example 3: Treat a logarithm as a product.
Example 4: Integrate an inverse-trigonometric function. Choose \(u=\arctan x\) and \(dv=dx\):
Example 5: Solve a cyclic integral. Let \(I=\int e^x\cos xdx\). Two applications of parts give \(I=e^x(\sin x+\cos x)-I\), so
Example 6: Use definite integration by parts.
Example 7: Substitute before applying parts. For \(\int x^3e^{x^2}dx\), let \(w=x^2\), so \(x^3dx=w\,dw/2\). Then
7. AP Reasoning Routine
Identify the accumulating quantity and units, preserve bounds, choose a valid integration technique, and check answers by differentiation.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Use integration by parts and show the choices of \(u\) and \(dv\).
(a) Evaluate \(\int x\cos(2x)dx\).
(b) Evaluate \(\int x^2e^x dx\).
(c) Evaluate \(\int x\ln xdx\) for \(x>0\).
(d) Evaluate \(\int\ln(3x)dx\) for \(x>0\).
(e) Evaluate \(\int\arcsin xdx\) on \((-1,1)\).
(f) Evaluate \(\int_0^{\pi/2}x\sin xdx\).
(g) Evaluate \(\int_1^e\ln xdx\).
(h) Evaluate the cyclic integral \(\int e^x\sin xdx\).
(i) Explain why choosing \(u=x^3\) and \(dv=e^{x^2}dx\) is ineffective for \(\int x^3e^{x^2}dx\), and identify a better first step.
(j) Evaluate \(\int xe^{-x}dx\).
Check the solution
(a) Take \(u=x\) and \(dv=\cos(2x)dx\). The result is \(\frac{x}{2}\sin(2x)+\frac14\cos(2x)+C\).
(b) Apply parts twice to obtain \(e^x(x^2-2x+2)+C\).
(c) Take \(u=\ln x\) and \(dv=x\,dx\). The result is \(\frac{x^2}{2}\ln x-\frac{x^2}{4}+C\).
(d) Take \(u=\ln(3x)\), \(dv=dx\). Since \(du=dx/x\), the result is \(x\ln(3x)-x+C\).
(e) Take \(u=\arcsin x\), \(dv=dx\). The result is \(x\arcsin x+\sqrt{1-x^2}+C\).
(f) \([-x\cos x+\sin x]_0^{\pi/2}=1\).
(g) \([x\ln x-x]_1^e=1\).
(h) Applying parts twice and solving for the repeated integral gives \(\frac12e^x(\sin x-\cos x)+C\).
(i) The proposed \(dv\) has no elementary antiderivative. First set \(w=x^2\), which transforms the integral to \(\frac12\int we^w dw\), and then use parts.
(j) With \(u=x\) and \(dv=e^{-x}dx\), the result is \(-xe^{-x}-e^{-x}+C=-(x+1)e^{-x}+C\).