Concepts: integration by parts and special integrals

Integrals22 min readLesson 4 of 8

4.1 The by-parts rule Integration by parts reverses the product rule. The integral of u times the derivative of v equals u v minus the integral of v times the derivative of u. The whole art is choosing which factor to call u: the rule of thumb is the LIATE order, inverse trigonometric, logarithmic, algebraic, trigonometric, exponential, taking whichever appears earlier in that list. The reason is that the method moves a derivative onto u and an integral onto v, so u should be something that becomes simpler when differentiated and v something that is easy to integrate. Some integrals need two rounds of by parts, and the original integral reappears on the right; it is then moved to the left and solved like an equation. The exponential times sine and exponential times cosine integrals are the standard examples of this. The method also produces the reduction formulae that handle powers of sine and cosine, though the board paper rarely asks for them directly.

\int u v d x = u \int v d x - \int (\frac{d u}{d x}) (\int v d x) d x

By parts with the second factor integrated first

\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C

The exponential pattern that saves a full by-parts round

4.2 Special integrals Four forms come up so often that the board expects them known, together with their derivations. The integral of the reciprocal of a quadratic with no real roots gives an inverse tangent after completing the square. The integral of the reciprocal of the square root of a quadratic gives an inverse sine after the same completion. The two square-root forms, root of a squared minus x squared and root of x squared plus or minus a squared, are handled by parts once and produce a logarithmic term. Rather than memorising the four answers blindly, remember the completing-the-square step; it reduces all four to one of the standard forms in the table. Then the answer follows from the formula sheet.

\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} \frac{x}{a} + C, \int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} \frac{x}{a} + C

The two completed-square forms

\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a} + C

Circle area integral by parts

\int x^{2} + a^{2} d x = \frac{x}{2} x^{2} + a^{2} + \frac{a ^{2}}{2} lo g x + x^{2} + a^{2} + C

Hyperbolic-looking logarithm from by parts

4.3 Worked examples Example 1: algebraic times exponential. Integrate x e to the x. Take u equal to x and v prime equal to e to the x, so the integral is x e to the x minus the integral of e to the x, which is x e to the x minus e to the x plus C. Differentiating confirms the answer. Example 2: algebraic times inverse sine. Integrate x sine inverse x. Take u equal to sine inverse x and v prime equal to x. The first term is half x squared sine inverse x, and the remaining integral is half the integral of x squared over the square root of one minus x squared. Substituting t equal to one minus x squared turns that into a power of t, and the final answer is half x squared sine inverse x minus one quarter sine inverse x plus one quarter x root of one minus x squared, plus C. Example 3: the exponential pattern. Integrate e to the x times sine x. Two rounds of by parts return the same integral with a minus sign, and solving the resulting equation gives half e to the x times sine x minus cosine x, plus C. The check by differentiation confirms the sign pattern.

\int x e^{x} d x = (x - 1) e^{x} + C

One round of by parts

\int e^{x} sin x d x = \frac{e ^{x}}{2} (sin x - cos x) + C

Two rounds and an equation

4.4 Common mistakes and exam tip Choosing u badly turns a simple integral into a harder one; the LIATE order prevents that. Dropping the minus sign in the second term is the most common arithmetic slip. In the special integrals, forgetting to complete the square and using a formula with the wrong constant a gives a wrong scale factor. In an exam, state the choice of u and v prime before substituting, so the examiner can follow the method even when the final algebra slips.

4.5 Inverse trigonometric and logarithmic integrands When the integrand is a lone inverse trigonometric function or a lone logarithm, take that function as u and take one as v prime. For the inverse tangent, the first round produces a rational integral that is finished by substitution. For the logarithm, the derivative is one over x, which turns the remaining integral into a simple power. This is the standard opening for questions such as the integral of the inverse sine of x, which appears in the exercise set.

4.6 A recurring reduction pattern Repeated by parts reduces the power of x in front of an exponential or a trigonometric function by one each time. For x squared e to the x, the first round leaves an integral of x e to the x, and the second round finishes it. Students who set out the two rounds in a table rather than in one long line make fewer sign errors, and the examiner can see the method clearly.

4.7 The independent check on a by-parts answer After any by-parts calculation, differentiate the answer and confirm that the original integrand returns. The check is more useful here than anywhere else in the chapter because the method has two places for a sign error: the choice of v prime and the sign in front of the second integral. A quick differentiation catches both, and it also confirms that the constant of integration has not been lost along the way.