Apply integration by parts to evaluate integrals of products, including repeated application and the recovery of the original integral

What this dot point is asking

NESA wants you to integrate products of functions using integration by parts, the reverse of the product rule. You must choose the parts sensibly, apply the method more than once where the algebraic factor is a higher power, and handle the case where the original integral reappears so that it can be recovered algebraically. The same machinery, applied to an integral that depends on an integer $n$, produces a reduction formula, which is one of the most reliably examined Extension 2 question types.

The formula and why it works

The product rule says $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$. Integrate both sides with respect to $x$. The left side integrates to $uv$ (integration undoes the derivative), and the right side splits into two integrals:

Rearranging to isolate the integral we want gives integration by parts:

The method does not evaluate an integral directly; it trades one integral for another. That trade is only worth making if the new integral $\int v\,\frac{du}{dx}\,dx$ is genuinely easier than the one you started with. Everything therefore hinges on the split: $u$ should become simpler when differentiated, and $\dfrac{dv}{dx}$ should be a factor you can actually integrate. A bad split leaves you with a harder integral, which is the signal to swap your choices and start again.

Choosing the parts: LIATE

A reliable order of preference for choosing $u$ is LIATE:

• Logarithmic ($\ln x$)
• Inverse trigonometric ($\arcsin x$, $\arctan x$)
• Algebraic ($x^n$)
• Trigonometric ($\sin x$, $\cos x$)
• Exponential ($e^x$)

Whatever appears earlier in this list is usually the better choice for $u$, and whatever is later becomes $\dfrac{dv}{dx}$. The list is not a law of nature, it is a heuristic ranked by how the function behaves under differentiation. Logarithms and inverse trig functions are at the top because they are awkward to integrate but differentiate to simple algebraic expressions, so you always want them as $u$. Exponentials are at the bottom because they integrate as easily as they differentiate, so they make a painless $\dfrac{dv}{dx}$. For example, in $\int x\ln x\,dx$ the logarithm ranks above the algebraic term, so $u = \ln x$ and $\dfrac{dv}{dx} = x$. In $\int x e^x\,dx$ the algebraic term ranks above the exponential, so $u = x$ and $\dfrac{dv}{dx} = e^x$.

A first example

Evaluate $\int x e^x\,dx$. By LIATE the algebraic factor outranks the exponential, so let $u = x$ and $\dfrac{dv}{dx} = e^x$. Then $\dfrac{du}{dx} = 1$ and $v = e^x$. Applying the formula:

The differentiation step has collapsed the $x$ to a constant $1$, leaving a standard integral. That is exactly the simplification a good split is designed to produce.

Repeated application

When the algebraic factor is a higher power, one pass is not enough, so apply the method repeatedly. For $\int x^2 e^x\,dx$, first take $u = x^2$, $\dfrac{dv}{dx} = e^x$:

The remaining integral $\int x e^x\,dx$ is the one we just did, so substitute that result: $x^2 e^x - 2(x - 1)e^x + C = (x^2 - 2x + 2)e^x + C$. Each application lowers the power of $x$ by one, so an integral with $x^n$ against an exponential or a sine or cosine needs $n$ applications until the power vanishes. Keep $u$ as the algebraic factor every single time; switching roles midway sends you in a circle.

The boomerang (recovery) case

For integrals such as $\int e^x \cos x\,dx$ neither factor ever simplifies, since differentiating $e^x$ gives $e^x$ and differentiating $\cos x$ cycles through trig functions. Two applications of integration by parts return a multiple of the original integral. You then solve for it algebraically. This is the boomerang: the integral comes back to you, letting you collect like terms and divide. The one rule that makes it work is consistency, you must take $u$ to be the same type of function (say the trig factor, or the exponential) in both applications. If you swap roles on the second pass, the two steps cancel and you return to where you started with nothing gained.

Integrating a lone logarithm or inverse trig

Some integrands look like a single function with nothing obvious to split, yet integration by parts still applies. To find $\int \ln x \, dx$, write the integrand as $1 \cdot \ln x$, take $u = \ln x$ and $\dfrac{dv}{dx} = 1$, so $v = x$. Then

The same trick handles $\int \arctan x \, dx$ and $\int \arcsin x \, dx$: the "second function" is an invisible $1$, and the leftover integral is finished by a quick substitution or a standard form. This is exactly why logarithmic and inverse-trigonometric functions sit at the top of LIATE, since they have no elementary antiderivative of their own but differentiate to something simple.

Definite integrals by parts

For a definite integral the formula carries the limits on every term:

Evaluate the boundary term $\big[uv\big]_a^b$ at both limits, then deal with the remaining definite integral. Watch for boundary terms that vanish, which is a common and deliberate simplification when a limit is $0$ or $\tfrac{\pi}{2}$ (so that $x$, $\sin\theta$ or $\cos\theta$ is zero there). Keeping the limits attached to every term throughout avoids the classic error of evaluating only the final integral and forgetting the $\big[uv\big]_a^b$ piece.

Reduction formulae

Repeated integration by parts on an integral that depends on an integer $n$ produces a reduction (or recurrence) formula, expressing $I_n$ in terms of $I_{n-1}$ or $I_{n-2}$, as in the $J_n = \frac{n-1}{n} J_{n-2}$ result above. These are a favourite Extension 2 question type because they reward a clean, structured argument. The method is always the same shape: split off one factor, integrate by parts, show the boundary term vanishes (or simplify it), and use a Pythagorean identity to bring the original integral back so you can collect terms. Once the recurrence is established you apply it repeatedly down to a known base case such as $J_0 = \tfrac{\pi}{2}$ or $J_1 = 1$, which the later parts of the question usually ask you to use.

How exam questions ask about integration by parts

• "Find $\int \ldots$" with an obvious product ($x e^x$, $x\cos x$, $x\ln x$). One or two passes of parts; choose $u$ by LIATE, show $\dfrac{du}{dx}$ and $v$, and finish with $+C$.
• "Which expression equals $\int \ldots$ after one application?" A multiple-choice probe of a single step. Differentiate the algebraic factor, integrate the exponential or trig factor, keep the minus sign, and do not touch the surviving integral.
• "Find $\int e^{x}\cos x\,dx$"-type integrals. The boomerang: integrate by parts twice with a consistent choice of $u$, recover the original integral, then collect and divide.
• "Show that $I_n = \ldots$" (a reduction formula). Split off one factor, integrate by parts, vanish the boundary term, apply a Pythagorean identity to recover $I_n$, and rearrange. The marks are for the structure as much as the answer.
• "Hence evaluate $I_4$ (or similar)." Use the reduction formula you just proved, applying it repeatedly down to the base case $I_0$ or $I_1$.
• "Find $\int \ln x\,dx$ / $\int \arctan x\,dx$" (a lone function). Write the integrand as $1 \cdot f(x)$, take $u = f(x)$ and $\dfrac{dv}{dx} = 1$.