What is sign slip when v involves a trig function?

Integrating sin x gives -cos x; that minus then interacts with the formula's minus. Track signs carefully.

SASpecialist MathematicsSyllabus dot point

How does reversing the product rule let us integrate products such as x times e to the x?

Evaluate integrals using integration by parts, including choosing u and dv and applying the method more than once where needed.

Integration by parts as the reverse of the product rule, choosing u and dv with the LIATE guide, repeated application, and the standard trick for integrating the natural logarithm.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-29

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What this dot point is asking
The formula
Choosing u and dv
A first example
Integrating the logarithm
Applying the method twice

What this dot point is asking

You need to evaluate integrals using integration by parts, choosing $u$ and $dv$ sensibly and applying the method repeatedly when required.

The formula

Starting from the product rule $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$ and integrating both sides gives

Choosing u and dv

A first example

Integrating the logarithm

A classic case is $\int \ln x\,dx$ , where there is seemingly no product. Take $u = \ln x$ and $dv = dx$ :

\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C.

Applying the method twice

When the algebraic factor is a higher power, one pass reduces it but leaves another integral by parts.

Applying integration by parts twice

Find $\displaystyle\int x^{2}\sin x\,dx$

Let $u = x^2$ , $dv = \sin x\,dx$ , so $du = 2x\,dx$ and $v = -\cos x$ :

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

Now do $\int x\cos x\,dx$ by parts: $u = x$ , $dv = \cos x\,dx$ , so $du = dx$ , $v = \sin x$ :

\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x.

Substitute back:

\int x^2\sin x\,dx = -x^2\cos x + 2\big(x\sin x + \cos x\big) + C = -x^2\cos x + 2x\sin x + 2\cos x + C.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2017 SACE Stage 23 marksUse integration by parts to find the integral of x e^(2x) with respect to x.

Show worked answer →

Integration by parts: integral of u dv = u v - integral of v du.

Choose u = x (so du = dx) and dv = e^(2x) dx (so v = (1/2) e^(2x)), picking the algebraic factor as u so it differentiates away. [1 mark]

Apply the formula:
integral of x e^(2x) dx = x (1/2) e^(2x) - integral of (1/2) e^(2x) dx
= (1/2) x e^(2x) - (1/2) integral of e^(2x) dx. [1 mark]

The remaining integral is (1/2) e^(2x), so:
= (1/2) x e^(2x) - (1/2)(1/2) e^(2x) + c = (1/2) x e^(2x) - (1/4) e^(2x) + c, where c is a constant. [1 mark]

2025 SACE Stage 23 marksGiven that the integral of x e^(2x) dx = (1/2) x e^(2x) - (1/4) e^(2x) + c, use integration by parts to show that the integral of x^2 e^(2x) dx = e^(2x)((1/2)x^2 - (1/2)x + 1/4) + k, where k is a constant.

Show worked answer →

Apply integration by parts to integral of x^2 e^(2x) dx. Choose u = x^2 (so du = 2x dx) and dv = e^(2x) dx (so v = (1/2) e^(2x)). [1 mark]

integral of x^2 e^(2x) dx = x^2 (1/2) e^(2x) - integral of (1/2) e^(2x) (2x) dx
= (1/2) x^2 e^(2x) - integral of x e^(2x) dx. [1 mark]

Substitute the given result for integral of x e^(2x) dx:
= (1/2) x^2 e^(2x) - [(1/2) x e^(2x) - (1/4) e^(2x)] + k
= (1/2) x^2 e^(2x) - (1/2) x e^(2x) + (1/4) e^(2x) + k
= e^(2x)((1/2) x^2 - (1/2) x + 1/4) + k. [1 mark]