Integration by parts converts a product integral into something simpler by choosing:
∫ u dv = u·v − ∫ v du
- Pick u so that du becomes simpler.
- Pick dv so that v is easy to integrate.
- For x^n times trig or exponential, repeat until the power drops to zero.
- Select a problem type that matches your integrand.
- Enter the required parameters, like n, a, and b.
- Press Calculate to view the integral result above the form.
- Review the chosen u, dv, and the step list.
- Use Download CSV or Download PDF to save the output.
Tip: When in doubt, choose u using the LIATE idea (log, inverse trig, algebraic, trig, exponential).
| Example integrand | Suggested u | Result summary |
|---|---|---|
| ∫ x^2 · e^(1x) dx | u = x^2 | e^x(x^2 − 2x + 2) + C |
| ∫ x^3 · sin(2x) dx | u = x^3 | Combination of sin(2x), cos(2x), and polynomials + C |
| ∫ x^1 · ln(x) dx | u = ln(x) | (x^2 ln(x))/2 − x^2/4 + C |
What this calculator solves
This tool targets integrals where the integrand is a product of two functions. Typical classroom cases include x^n·e^(ax), x^n·sin(ax), x^n·cos(ax), x^n·ln(x), and e^(ax)·sin(bx) or cos(bx). These patterns appear in calculus, signals, and probability work. In typical exams, these are common when substitution fails, because neither factor is a clean derivative of the other alone.
Why integration by parts works
Integration by parts comes from the product rule: d(uv)=u dv+v du. Rearranging gives ∫u dv = u·v − ∫v du. The “win” happens when du is simpler than u, so the remaining integral becomes easier than the original product integral.
Choosing u and dv efficiently
A practical guide is LIATE: logarithm, inverse trig, algebraic, trig, exponential. Prefer u from the left side of that list. For example, with x·ln(x), choose u=ln(x) so du=1/x dx, while dv=x dx integrates smoothly to v=x^2/2.
Polynomial products and repeated steps
When a polynomial multiplies trig or an exponential, each by-parts step reduces the polynomial degree by one. If n=6, you may need up to six iterations before the polynomial vanishes. The calculator keeps n capped for readable output, but the same reduction idea scales to larger n on paper. Each step also introduces a boundary term u·v that must be simplified.
Exponential–trigonometric loops
For e^(ax)·sin(bx) or e^(ax)·cos(bx), two rounds of by parts bring the original integral back. You then solve a simple linear equation to isolate the integral. The closed form uses a^2+b^2 in the denominator, which also helps spot algebra mistakes.
Quick sanity checks
Differentiate your final expression mentally: e^(ax) factors should remain, trig signs should alternate, and polynomial terms should match the degree pattern. For ln(x) cases, expect a dominant x^(n+1)·ln(x)/(n+1) term plus a smaller power correction. These checks catch missing constants and sign slips.
Study workflow with exports
After calculating, download CSV to store integrands and results in a practice sheet. Use PDF to print worked steps for revision. A good routine is ten problems per session: vary n, switch sin to cos, and change a or b to see how coefficients scale across solutions. Compare your PDF answers with manual work to build confidence fast.
Frequently asked questions
What is the main formula used here?
It applies ∫u dv = u·v − ∫v du, derived from the product rule. The calculator displays chosen u, dv, du, and v so you can track each substitution step clearly.
How do I pick u and dv quickly?
Use LIATE: choose u as logarithm first, then inverse trig, algebraic, trig, and exponential. Pick dv as the remaining factor so v is easy to integrate.
Why does n have a limit in the polynomial options?
Repeated by-parts steps grow the expression quickly. Limiting n keeps results readable on one page, while still covering the most common homework and exam examples.
Does the tool verify my answer?
It provides a structured closed form for supported templates and shows the parts selection. For best verification, differentiate the final result and confirm it matches the original integrand.
What happens in the e^(ax)·sin(bx) and e^(ax)·cos(bx) cases?
Two integrations by parts recreate the starting integral. You then solve a linear equation, leading to a denominator a^2+b^2 and a clean combination of sine and cosine terms.
Can I use the custom single-step mode for any functions?
Yes for planning: enter u(x) and dv, and it prints the by-parts identity with placeholders. You must still compute v and du yourself if the functions are not in the built-in templates.