Tabular Integration Tool

Math Calculus • Integrals

Written by STEM Calculators Team Published December 28, 2025 Updated February 24, 2026

9. Tabular Integration Tool

Applies the tabular integration by parts (DI method) for products like polynomial × trig/exp. Supports definite bounds, shows the DI table, and plots \(f(x)\), \(F(x)+C\) with shaded area in definite mode.

Inputs

Mode

Auto builds the DI table and stops when the derivative becomes 0.

Integral mode

Definite mode evaluates \(F(b)-F(a)\) and shades \([a,b]\) on the graph.

Polynomial \(P(x)\)

Polynomial only. Examples: x^3, 3x^4-2x+7, (1/2)x^2-3, pi x^2 + 1.

Partner function \(g(kx)\)

Used as \(g(kx)\).

Argument coefficient \(k\)

Must be nonzero. Supports pi, e, + − * / ^.

Max DI rows

Auto stops earlier when derivative becomes 0.

Plot constant \(C\) for \(F(x)+C\)

Does not change \(F'(x)\).

Options Show steps Show DI table Show grid Shade \([a,b]\) (definite) Show \(f(x)\) Show \(F(x)+C\)

Presets

Click to load and evaluate.

Ready

Graph

Drag to pan • wheel/pinch to zoom. In definite mode, shaded region shows the signed area on \([a,b]\).

Center \(c\)

Half-width \(w\)

Legend

\(f(x)\) \(F(x)+C\) shaded \([a,b]\)

x: 0, y: 0, zoom(px/unit): 60

Result

Click Evaluate to build the DI table and compute the integral.

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9. Tabular Integration Tool — Theory

The tabular integration (DI-column method) is a fast way to perform repeated integration by parts for products like \(P(x)\cdot \sin(kx)\), \(P(x)\cdot \cos(kx)\), or \(P(x)\cdot e^{kx}\), especially when \(P(x)\) is a polynomial (its derivatives eventually become 0).

1) Core idea

The method is basically integration by parts done repeatedly, but organized in a table so you don’t rewrite the same steps. Suppose:

\[ \int P(x)\,g(kx)\,dx \]

\(D\)-column (Differentiate): repeatedly differentiate \(P(x)\).
\(I\)-column (Integrate): repeatedly integrate \(g(kx)\).
Alternate signs: \(+,-,+,-,\dots\)
Stop when the derivative column reaches 0 (this happens for polynomials).

If \(D_0=P, D_1=P', D_2=P'',\dots\) and \(I_0=g, I_1=\int g, I_2=\int\!\int g,\dots\), then:

\[ \int P(x)\,g(kx)\,dx \;=\; \sum_{i=0}^{n-1} (-1)^i\,D_i(x)\,I_{i+1}(x)\;+\;C \]

The tool automates this sum and shows the DI table. When definite bounds are chosen, it evaluates \(F(b)-F(a)\) and shades \([a,b]\) on the graph.

2) How to use the table (DI method)

Put \(P(x)\) at the top of the D column and keep differentiating down until you hit 0.
Put \(g(kx)\) at the top of the I column and keep integrating down the same number of rows.
Multiply diagonally: \(D_0\cdot I_1, D_1\cdot I_2, D_2\cdot I_3,\dots\)
Apply alternating signs: \(+,-,+,-,\dots\)
Add them up and append \(+C\).

This is exactly what repeated integration by parts produces, just formatted to be fast and less error-prone.

3) I-column patterns (very important)

When the partner is \(g(kx)\), each integration introduces factors of \(\frac{1}{k}\). That’s why the tool asks for argument coefficient \(k\).

Partner	\(I_0\)	\(I_1=\int I_0\,dx\)	\(I_2=\int I_1\,dx\)	\(I_3=\int I_2\,dx\)
\(\cos(kx)\)	\(\cos(kx)\)	\(\frac{\sin(kx)}{k}\)	\(-\frac{\cos(kx)}{k^2}\)	\(-\frac{\sin(kx)}{k^3}\)
\(\sin(kx)\)	\(\sin(kx)\)	\(-\frac{\cos(kx)}{k}\)	\(-\frac{\sin(kx)}{k^2}\)	\(\frac{\cos(kx)}{k^3}\)
\(e^{kx}\)	\(e^{kx}\)	\(\frac{e^{kx}}{k}\)	\(\frac{e^{kx}}{k^2}\)	\(\frac{e^{kx}}{k^3}\)

If \(k=1\), these fractions simplify. If \(k\neq1\), forgetting the \(\frac{1}{k}\) factors is the #1 mistake.

4) Worked example

Example: compute \(\int x^3\cos(x)\,dx\).

Build the DI table:

Row	D (differentiate)	I (integrate)	Sign	Diagonal product
0	\(x^3\)	\(\cos x\)	\(+\)	\(x^3\cdot \sin x\)
1	\(3x^2\)	\(\sin x\)	\(-\)	\(-3x^2\cdot (-\cos x)\)
2	\(6x\)	\(-\cos x\)	\(+\)	\(+6x\cdot (-\sin x)\)
3	\(6\)	\(-\sin x\)	\(-\)	\(-6\cdot (\cos x)\)
4	\(0\)	\(\cos x\)	stop	—

Now add the diagonal products:

\[ \int x^3\cos x\,dx = x^3\sin x \;-\;3x^2(-\cos x) \;+\;6x(-\sin x) \;-\;6(\cos x) \;+\;C \]

Clean it up:

\[ \int x^3\cos x\,dx = x^3\sin x +3x^2\cos x -6x\sin x -6\cos x +C \]

5) Definite integrals

In definite mode, once the tool builds an antiderivative \(F(x)\), it computes:

\[ \int_a^b f(x)\,dx = F(b)-F(a) \]

The graph shades the region between the curve \(y=f(x)\) and the \(x\)-axis from \(x=a\) to \(x=b\). If \(f(x)\) is negative on part of the interval, that portion is shaded as negative area (signed area).

If you swap the bounds (enter \(a>b\)), the value changes sign automatically: \(\int_a^b = -\int_b^a\).

6) Tips & pitfalls

Don’t forget \(k\): for \(\sin(kx)\), \(\cos(kx)\), \(e^{kx}\), each integration contributes a factor \(\frac{1}{k}\).
Best use-case: polynomial × (sin/cos/exp). The derivative column reaches 0 quickly.
Not ideal: products where the derivative column doesn’t terminate (like \(\ln x\) or \(\arctan x\)); then you typically use standard integration by parts (or special techniques).
Sign pattern matters: the alternating \((+,-,+,-,\dots)\) is essential. One sign error breaks the entire result.
Graph view: if you don’t see the curve, press Reset view or increase the half-width \(w\). Use zoom to inspect oscillations.

Frequently Asked Questions

What is the tabular method for integration?

The tabular method is a shortcut for repeated integration by parts. It organizes successive derivatives of one factor and successive integrals of the other factor in a table, then combines products with alternating signs.

When should I use tabular integration instead of standard integration by parts?

It is most useful when integration by parts must be applied multiple times, especially for a polynomial times an exponential, sine, or cosine. The polynomial eventually differentiates to 0, which makes the table terminate cleanly.

How does the sign pattern work in the DI table?

After choosing the factor to differentiate and the factor to integrate, the method uses alternating signs +, -, +, - down the diagonal products. These signed products are summed to form the main part of the result.

Can the tabular method be used for definite integrals?

Yes. After forming an antiderivative from the DI table, the definite value is computed by evaluating it at the bounds and taking F(b) - F(a).

What if neither factor becomes simpler when differentiated or integrated?

Tabular integration is not ideal in that case because the table may not terminate or may not simplify the remaining integral. A different technique such as substitution, partial fractions, or a trig identity approach may be more appropriate.

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