Integration by Parts: The "DI" Method (Tabular Integration)

🎯 The Strategy

When you need to apply Integration by Parts (IBP) multiple times, the standard $u \cdot v$ formula becomes tedious and error-prone. The DI Method (also called Tabular Integration) provides a systematic, fast approach.

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📐 The Standard IBP Formula

$\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx$

Or in the LIATE form: $\int u \, dv = uv - \int v \, du$

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🔄 The DI Method Setup

Create a table with three columns:

Rules:

1. D column: Keep differentiating until you get 0
2. I column: Keep integrating
3. Sign column: Alternates starting with +
4. Multiply diagonally and add all terms

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📝 JEE Main Previous Year Examples

Example 1: JEE Main 2025

Question: Let $\int x^3 \sin x \, dx = g(x) + C$, where $C$ is the constant of integration. If $8\left(g\left(\frac{\pi}{2}\right) + g'\left(\frac{\pi}{2}\right)\right) = \alpha\pi^3 + \beta\pi^2 + \gamma$, where $\alpha, \beta, \gamma \in \mathbb{Z}$, then $\alpha + \beta - \gamma$ equals:

Solution using DI Method:

Result: Multiply diagonally and add: $g(x) = x^3(-\cos x) - 3x^2(-\sin x) + 6x(\cos x) - 6(\sin x) + C$ $= -x^3\cos x + 3x^2\sin x + 6x\cos x - 6\sin x + C$

Now calculate:

• $g\left(\frac{\pi}{2}\right) = -\frac{\pi^3}{8}(0) + 3\cdot\frac{\pi^2}{4}(1) + 6\cdot\frac{\pi}{2}(0) - 6(1) = \frac{3\pi^2}{4} - 6$

• $g'(x) = -3x^2\cos x + x^3\sin x + 6x\sin x + 3x^2\cos x + 6\cos x - 6x\sin x - 6\cos x$ $= x^3\sin x$ (after simplification)

• $g'\left(\frac{\pi}{2}\right) = \frac{\pi^3}{8}(1) = \frac{\pi^3}{8}$

So: $8\left(g\left(\frac{\pi}{2}\right) + g'\left(\frac{\pi}{2}\right)\right) = 8\left(\frac{3\pi^2}{4} - 6 + \frac{\pi^3}{8}\right)$ $= 6\pi^2 - 48 + \pi^3 = \pi^3 + 6\pi^2 - 48$

Therefore: $\alpha = 1$, $\beta = 6$, $\gamma = -48$

Answer: $\alpha + \beta - \gamma = 1 + 6 - (-48) = 55$

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Example 2: JEE Main 2020

Question: Evaluate $\int x^2 \sin 2x \, dx$

Solution using DI Method:

Result: $\int x^2 \sin 2x \, dx = x^2\left(-\frac{1}{2}\cos 2x\right) - 2x\left(-\frac{1}{4}\sin 2x\right) + 2\left(\frac{1}{8}\cos 2x\right) + C$

$= -\frac{x^2}{2}\cos 2x + \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x + C$

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Example 3: JEE Main 2018

Question: $\int x \sin^2 x \, dx = ?$

Solution:

First, use identity: $\sin^2 x = \frac{1 - \cos 2x}{2}$

$\int x \sin^2 x \, dx = \int x \cdot \frac{1 - \cos 2x}{2} \, dx = \frac{1}{2}\int x \, dx - \frac{1}{2}\int x \cos 2x \, dx$

For the second integral, use DI Method:

$\int x \cos 2x \, dx = \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x + C$

Final Answer: $\int x \sin^2 x \, dx = \frac{x^2}{4} - \frac{1}{2}\left(\frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x\right) + C$ $= \frac{x^2}{4} - \frac{x}{4}\sin 2x - \frac{1}{8}\cos 2x + C$

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Example 4: JEE Main 2019

Question: $\int (\log x)^2 \, dx = ?$

Solution using DI Method:

Here $u = (\log x)^2$ and $v = 1$

This doesn't terminate nicely with the table. Better to use standard IBP:

$\int (\log x)^2 dx = x(\log x)^2 - \int x \cdot \frac{2\log x}{x} dx$ $= x(\log x)^2 - 2\int \log x \, dx$

For $\int \log x \, dx$, use DI:

$\int \log x \, dx = x\log x - \int x \cdot \frac{1}{x} dx = x\log x - x + C$

Final Answer: $\int (\log x)^2 dx = x(\log x)^2 - 2(x\log x - x) + C$ $= x(\log x)^2 - 2x\log x + 2x + C$ $= x[(\log x)^2 - 2\log x + 2] + C$

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Example 5: JEE Main 2017

Question: $\int e^x \sin x \, dx$

Solution:

This is a cyclic case - differentiation doesn't terminate!

Let $I = \int e^x \sin x \, dx$

From the table: $I = e^x(-\cos x) - e^x(-\sin x) + \int e^x(-\sin x) dx$ $I = -e^x\cos x + e^x\sin x - I$ $2I = e^x(\sin x - \cos x)$ $I = \frac{e^x(\sin x - \cos x)}{2} + C$

Alternative form: $I = \frac{e^x}{\sqrt{2}}\sin\left(x - \frac{\pi}{4}\right) + C$

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🔑 LIATE Rule for Choosing u

When deciding which function to differentiate (D) and which to integrate (I), use LIATE:

Choose the function higher in the list as 'u' (to differentiate).

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⚠️ Special Cases

1. Cyclic Integrals

When the original integral reappears: $\int e^x \sin x \, dx, \quad \int e^x \cos x \, dx$

Strategy: Let $I$ = integral, solve the equation for $I$

2. Reduction Formulas

For $\int \sin^n x \, dx$ or $\int x^n e^x \, dx$, the table helps derive reduction formulas.

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⏱️ Time Comparison

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📌 Practice Problems

1. $\int x^4 e^x \, dx$

2. $\int x^2 \cos 3x \, dx$

3. $\int x^3 \ln x \, dx$

4. $\int e^{2x} \cos 3x \, dx$ (Cyclic case)

5. $\int x^2 \tan^{-1} x \, dx$

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💡 Pro Tips

1. Always differentiate the polynomial (it eventually becomes 0)
2. For log functions, put them in the D column (LIATE rule)
3. For cyclic integrals, stop after two rows and solve algebraically
4. Check signs carefully - alternating +, -, +, -, ...
5. Verify by differentiation if time permits