🔶 Integration by Parts — The Product Rule in Reverse

Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)
Published: October 2025
Category: Calculus | Integration Techniques

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🔹 What Is Integration by Parts?

When differentiation involves a product rule, integration involves its reverse — and that’s exactly what integration by parts is.

It allows us to integrate the product of two functions when direct integration isn’t straightforward.

If you know how to differentiate a product: ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}dxd​(uv)=udxdv​+vdxdu​

Then, rearranging gives the formula for integration by parts: ∫u dv=uv−∫v du\int u \, dv = uv – \int v \, du∫udv=uv−∫vdu

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🔹 The Formula

∫u dv=uv−∫v du\boxed{\int u \, dv = uv – \int v \, du}∫udv=uv−∫vdu​

Here:

• uuu: the function we differentiate
• dvdvdv: the function we integrate

Choosing which term is uuu and which is dvdvdv is the art of this technique.

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🔹 The LIATE Rule — How to Choose uuu

A simple memory rule helps you decide what to differentiate (uuu):

Priority	Function Type	Example
1	Logarithmic	ln⁡x\ln xlnx
2	Inverse Trig	tan⁡−1x\tan^{-1}xtan−1x
3	Algebraic	x2,x,3x3x^2, x, 3x^3×2,x,3×3
4	Trigonometric	sin⁡x,cos⁡x\sin x, \cos xsinx,cosx
5	Exponential	ex,axe^x, a^xex,ax

Choose uuu according to the earliest function type appearing in this list.

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🔹 Step-by-Step Example 1 — ∫xex dx\int x e^x \, dx∫xexdx

Following LIATE, u=xu = xu=x, dv=ex dxdv = e^x \, dxdv=exdx.

Differentiate and integrate: du=dx,v=exdu = dx, \quad v = e^xdu=dx,v=ex

Substitute into formula: ∫xex dx=xex−∫ex dx=ex(x−1)+C\int x e^x \, dx = x e^x – \int e^x \, dx = e^x(x – 1) + C∫xexdx=xex−∫exdx=ex(x−1)+C

✅ Result: ex(x−1)+Ce^x(x – 1) + Cex(x−1)+C

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🔹 Example 2 — ∫xcos⁡x dx\int x \cos x \, dx∫xcosxdx

Let u=xu = xu=x, dv=cos⁡x dxdv = \cos x \, dxdv=cosxdx du=dx,v=sin⁡xdu = dx, \quad v = \sin xdu=dx,v=sinx

Then: ∫xcos⁡x dx=xsin⁡x−∫sin⁡x dx=xsin⁡x+cos⁡x+C\int x \cos x \, dx = x \sin x – \int \sin x \, dx = x \sin x + \cos x + C∫xcosxdx=xsinx−∫sinxdx=xsinx+cosx+C

✅ Result: xsin⁡x+cos⁡x+Cx \sin x + \cos x + Cxsinx+cosx+C

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🔹 Example 3 — ∫ln⁡x dx\int \ln x \, dx∫lnxdx

Even though this looks like one function, treat it as 1⋅ln⁡x1 \cdot \ln x1⋅lnx.

Let u=ln⁡x, dv=dxu = \ln x, \ dv = dxu=lnx, dv=dx du=1xdx, v=xdu = \frac{1}{x} dx, \ v = xdu=x1​dx, v=x

Then: ∫ln⁡x dx=xln⁡x−∫x⋅1xdx=xln⁡x−x+C\int \ln x \, dx = x \ln x – \int x \cdot \frac{1}{x} dx = x \ln x – x + C∫lnxdx=xlnx−∫x⋅x1​dx=xlnx−x+C

✅ Result: x(ln⁡x−1)+Cx(\ln x – 1) + Cx(lnx−1)+C

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🔹 Example 4 — ∫exsin⁡x dx\int e^x \sin x \, dx∫exsinxdx (Repeating Integrals)

This integral requires repeated application of the formula.

Let u=sin⁡x, dv=exdxu = \sin x, \ dv = e^x dxu=sinx, dv=exdx. du=cos⁡xdx, v=exdu = \cos x dx, \ v = e^xdu=cosxdx, v=ex ∫exsin⁡xdx=exsin⁡x−∫excos⁡xdx\int e^x \sin x dx = e^x \sin x – \int e^x \cos x dx∫exsinxdx=exsinx−∫excosxdx

Now integrate ∫excos⁡xdx\int e^x \cos x dx∫excosxdx by parts again: =exsin⁡x−[excos⁡x−∫ex(−sin⁡x)dx]= e^x \sin x – [e^x \cos x – \int e^x (-\sin x) dx]=exsinx−[excosx−∫ex(−sinx)dx] =exsin⁡x−excos⁡x+∫exsin⁡xdx= e^x \sin x – e^x \cos x + \int e^x \sin x dx=exsinx−excosx+∫exsinxdx

Now notice — the original integral appears again: I=ex(sin⁡x−cos⁡x)+II = e^x (\sin x – \cos x) + II=ex(sinx−cosx)+I ⇒∫exsin⁡xdx=12ex(sin⁡x−cos⁡x)+C\Rightarrow \int e^x \sin x dx = \frac{1}{2} e^x (\sin x – \cos x) + C⇒∫exsinxdx=21​ex(sinx−cosx)+C

✅ Result: ex2(sin⁡x−cos⁡x)+C\frac{e^x}{2}(\sin x – \cos x) + C2ex​(sinx−cosx)+C

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🔹 Common Pitfalls

1. ❌ Choosing uuu and dvdvdv incorrectly — follow LIATE.
2. ❌ Forgetting the minus sign in the formula.
3. ❌ Not simplifying terms before applying the rule.
4. ❌ Forgetting constants of integration (especially in long solutions).

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🔹 Advanced Applications

Integration by Parts is the foundation for:

• Integrating products (e.g., xex,xsin⁡xx e^x, x \sin xxex,xsinx)
• Reduction formulas (IB/A Level advanced integration)
• Deriving Laplace transforms and Fourier series
• STEP & MAT integration challenges

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🌟 Why Integration by Parts Matters

It’s a simple formula that captures the duality between differentiation and integration — and teaches how structure in calculus can simplify complexity.

Every advanced integration method (from differential equations to series expansions) builds upon this idea.

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📘 Learn Beyond the Formula

At Math By Rishabh, students don’t just memorize formulas — they understand where they come from and how to adapt them.

In the Mathematics Elevate Mentorship, you’ll:
✅ Learn how to select uuu intelligently,
✅ Master multi-step integrations,
✅ Develop problem intuition for IB, AP, and STEP exams.

🚀 Transform your understanding of calculus.
👉 Book your personalized mentorship session now at MathByRishabh.com