Integration by Parts

How do you integrate x\cos x? Substitution does not help — there is no inner function whose derivative is sitting there waiting to be absorbed. You cannot split it into two simpler integrals, because x and \cos x are not glued by addition, they are glued by multiplication.

What you need is a rule for integrating a product. One exists. It is called integration by parts, and it comes from running the product rule for derivatives backward.

The product rule says that for any two differentiable functions u and v,

Integrate both sides with respect to x. The left side integrates trivially — the integral of a derivative is the function itself:

Now solve for one of the integrals on the right. Say you want \int uv'\,dx. Move \int u'v\,dx to the other side:

That is the integration-by-parts formula. It says: if you can identify one factor of your integrand as u and another as v', then you can trade \int uv'\,dx for \int u'v\,dx. You win if the new integral is easier than the old one.

The formula, labelled

Let me rewrite the formula with explicit labels, since the bookkeeping is the part that trips students up.

Reading the formula: you start with a product u \cdot v' under an integral sign. You differentiate u (to get u') and integrate v' (to get v). The new integral has u' \cdot v in it. If u' is simpler than u, you win. If it is not, you picked wrong — swap the roles and try again.

Let me run it on the first problem. Take \int x\cos x\,dx. Identify the two factors: x and \cos x. You have to choose which one is u and which one is dv.

Try 1. Let u = x, dv = \cos x\,dx. Then du = dx and v = \sin x. Plug into the formula:

That worked. The integral on the right, \int \sin x\,dx, is trivial because differentiating u = x collapsed it to du = dx, removing the polynomial factor entirely.

Try 2. What if you had made the opposite choice? Let u = \cos x, dv = x\,dx. Then du = -\sin x\,dx and v = x^2/2. The formula gives

Worse. The new integral has x^2 in it instead of x. You made the polynomial bigger instead of smaller. The technique is still valid — the formula is still true — but you picked the wrong u, so the new integral is harder than the old one.

The lesson: the choice of u matters. You want the factor that gets simpler when differentiated to be u, and the factor that integrates cleanly to be dv.

The ILATE rule

There is a mnemonic that tells you which factor to pick as u in almost every textbook problem. It is called ILATE (or LIATE in some books — same rule, letters reordered). It gives a priority order:

• I — Inverse trigonometric functions (\sin^{-1}, \cos^{-1}, \tan^{-1}, \ldots)
• L — Logarithmic functions (\ln x, \log x)
• A — Algebraic functions (polynomials like x^n)
• T — Trigonometric functions (\sin x, \cos x, \tan x, \ldots)
• E — Exponential functions (e^x, a^x)

Rule: pick u to be the factor highest on the ILATE list; the other factor (plus dx) becomes dv.

So for \int x\cos x\,dx: x is algebraic (A), \cos x is trigonometric (T). A is higher on the list than T, so u = x. That is the choice that worked above.

For \int x^2 e^x\,dx: x^2 is algebraic (A), e^x is exponential (E). A beats E, so u = x^2. You will see this one in Example 1.

For \int x\ln x\,dx: \ln x is logarithmic (L), x is algebraic (A). L beats A, so u = \ln x. You will see this one in Example 2.

For \int \tan^{-1}(x)\,dx: there seems to be only one factor, not a product. Trick: multiply mentally by 1 and call the other factor dv = 1\,dx, so v = x. Then u = \tan^{-1}(x) — I is the top of the ILATE list, so it is u. This is a standard application called integration by parts with a missing factor, and it is the only way to integrate \sin^{-1}, \cos^{-1}, \tan^{-1}, and \ln x by themselves.

Why does ILATE work? Because differentiating factors near the top of the list simplifies them sharply (inverse trig and log become algebraic; polynomials lose a degree), while integrating factors near the bottom is easy (exponentials and basic trig integrate to themselves up to signs). You want differentiation to do the hard work and integration to do the easy work.

The first worked example

Here is the full version, with all the bookkeeping on display.

Check by differentiating the answer:

Back to the integrand. The two applications of by parts were the right move, and the ILATE choice was the right choice at each step.

The second worked example

The second example brings in a different flavour — a logarithm, which is what ILATE flags as the u candidate.

Sanity check by differentiating: \frac{d}{dx}\left[\frac{x^2 \ln x}{2}\right] = x\ln x + \frac{x^2}{2}\cdot\frac{1}{x} = x\ln x + \frac{x}{2}, and \frac{d}{dx}\left[-\frac{x^2}{4}\right] = -\frac{x}{2}. The two \frac{x}{2} pieces cancel, leaving x\ln x. Correct.

The "missing factor" trick

Some integrands have only one obvious factor. The trick is to multiply by 1 and use that 1 as dv.

Integrating \ln x by itself. Take \int \ln x\,dx. There is only one factor — \ln x — but you can write the integrand as \ln x \cdot 1 and apply by parts with u = \ln x and dv = 1\,dx:

Three lines of work, a standard result. This is the only way to handle \int \ln x\,dx — you cannot guess an antiderivative, and substitution will not help.

Integrating \tan^{-1}(x) by itself. Same trick: u = \tan^{-1}(x), dv = dx, so du = \frac{1}{1 + x^2}\,dx and v = x:

The second integral is a substitution — set w = 1 + x^2, so dw = 2x\,dx — and you get \frac{1}{2}\ln(1 + x^2). Final answer:

Every inverse trigonometric function and the logarithm are integrated this way. Without the missing-factor trick, you would not have anywhere to start.

Repeated application — the polynomial-times-exponential family

When the integrand is x^n \cdot e^{ax} (or x^n \cdot \sin(ax), or x^n \cdot \cos(ax)), each application of by parts knocks one power off the polynomial. After n applications, the polynomial is gone and you have an integral you can finish. This is why the ILATE rule always puts A above T and E: the A-factor is the one that eventually vanishes.

There is even a shortcut called tabular integration that bookkeepers use for this family. You write two columns — one that you keep differentiating (u and its derivatives), the other that you keep integrating (dv and its antiderivatives) — and you pair up diagonals with alternating signs. For \int x^3 e^x\,dx:

Multiply each row across and apply the sign, then add them up:

One step, no recursion. This only works when the u-column terminates at zero (which happens precisely for polynomials). When it doesn't — say, for e^x \sin x — you need the cyclic trick from the next article.

Common confusions

• "I always pick the simpler factor as u." The opposite. You pick the factor you want to simplify — the one that gets easier when differentiated. A log or inverse trig is hard-looking on its own but simple after you differentiate it. A polynomial is simple on its own but still a polynomial after integration. Pick the hard one as u.

• "The formula has a +, not a -." The formula is \int u\,dv = uv - \int v\,du — with a minus sign. Getting the sign wrong is the single most common way to botch an application of by parts.

• "I need to add +C after finding v." No. Any constant you add to v in the middle of the computation cancels out. Wait until the final answer to tack on +C.

• "I can only use by parts when the integrand is a product of exactly two factors." You can also use it when it looks like one factor — the missing-factor trick above. And you can use it on an integrand that doesn't look like a product by strategically splitting it: \int \sin^{-1}(x)\,dx = \int 1\cdot\sin^{-1}(x)\,dx and so on.

• "After one application of by parts, if the integral is not trivial, I picked wrong." Sometimes you need two or three applications. If each application is reducing the integrand's complexity in a measurable way, you are on the right track. If an application makes the integrand more complicated, that is when you know you picked wrong.

Going deeper

By parts for definite integrals

The by-parts formula has a definite-integral version that saves you from writing back-substitutions. If the limits are a and b,

where [uv]_a^b means u(b)v(b) - u(a)v(a). As an example, \int_0^1 x e^x\,dx. Take u = x, dv = e^x\,dx, so du = dx and v = e^x:

Clean.

The cyclic trick — when the integral reappears

The mnemonic says a polynomial factor eventually dies off. But some by-parts chains never terminate — they loop. The classic example is \int e^x \sin x\,dx. Let u = \sin x, dv = e^x\,dx:

The new integral \int e^x \cos x\,dx is a new problem. Apply by parts again, u = \cos x, dv = e^x\,dx:

Now substitute that back into the first equation:

The original integral has reappeared on the right. But now you can treat it as an algebraic unknown and solve:

This trick — pushing through by parts twice until the integral returns, then solving algebraically — is the whole content of the next article. It is how you handle e^{ax}\sin(bx) and e^{ax}\cos(bx) in general.

When ILATE lies

The rule works for almost every problem you will meet in JEE. It fails occasionally on exotic integrands, usually when both factors are from the same family (two algebraic factors, say). In those cases you have to think about which factor integrates more pleasantly as dv and just try it.

A small example: \int x^3 / \sqrt{1 - x^2}\,dx. Both x^3 and \frac{1}{\sqrt{1-x^2}} are algebraic, so ILATE gives no guidance. Here the right move is substitution (u = 1 - x^2), not by parts at all. By parts is not always the answer — it is one technique among several, and picking the right technique is a skill that comes with practice.

Where this leads next

• Integration by Parts - Special Forms — the e^{ax}\sin(bx) and e^{ax}\cos(bx) cyclic trick, the e^x(f + f') shortcut, and reduction formulas.
• Special Integrals - Part 3 — the three \sqrt{\text{quadratic}} formulas, which are by-parts derivations in disguise.
• Integration by Substitution — the other half of your integration toolbox; when in doubt, try substitution first.
• Definite Integration Techniques — how by parts interacts with limits of integration and why the [uv] bracket makes life easier.
• Reduction Formulas — using by parts recursively to integrate \sin^n x and \cos^n x for arbitrary n.