Integration by parts

What is integration by parts?

In plain words: the integral of a product is (first function) × (integral of the second), minus the integral of (derivative of the first) × (integral of the second). NCERT derives it on pages 259–260 by integrating the product rule d/dx (uv) = u dv/dx + v du/dx and rearranging. Integration by parts trades one integral for another. It helps when differentiating the first function makes it simpler (a power of x drops a degree, log x becomes 1/x) while the second function is easy to integrate.

Choosing the first function

NCERT Example 17 (page 260): for ∫x cos x dx, take f(x) = x and g(x) = cos x. Then

∫x cos x dx = x sin x − ∫sin x dx

= x sin x + cos x + C.

With the choice the other way round, the new integral is ∫sin x × (x²/2) dx, which is harder: the choice matters. NCERT’s Remarks (pages 260–261):

• Take a power of x or a polynomial as the first function, unless the other function is an inverse trigonometric or a logarithmic function; then that one is first.
• No constant is needed when integrating the second function: it cancels out in the end.

Beyond the NCERT text, a common classroom mnemonic for the order is ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential): the earlier one is usually the first function. It is a guide, not a rule.

When there is no second function, use 1. NCERT Example 18 (page 261): ∫log x dx = log x × x − ∫(1/x) × x dx = x log x − x + C. NCERT Example 19: ∫xe^x dx = xe^x − e^x + C. NCERT Example 20 (pages 261–262) takes sin^⁻¹ x first in ∫x sin^⁻¹ x/√(1 − x²) dx; the integral of the second function is −√(1 − x²), and the answer is x − √(1 − x²) sin^⁻¹ x + C.

When the integral comes back. NCERT Example 21 (page 262): for I = ∫e^x sin x dx, two integrations by parts give I = −e^x cos x + e^x sin x − I. Solving for I:

∫e^x sin x dx = (e^x/2)(sin x − cos x) + C.

NCERT slip (page 260): Remark (i) says the method does not work for ∫√x sin x dx because “there does not exist any function whose derivative is √x sin x”. An anti derivative does exist (the area function of NCERT Section 7.8 (page 267) gives one for any continuous function); the elementary integration table is insufficient here. In fact, parts followed by u = √x gives −√x cos x + ∫cos (u^²) du, a Fresnel-type integral beyond this chapter. Existence of an anti derivative and an elementary formula for it are different questions; NCERT itself notes on page 234 that some anti derivatives are not expressible in terms of elementary functions. In an exam, say that this integral cannot be found in terms of elementary functions.

The type ∫e^ˣ[f(x) + f′(x)] dx

NCERT Section 7.6.1 (pages 262–263) integrates e^xf(x) by parts; the term ∫f′(x)e^x dx that appears cancels the second half of the integrand:

So look for a function and its derivative inside the bracket. NCERT Example 22 (page 263): in ∫e^x(tan^⁻¹ x + 1/(1 + x²)) dx, f(x) = tan^⁻¹ x, so the integral is e^x tan^⁻¹ x + C. In ∫(x² + 1)e^x/(x + 1)² dx, write x² + 1 = (x² − 1) + 2: the bracket is (x − 1)/(x + 1) + 2/(x + 1)², a function and its derivative, so the integral is e^x(x − 1)/(x + 1) + C.

Three integrals with a square root

Take a > 0 and work on intervals where the roots are real: |x| > a for √(x^² − a^²), |x| < a for √(a^² − x^²), and all real x for √(x^² + a^²). Integrating by parts with 1 as the second function gives three more standard results (NCERT Section 7.6.2, pages 264–265):

For the first, NCERT writes I = x√(x² − a²) − ∫x²/√(x² − a²) dx, splits x² = (x² − a²) + a², and finds I on the right again, so 2I equals x√(x² − a²) minus a² log |x + √(x² − a²)|, from NCERT 7.4 (4), page 244.

A quadratic under the root is first written as a perfect square plus or minus a constant. NCERT Example 23 (page 265): x² + 2x + 5 = (x + 1)² + 2², so

∫√(x² + 2x + 5) dx = ½(x + 1)√(x² + 2x + 5)

+ 2 log |x + 1 + √(x² + 2x + 5)| + C.

NCERT Example 24 (pages 265–266): 3 − 2x − x² = 4 − (x + 1)², so the integral of its square root is ½(x + 1)√(3 − 2x − x²) + 2 sin^⁻¹ ((x + 1)/2) + C.

Solved examples

Common mistakes to avoid

Quick revision summary

• ∫f g dx = f ∫g dx − ∫(f′ ∫g dx) dx: choose as f the function that becomes simpler when differentiated; with no second function, use 1.
• If the integral reappears, as in ∫e^x sin x dx, collect it on one side and solve.
• ∫e^x[f(x) + f′(x)] dx = e^xf(x) + C.
• The three square-root integrals come from parts with 1 as the second function; a quadratic under the root is first completed to a square.

Questions students ask

ILATE or LIATE: which is correct for integration by parts?

ILATE and LIATE are both fine guides to the first function in integration by parts: they differ only in the order of inverse trigonometric and logarithmic functions, which rarely meet in one integrand. Neither is NCERT’s, whose Remark (iii) (page 261) only says when a power of x comes first, and neither is absolute: the second function must be one you can integrate. In ∫x³e^x2 dx, ILATE’s x³ first leaves e^x2, which no method of the chapter integrates; x² first and xe^x2 second gives ½e^x2(x² − 1) + C.

What is the tabular (DI) method of integration by parts, and is it Bernoulli’s formula?

The tabular or DI method is integration by parts repeated in one line; as a formula it is what some Indian textbooks call Bernoulli’s formula (beyond the NCERT text, reasoned by repeating parts): ∫uv dx is uv₁ − u′v₂ + u″v₃ − …, with v₁, v₂, v₃ the first, second and third integrals of v, stopping when a derivative of u is 0. With u = x³ and v = e^x, ∫x³e^x dx is e^x times (x³ − 3x² + 6x − 6), plus C. Terminating this way needs a polynomial u, whose derivatives run out. More generally the table can stop with a remainder integral, or reveal a cycle: with e^x sin x, solve for the recurring integral instead of waiting for a zero derivative.

What is the difference between integration by parts and the product rule?

Integration by parts is the product rule integrated: from (uv)′ = u′v + uv′, integrating and rearranging gives ∫uv′ dx = uv − ∫u′v dx, which is NCERT’s formula with u as the first function. The product rule always finishes a derivative, but integration by parts only trades one integral for another, so it helps only when ∫u′v dx is easier than the integral you started with. The product rule is also the quickest check of an answer: for ∫xe^−x dx, whose answer is −(x + 1)e^−x + C, differentiating gives −e^−x + (x + 1)e^−x, which is xe^−x.

Does it matter which function is taken first in ∫e^x sin x dx?

In ∫e^x sin x dx either function can be taken first, since both are easy to integrate and neither gets simpler when differentiated, but the second integration by parts must keep the same kind of function first. With e^x first, I = −e^x cos x + J, where J = ∫e^x cos x dx; if cos x is now taken first, J comes out as e^x cos x + I, and everything cancels to I = I, which says nothing. Keeping the same kind first both times brings back −I, so that I can be solved for, as in NCERT Example 21 (page 262); the same steps give ∫e^x cos x dx as (e^x/2)(sin x + cos x), plus C.

Is there an e^kx version of the ∫e^x[f + f′] dx rule?

The e^x[f + f′] rule has a version for any constant k (beyond the NCERT text, reasoned from the product rule): the derivative of e^kxf(x) is e^kx[kf(x) + f′(x)], so the integral of e^kx[kf(x) + f′(x)] is e^kxf(x) + C, and k = 1 is NCERT’s result. For ∫e^3x(3x² + 2x) dx, f(x) = x² fits, so the answer is x²e^3x + C. With k = −1 the bracket is f′(x) − f(x), as in ∫e^−x(cos x − sin x) dx, which is e^−x sin x + C.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find ∫x sin x dx and ∫x sin 3x dx (NCERT Exercise 7.6, Q1 and Q2, page 263).

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   −x cos x + ∫cos x dx = −x cos x + sin x + C. And −(x/3) cos 3x + (1/3)∫cos 3x dx = −(x/3) cos 3x + (1/9) sin 3x + C.

2. Q2. Find ∫x log x dx and ∫x² log x dx (NCERT Exercise 7.6, Q4 and Q6, page 263).

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   With log x first: (x²/2) log x − ∫(x/2) dx = (x²/2) log x − x²/4 + C, and (x³/3) log x − ∫(x²/3) dx = (x³/3) log x − x³/9 + C.

3. Q3. Find ∫e^x(sin x + cos x) dx and ∫xe^x/(1 + x)² dx (NCERT Exercise 7.6, Q16 and Q17, page 264).

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   The first is e^x[f + f′] with f = sin x: e^x sin x + C. For the second, x/(1 + x)² = 1/(1 + x) − 1/(1 + x)², which is f + f′ with f = 1/(1 + x): e^x/(1 + x) + C.

4. Q4. Find ∫√(4 − x²) dx (NCERT Exercise 7.7, Q1, page 266).

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   With a = 2 in the third square-root formula: (x/2)√(4 − x²) + 2 sin^⁻¹ (x/2) + C.