Indefinite integration for NDA

Integration is differentiation run backwards, and that is exactly how NDA tests it. The official syllabus asks for integration as the inverse of differentiation, substitution, integration by parts, and standard integrals of algebraic, trigonometric, exponential and hyperbolic functions. Most questions give you an integrand and four answers. You rarely need to integrate from scratch: often the fastest route is to differentiate the options and see which one gives back the integrand.

The chapter is worth the effort because it feeds two others. Definite integrals, areas and differential equations all start with an antiderivative, so a weak integration table costs you marks in three places.

Typical question types are:

• a direct standard integral, sometimes dressed up with a constant or a shift (ax + b);
• a substitution hidden in plain sight, such as f′(x)/f(x);
• integration by parts, usually x eˣ, x sin x, x ln x or ln x alone;
• the eˣ[f(x) + f′(x)] pattern;
• forms like 1/(x² + a²) that give inverse trigonometric or log answers;
• "if ∫f(x) dx = g(x) + C, find f" questions, which are really differentiation.

The standard table (add + C to every answer)

Strictly, the log answers carry a modulus: ∫(1/x) dx = ln|x| + C. Options in NDA papers sometimes drop it; don't let that throw you.

Forms that give inverse trig and log answers (a > 0)

With a = 1 you get the three you already know: tan⁻¹x, sin⁻¹x and, for 1/(x√(x² − 1)), sec⁻¹x.

Substitution

Substitution works when the integrand contains a function and (a constant times) its derivative. Put u = the inner function, replace du, integrate in u, then put x back.

Take ∫2x cos(x²) dx. With u = x², du = 2x dx, and the integral becomes ∫cos u du = sin u = sin(x²) + C.

Two patterns are so common that you should recognise them on sight:

• ∫f′(x)/f(x) dx = ln|f(x)| + C. If the numerator is the derivative of the denominator, the answer is a log. This is why ∫tan x dx = ∫sin x/cos x dx = −ln|cos x| = ln|sec x|.
• ∫[f(x)]ⁿ f′(x) dx = [f(x)]ⁿ⁺¹/(n + 1) + C for n ≠ −1. For example, ∫sin²x cos x dx = sin³x/3 + C.

Trigonometric powers

For sin²x and cos²x, lower the power first with the double-angle identities: sin²x = (1 − cos 2x)/2 and cos²x = (1 + cos 2x)/2. For odd powers such as cos³x sin x, substitute u = cos x.

Integration by parts

∫u dv = uv − ∫v du. In words: first function × integral of the second, minus the integral of (derivative of the first × integral of the second).

To choose the first function (the one you differentiate), use the order ILATE: Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential. Whichever comes earlier is u.

• ∫x eˣ dx: u = x, so x eˣ − ∫eˣ dx = eˣ(x − 1) + C.
• ∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x + C.
• ∫ln x dx: treat it as ∫(ln x)(1) dx, so x ln x − ∫x(1/x) dx = x ln x − x + C.

The eˣ shortcut. ∫eˣ[f(x) + f′(x)] dx = eˣ f(x) + C. It comes from parts, and it turns a two-step question into one line. ∫eˣ(sin x + cos x) dx = eˣ sin x + C, because cos x is the derivative of sin x.

Partial fractions (the simple cases)

A rational function with a factorised denominator splits into simpler fractions:

• 1/[x(x + 1)] = 1/x − 1/(x + 1), so the integral is ln|x| − ln|x + 1| = ln|x/(x + 1)| + C.
• 1/[(x − a)(x − b)] = [1/(a − b)] × [1/(x − a) − 1/(x − b)] when a ≠ b.

If the numerator's degree is not lower than the denominator's, divide first.

Worked NDA-style MCQs

Q1. ∫(2x + 3)/(x² + 3x + 5) dx equals:
(a) ln|x² + 3x + 5| + C (b) (2x + 3) ln x + C (c) 1/(x² + 3x + 5) + C (d) tan⁻¹(x + 3) + C

The numerator is exactly the derivative of the denominator, so the f′/f rule applies. Answer: (a).

Q2. ∫eˣ(1/x − 1/x²) dx equals:
(a) eˣ/x² + C (b) eˣ/x + C (c) eˣ ln x + C (d) −eˣ/x + C

With f(x) = 1/x, f′(x) = −1/x², so the integrand is eˣ[f + f′]. The answer is eˣ f(x) = eˣ/x + C. Answer: (b).

Q3. ∫x ln x dx equals:
(a) x ln x − x + C (b) (x²/2) ln x + C (c) (x²/2) ln x − x²/4 + C (d) x² ln x − x²/2 + C

By ILATE, the log is u and x is dv. So (x²/2) ln x − ∫(x²/2)(1/x) dx = (x²/2) ln x − x²/4 + C. Option (b) forgets the second term. Answer: (c).

Q4. ∫dx/(x² + 4) equals:
(a) tan⁻¹(x/2) + C (b) 2 tan⁻¹(x/2) + C (c) (1/2) tan⁻¹(x/2) + C (d) (1/4) tan⁻¹x + C

Here a = 2, so (1/a) tan⁻¹(x/a) = (1/2) tan⁻¹(x/2). Answer: (c).

Q5. ∫sin²x dx equals:
(a) sin³x/3 + C (b) x/2 − (sin 2x)/4 + C (c) x/2 + (sin 2x)/4 + C (d) −cos²x + C

Write sin²x = (1 − cos 2x)/2 and integrate: x/2 − (sin 2x)/4. Option (c) is ∫cos²x. Answer: (b).

Q6. ∫dx/(eˣ + e⁻ˣ) equals:
(a) ln(eˣ + e⁻ˣ) + C (b) tan⁻¹(eˣ) + C (c) sin⁻¹(eˣ) + C (d) eˣ − e⁻ˣ + C

Multiply top and bottom by eˣ: ∫eˣ dx/(e²ˣ + 1). Put u = eˣ, du = eˣ dx, giving ∫du/(u² + 1) = tan⁻¹u. Answer: (b).

Common mistakes

• Dropping the + C. Options without it can still be right, but "if F(0) = 2" questions need it.
• Not dividing by the inner coefficient in ∫f(ax + b) dx.
• Sign errors with sin and cos. ∫sin x dx = −cos x; the derivative of cos x is −sin x. Say it aloud once.
• Applying the power rule to 1/x. n = −1 is the one exception, and its answer is ln|x|.
• Picking the wrong u in by parts. Differentiating eˣ or integrating ln x first makes the problem harder, not easier.

Practice set

1. ∫sec x (sec x + tan x) dx = (a) tan x + sec x + C (b) sec²x + C (c) tan x − sec x + C (d) ln|sec x| + C
2. ∫cos³x sin x dx = (a) cos⁴x/4 + C (b) −cos⁴x/4 + C (c) sin⁴x/4 + C (d) −sin⁴x/4 + C
3. ∫dx/[x(x + 1)] = (a) ln|x(x + 1)| + C (b) ln|(x + 1)/x| + C (c) ln|x/(x + 1)| + C (d) 1/(x + 1) + C
4. ∫x eˣ dx = (a) x eˣ + C (b) eˣ(x + 1) + C (c) eˣ(x − 1) + C (d) x²eˣ/2 + C
5. ∫ln x dx = (a) 1/x + C (b) x ln x + C (c) x ln x − x + C (d) x ln x + x + C
6. ∫tan x dx = (a) sec²x + C (b) ln|sec x| + C (c) ln|cos x| + C (d) −ln|sec x| + C
7. ∫dx/√(9 − x²) = (a) sin⁻¹(x/3) + C (b) (1/3) sin⁻¹(x/3) + C (c) 3 sin⁻¹x + C (d) cos⁻¹(3x) + C
8. ∫eˣ(sin x + cos x) dx = (a) eˣ cos x + C (b) eˣ sin x + C (c) −eˣ cos x + C (d) eˣ(sin x − cos x) + C

Answers:

1. (a). Expand: sec²x + sec x tan x, which integrate to tan x and sec x.
2. (b). Put u = cos x, du = −sin x dx: −∫u³ du = −u⁴/4.
3. (c). Partial fractions 1/x − 1/(x + 1).
4. (c). By parts with u = x.
5. (c). By parts with u = ln x and dv = dx.
6. (b). −ln|cos x| = ln|sec x|; option (c) has the wrong sign.
7. (a). a = 3; this form has no 1/a factor, unlike the tan⁻¹ form.
8. (b). eˣ[f + f′] with f = sin x.

What to do next

• Write both tables from memory, then check them against this page. Repeat until they come out clean twice in a row.
• Differentiate every answer in the practice set to confirm it. This habit is the fastest check in the exam hall.
• Move on to definite integrals and area, where these antiderivatives meet limits and properties.
• If any derivative in this page slowed you down, revise differentiation first.