Integral of sin x Explained — Derivation, Examples & Quizzes

Q: Why is there a minus sign in ∫ sin x dx?

Because d/dx[cos x] = −sin x. To get +sin x back, the antiderivative must be −cos x, so the two minus signs cancel.

Q: How do I remember the sign?

Say 'sine to negative cosine.' The derivative goes sine → cosine (no minus); the integral goes sine → −cosine (minus). The minus belongs to the integral direction.

Where it comes from

The problem is the reverse of d/dx [cos x] = −sin x. The derivative of cosine is negative sine — so to get positive sine back, the antiderivative must carry a minus sign of its own.

The tempting sign-drop:

Before reading on: d/dx [cos x] = −sin x. The naive guess cos x + C differentiates to the negative of what we want. What is the smallest possible fix — and why does it work?

\int sin x dx = cos x + C ?? the tempting — and wrong — guess

Kill it by differentiating the claimed answer: d/dx [cos x + C] = −sin x ≠ sin x. The guess produces the negative of what we want. The repair is a single minus sign:

d/dx [−cos x]

=

−(−sin x) = sin x

The derivative contributes one minus; the leading minus contributes another. Two negatives make the positive sine. That is the entire reason for the sign.

Intuition for the shape: sin x is positive on (0, π), so its antiderivative must be increasing there — and −cos x indeed climbs from −1 to 1 on that interval. The sign is not a bookkeeping accident; it is the geometry talking.

One anchor to memorize: the area under a single hump of sine is exactly 2:

\int 0 π sin x dx = [-cos x] 0 π = (-cos π) - (-cos 0) = 1 + 1 = 2 a clean number worth knowing cold Say it: the integral from zero to pi of sine x d x equals two

Derivation

Guess-and-verify once more: propose F(x) = −cos x and differentiate. The double negative does all the work.

F(x)

=

−cos x

Step 1 — the guess. Since d/dx [cos x] = −sin x, negating the cosine should flip the sign of the result.

F′(x)

=

− · d/dx [cos x]

Step 2 — constant multiple. The leading minus factors out of the derivative.

=

−(−sin x) = sin x

Step 3 — the double negative. The derivative’s minus meets the leading minus: (−1)·(−1) = +1. So F′ = sin x, and ∫ sin x dx = −cos x + C. ∎

Why +C is the whole story: any two antiderivatives of sin x differ by a constant (zero derivative ⇒ constant), so −cos x + C captures every one of them.

How to use it

Before reading on: the sine hump from 0 to π is symmetric with peak 1. Before the reveal: is its area closer to 1, 2, or 3? Commit to a number, then see how the integral settles it.

The procedure, every time:

Confirm the integrand is sin x alone (times constants). If the argument is 3x or x², the chain rule left a fingerprint — see below.
Write −cos x + C. Minus sign first, then cosine.
Constants factor out: ∫ 5 sin x dx = −5 cos x + C.
Check by differentiating: does −(−sin x) give back sin x? If yes, done.

The chain-rule fingerprint

∫ sin(4x) dx is not −cos(4x) + C: differentiating −cos(4x) gives 4 sin(4x) — four times too big. Divide by 4: ∫ sin(4x) dx = −cos(4x)/4 + C. This is u-substitution in disguise (u = 4x).

Common mistake: dropping the minus: ∫ sin x dx = cos x + C. Differentiate to check: −sin x ≠ sin x. Say it while you write: “sine to negative cosine.”

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: ∫ sin x dx

Recognize the pattern. Integrand is sin x with argument exactly x.
Write the answer: −cos x + C. (Why the minus? d/dx [cos x] = −sin x, so the antiderivative needs its own minus to flip it back.)
Check: d/dx [−cos x] = −(−sin x) = sin x. Matches ✓

Your turn: Compute ∫ 3 sin x dx.

Answer: −3 cos x + C

Factor the 3: 3·∫ sin x dx = −3 cos x + C. Check: d/dx [−3 cos x] = −3(−sin x) = 3 sin x. ✓

Example 2 — definite: ∫₀^π/2 sin x dx

Antiderivative: −cos x.
Evaluate: (−cos(π/2)) − (−cos 0) = 0 − (−1) = 1.
Sanity check: a quarter-hump of sine — the full hump has area 2, and this is the left half of it, but the hump is symmetric, so half of 2 is 1. ✓

Common mistake: (−cos(π/2)) − (−cos 0) = 0 − 1 = −1 — dropping the inner minus on −cos 0 = −1. Write the parentheses every time.

Your turn: Compute ∫_π/2^π sin x dx.

Answer: 1

Antiderivative −cos x: (−cos π) − (−cos(π/2)) = 1 − 0 = 1. Sanity: the right half of the symmetric hump of total area 2. ✓

Example 3 — term by term: ∫ (sin x + cos x) dx

Split: ∫sin x dx + ∫cos x dx.
First piece: −cos x. Second piece: sin x (the cosine integral — no minus there).
Combine: −cos x + sin x + C.
Check: d/dx gives sin x + cos x. Matches ✓

Common mistake: giving both terms a minus (−cos x − sin x). The minus belongs to sine’s integral only — say “sine to negative cosine; cosine to sine.”

Your turn: Compute ∫ (sin x − cos x) dx.

Answer: −cos x − sin x + C

Split: ∫ sin x dx = −cos x; ∫ cos x dx = sin x, so minus that is −sin x. Check: d/dx gives sin x − cos x. ✓

Example 4 — initial value: find F with F′ = sin x and F(π) = 0

General antiderivative: F(x) = −cos x + C.
Use the condition: F(π) = −cos π + C = 1 + C = 0, so C = −1.
Answer: F(x) = −cos x − 1.
Check: F′ = sin x ✓; F(π) = 1 − 1 = 0 ✓.

Your turn: Find F with F′ = sin x and F(π/2) = 0.

Answer: F(x) = −cos x

General antiderivative: F(x) = −cos x + C. Condition: F(π/2) = −cos(π/2) + C = 0 + C = 0, so C = 0. Check: F′ = sin x ✓; F(π/2) = 0 ✓.

Memorization tips

Say it aloud: “sine to negative cosine.” Stress the word negative — that is the entire content of the rule.
The area-2 anchor: ∫₀^π sin x dx = 2. If your antiderivative gives anything else on [0, π], the sign is wrong.
Pair the directions: derivative: sine → cosine (no minus). Integral: sine → minus cosine. The minus lives on the integral side.
The 5-second check: differentiate your answer. Two negatives must appear and cancel — if only one appears, you dropped the sign.
Don’t mirror the derivative: d/dx [sin x] = cos x has no minus, which tempts ∫ sin x dx = cos x. The integral is not the derivative — it is its mirror with a twist.

Final challenge

Five mixed questions — signed areas, the sign trap, and a motion problem. Score 5/5 and the sine integral is yours.

How to learn a formula here

Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
Work the examples with the answers covered, then uncover one step at a time and compare.
Finish with the final challenge — five mixed questions including the classic traps.
Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the integral of sin x?

∫ sin x dx = −cos x + C. The minus sign survives the reversal: differentiating −cos x gives −(−sin x) = sin x.

Why is there a minus sign in ∫ sin x dx?

Because d/dx [cos x] = −sin x. To get +sin x back, the antiderivative must be −cos x, so the two minus signs cancel.

What is the area under sin x from 0 to π?

∫₀^π sin x dx = [−cos x]₀^π = (−cos π) − (−cos 0) = 1 + 1 = 2.

Is ∫ sin(2x) dx just −cos(2x) + C?

No — the chain rule demands ÷2: ∫ sin(2x) dx = −cos(2x)/2 + C. Differentiating −cos(2x) gives 2 sin(2x), twice what you want.

How do I remember the sign?

Say “sine to negative cosine.” The derivative goes sine → cosine (no minus); the integral goes sine → −cosine (minus). The minus belongs to the integral direction.

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