Why is there a negative sign in the -\cos x?

Because the derivative of \cos x is -\sin x (with a negative). To reverse this process and recover \sin x, the antiderivative must have a negative: \frac{d}{dx}(-\cos x) = \sin x. The negative sign is not optional; it comes directly from how trig derivatives work.

Do I always have to write the +C for indefinite integrals?

Yes. The C represents the entire family of antiderivatives; without it, you have found only one solution. When you compute a definite integral with limits, the constants cancel and disappear, but for indefinite integrals the C is mandatory.

What if the sine has a different argument like \sin(3x) or \sin(x + 1)?

This formula no longer applies directly. Use substitution: if u = 3x, then du = 3\,dx, so \int \sin(3x)\,dx = \frac{1}{3}\int \sin(u)\,du = -\frac{1}{3}\cos(3x) + C. The substitution method handles any shift or scaling inside the sine.

How is finding an integral different from taking a derivative?

Differentiation finds the rate of change of a function; integration reverses this to find the original function. They are inverse operations: if you differentiate -\cos x, you get \sin x, and if you integrate \sin x, you get back -\cos x (plus a constant). Use this integral when a problem gives you a rate involving sine and asks for the accumulated total.

Integral of Sin x

The antiderivative of sine is negative cosine plus an arbitrary constant, used to reverse differentiation of sine in calculus and physics applications.

\int \sin x\,dx = -\cos x + C

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What each symbol means

What Integral of Sin x takes

Symbol	Meaning
$x$	The angle variable measured in radians; if you read this in degrees instead you will get incorrect numerical answers since this formula assumes the sine function takes radian input.
$C$	The constant of integration representing all possible antiderivatives; if you forget it you have only found one particular solution, not the complete family of all antiderivatives.

When to use it

Use this integral when you need to find the antiderivative of sine or when solving differential equations containing sine.

Level

Usually taught in: Calculus I

Worked examples

1. Integrate a basic sine expression with a constant coefficient

Problem

Find

\int 3\sin x\,dx

$\int 3\sin x\,dx$
We need to integrate this expression with coefficient 3 in front of sine.
$= 3\int \sin x\,dx$
Factor out the constant 3 using the constant rule for integrals.
$= 3 \cdot (-\cos x) + C$
Apply the standard formula $\int \sin x\,dx = -\cos x + C$ .
$= -3\cos x + C$
Simplify by distributing the 3.

Answer: $-3\cos x + C$

The constant rule allows us to pull 3 outside the integral before applying the sine formula, then simplify the result.

2. Integrate a sum with negative coefficient and constant term

Problem

Evaluate

\int (-2\sin x + 5)\,dx

$\int (-2\sin x + 5)\,dx$
We need to integrate a sum containing a negative coefficient on sine and a constant term.
$= \int (-2\sin x)\,dx + \int 5\,dx$
Apply the sum rule: split this into two separate integrals.
$= -2\int \sin x\,dx + 5x + C$
Factor out $-2$ from the first integral; the constant 5 integrates directly to $5x$ .
$= -2 \cdot (-\cos x) + 5x + C$
Apply $\int \sin x\,dx = -\cos x + C$ to the first term.
$= 2\cos x + 5x + C$
Simplify: $-2 \cdot (-\cos x) = 2\cos x$ by multiplying the negatives.

Answer: $2\cos x + 5x + C$

This problem requires careful handling of the negative sign, using the sum rule to separate terms, and recognizing that constants integrate to linear terms; all three must be done correctly for the right answer.

3. Model periodic cooling using a sine rate of change

Problem

A temperature sensor measures oscillation with rate

\frac{dT}{dt} = -\sin t

where

T

is temperature in degrees and

t

is time in seconds. If the temperature is

T(0) = 20

t = 0

, find the temperature function

T(t)

$T(t) = \int (-\sin t)\,dt$
To find temperature as a function of time, integrate the given rate of change.
$= -\int \sin t\,dt$
Factor out the negative sign using the constant rule.
$= -(-\cos t) + C$
Apply the standard integral $\int \sin t\,dt = -\cos t + C$ .
$= \cos t + C$
Simplify the double negative: $-(-\cos t) = \cos t$ .
$T(0) = \cos(0) + C$
Use the initial condition $T(0) = 20$ to find the constant $C$ .
$20 = 1 + C$
Evaluate $\cos(0) = 1$ .
$C = 19$
Solve for $C$ by subtracting 1 from both sides.
$T(t) = \cos t + 19$
The complete solution describes how temperature varies over time.

Answer: $T(t) = \cos t + 19$

This application demonstrates how the sine integral solves differential equations modeling periodic motion; the initial condition selects which member of the antiderivative family matches the physical situation.

Common mistakes

Where Integral of Sin x usually goes wrong

The mistake	Why it is wrong	The fix
$\int \sin x\,dx = \cos x + C$	The derivative of $\cos x$ is $-\sin x$ (with a negative), so to reverse this you must use negative cosine, not positive cosine.	Remember: $\int \sin x\,dx = -\cos x + C$ ; the negative is essential because of how trig derivatives work.
$\int \sin x\,dx = -\cos x$	The constant $C$ is required because any constant disappears during differentiation, so infinitely many antiderivatives exist differing only by a constant.	Always include the constant of integration: $\int \sin x\,dx = -\cos x + C$ represents the complete family of antiderivatives.
$\int \sin(2x)\,dx = -\cos(2x) + C$	When sine has a coefficient inside like $\sin(2x)$ , you cannot apply this formula directly; the chain rule creates an extra factor that must be accounted for.	Use substitution: let $u = 2x$ so $du = 2\,dx$ , giving $\int \sin(2x)\,dx = \frac{1}{2}\int \sin(u)\,du = -\frac{1}{2}\cos(2x) + C$ .

Tips and when to use something else

Verify your antiderivative by differentiating: $\frac{d}{dx}(-\cos x + C) = \sin x$ confirms the formula is correct.
When integrating expressions like $5\sin x$ or $\sin x + 3$ , use linearity to factor out constants and split sums before applying this formula.
If you encounter $\sin(kx)$ where $k \ne 1$ , you must use substitution; this formula only works directly when the argument is exactly $x$ .
For products like $\sin x \cos x$ or complicated expressions, reach for integration by parts or trigonometric identities instead of this formula alone.

Related formulas

Integral of Cos xIntegrating cos x gives sin x plus a constant; use this when you need to find areas under cosine curves or solve differential equations.U-SubstitutionU-substitution simplifies integrals by transforming them into easier forms using a change of variables, ideal when you spot a function and its derivative.Indefinite IntegralThe indefinite integral finds all functions whose derivative equals a given function, expressed as an antiderivative plus an arbitrary constant.Integration by PartsIntegration by Parts is a calculus technique for evaluating integrals of products by transforming them into simpler forms when substitution fails.Fundamental Theorem of CalculusCompute the exact value of a definite integral using the antiderivative—connect area under a curve to the accumulation of change.

Frequently asked questions

Why is there a negative sign in the $-\cos x$ ?: Because the derivative of $\cos x$ is $-\sin x$ (with a negative). To reverse this process and recover $\sin x$ , the antiderivative must have a negative: $\frac{d}{dx}(-\cos x) = \sin x$ . The negative sign is not optional; it comes directly from how trig derivatives work.
Do I always have to write the $+C$ for indefinite integrals?: Yes. The $C$ represents the entire family of antiderivatives; without it, you have found only one solution. When you compute a definite integral with limits, the constants cancel and disappear, but for indefinite integrals the $C$ is mandatory.
What if the sine has a different argument like $\sin(3x)$ or $\sin(x + 1)$ ?: This formula no longer applies directly. Use substitution: if $u = 3x$ , then $du = 3\,dx$ , so $\int \sin(3x)\,dx = \frac{1}{3}\int \sin(u)\,du = -\frac{1}{3}\cos(3x) + C$ . The substitution method handles any shift or scaling inside the sine.
How is finding an integral different from taking a derivative?: Differentiation finds the rate of change of a function; integration reverses this to find the original function. They are inverse operations: if you differentiate $-\cos x$ , you get $\sin x$ , and if you integrate $\sin x$ , you get back $-\cos x$ (plus a constant). Use this integral when a problem gives you a rate involving sine and asks for the accumulated total.

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Reviewed 2026-09-18