How do you find the particular anti-derivative through a given point?

First anti-differentiate the gradient function to get the general form ending in plus C. Then substitute the coordinates of the given point and solve for C. For example if f dash equals 4 cosine 2x and the curve passes through the point pi over 4 and 3, then f equals 2 sine 2x plus C, and 3 equals 2 times sine of pi over 2 plus C gives C equals 1, so f equals 2 sine 2x plus 1.

How do you find the particular anti-derivative through a given point?

First anti-differentiate the gradient function to get the general form ending in plus C. Then substitute the coordinates of the given point and solve for C. For example if f dash equals 4 cosine 2x and the curve passes through the point pi over 4 and 3, then f equals 2 sine 2x plus C, and 3 equals 2 times sine of pi over 2 plus C gives C equals 1, so f equals 2 sine 2x plus 1.

Year 12 Methods (Unit 3 & 4) Anti-differentiation

Anti-differentiation of trigonometric functions

Q: What are the anti-derivatives of sine and cosine?

The two standard formulas are the integral of sine x is minus cosine x plus C, and the integral of cosine x is sine x plus C. Anti-differentiating sine introduces a minus sign, while anti-differentiating cosine does not. Both carry the constant of integration plus C because they are indefinite integrals.

Q: Why does the integral of sine have a minus sign?

Because differentiation and anti-differentiation are reverse processes. The derivative of cosine x is minus sine x, so to get sine x back when you differentiate you must start from minus cosine x. That is why the integral of sine x is minus cosine x plus C, while the integral of cosine x is simply sine x plus C.

Q: How do you anti-differentiate sin(kx) or cos(ax+b)?

Anti-differentiate as if the inner function were x, then divide by the coefficient of x. So the integral of cosine of kx is one over k times sine of kx plus C, and the integral of sine of ax plus b is minus one over a times cosine of ax plus b plus C. The plus b inside does not change the factor, because the inner function is still linear with slope a.

Q: Why do you always add plus C?

Because the derivative of any constant is zero, adding any constant to an anti-derivative gives another function with the same derivative. An indefinite integral therefore stands for a whole family of curves, and the constant of integration plus C represents that freedom. You drop it only once a boundary condition has fixed its value.

Q: How do you evaluate a definite integral of sine or cosine?

Anti-differentiate first, with no plus C needed, then substitute the upper and lower limits into the anti-derivative and subtract. For example the integral of sine x from 0 to pi equals minus cosine x evaluated from 0 to pi, which is minus cosine pi minus minus cosine 0, that is 1 minus minus 1, giving 2. Exact values such as cosine pi equals minus 1 make the arithmetic exact.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the two anti-derivatives of the trigonometric functions are \(\displaystyle\int \sin x\,dx=-\cos x+C\) and \(\displaystyle\int \cos x\,dx=\sin x+C\). Mind the minus sign on the \(\sin\) integral and always add the constant of integration \(+C\). For \(\sin(kx)\) or \(\cos(ax+b)\) apply the \(\dfrac{1}{a}\) factor, integrate a sum term by term using linearity, fix \(C\) from a boundary point, and evaluate definite integrals over exact-value limits.

Anti-differentiation (finding the indefinite integral) reverses differentiation: given a gradient function \(f'(x)\), you find a function \(f(x)\) whose derivative is \(f'(x)\), written \(\displaystyle\int f'(x)\,dx\).

For sine and cosine the two standard results are:

\(\displaystyle\int \sin x\,dx=-\cos x+C,\qquad \int \cos x\,dx=\sin x+C.\)

Anti-differentiating \(\sin\) introduces a minus sign (because \(\dfrac{d}{dx}\cos x=-\sin x\)), while \(\cos\) does not. Every indefinite integral carries the constant of integration \(+C\). When the angle is a linear function \(kx\) or \(ax+b\), divide by the coefficient of \(x\); by linearity, a sum of sine and cosine terms integrates term by term. All angles are measured in radians.

Key idea. \(\displaystyle\int \sin x\,dx=-\cos x+C\) and \(\displaystyle\int \cos x\,dx=\sin x+C\); for \(\cos(ax+b)\) or \(\sin(ax+b)\) divide by \(a\). A boundary point \((a,b)\) fixes \(C\).

Anti-derivatives of \(\cos x\): the family \(y=\sin x+C\)

\(\displaystyle\int_{0}^{\pi}\sin x\,dx=2\): the shaded area under one arch

The two standard anti-derivatives (angles in radians):

\[\int \sin x\,dx=-\cos x+C,\qquad \int \cos x\,dx=\sin x+C\]

\int \sin x d x = - \cos x + C

Linear inner function — divide by the coefficient \(a\) of \(x\):

\[\int \cos(ax+b)\,dx=\dfrac{1}{a}\sin(ax+b)+C,\qquad \int \sin(ax+b)\,dx=-\dfrac{1}{a}\cos(ax+b)+C\]

\int \cos (a x + b) d x = \frac{1}{a} \sin (a x + b) + C

Linearity — integrate term by term and pull out constant multiples:

\[\int \big(f(x)+g(x)\big)\,dx=\int f(x)\,dx+\int g(x)\,dx,\qquad \int k\,f(x)\,dx=k\int f(x)\,dx\]

Definite integral — anti-differentiate, then substitute the limits (no \(+C\)):

\[\int_{a}^{b} f(x)\,dx=\Big[F(x)\Big]_{a}^{b}=F(b)-F(a)\]

Mind the sign. The integral of \(\sin\) carries a minus (\(-\cos x\)); the integral of \(\cos\) does not (\(+\sin x\)). Differentiate your answer to check it returns the original integrand.

How to anti-differentiate sine and cosine

Apply the standard form. \(\displaystyle\int \sin x\,dx=-\cos x+C\) (minus) and \(\displaystyle\int \cos x\,dx=\sin x+C\) (no minus).
Handle a linear inner function. For \(\sin(ax+b)\) or \(\cos(ax+b)\), divide by the coefficient \(a\) of \(x\); the \(+b\) does not change that factor.
Use linearity. Integrate a sum term by term and take constant multiples out the front, keeping a single \(+C\).
Fix \(C\) if a point is given. Substitute the boundary point \((a,b)\) and solve for \(C\) to get the particular anti-derivative.
For a definite integral, anti-differentiate (no \(+C\)), then substitute the upper and lower limits and subtract, using exact values.

Check by differentiating. Differentiation is the reverse of what you just did, so \(\dfrac{d}{dx}\) of your anti-derivative should return the original integrand — a fast way to catch a sign slip or a missing factor.

Example 1 — Term by term

Find \(\displaystyle\int (3\cos x-2\sin x)\,dx\).

Solution

Integrate each term; \(-2\sin x\) gives \(-2(-\cos x)=+2\cos x\).

\(\displaystyle\int (3\cos x-2\sin x)\,dx\)	\(=\)	\(3\sin x-2(-\cos x)+C\)
	\(=\)	\(3\sin x+2\cos x+C\)

3 \sin x + 2 \cos x + C

Example 2 — The \(\tfrac{1}{k}\) factor

Find \(\displaystyle\int 8\cos 4x\,dx\).

Solution

Divide by the coefficient of \(x\), here \(k=4\).

\(\displaystyle\int 8\cos 4x\,dx\)	\(=\)	\(8\times\dfrac{1}{4}\sin 4x+C\)
	\(=\)	\(2\sin 4x+C\)

2 \sin 4 x + C

Example 3 — Particular anti-derivative

\(f'(x)=4\cos 2x\) and the curve passes through \(\left(\dfrac{\pi}{4},3\right)\). Find \(f(x)\).

Solution

Integrate, then substitute the point to find \(C\).

\(f(x)\)	\(=\)	\(2\sin 2x+C\)
\(3\)	\(=\)	\(2\sin\dfrac{\pi}{2}+C=2(1)+C\Rightarrow C=1\)
\(f(x)\)	\(=\)	\(2\sin 2x+1\)

f (x) = 2 \sin 2 x + 1

Example 4 — Definite integral

Evaluate \(\displaystyle\int_{0}^{\pi} \sin x\,dx\).

Solution

Anti-differentiate to \(-\cos x\), then substitute the limits.

\(\displaystyle\int_{0}^{\pi} \sin x\,dx\)	\(=\)	\(\Big[-\cos x\Big]_{0}^{\pi}\)
	\(=\)	\((-\cos\pi)-(-\cos 0)=1-(-1)=2\)

\int \sin x d x = 2

Common pitfalls

Watch the sign. \(\displaystyle\int \sin x\,dx=-\cos x+C\) (with a minus), while \(\displaystyle\int \cos x\,dx=\sin x+C\) (no minus). Swapping the sign is the most common slip.

Divide by the coefficient of \(x\). \(\displaystyle\int \cos 4x\,dx=\dfrac{1}{4}\sin 4x+C\), not \(\sin 4x\) and not \(4\sin 4x\); the inner factor \(4\) means you divide by \(4\).

Never forget \(+C\). An indefinite integral stands for a whole family of curves. Only drop the constant once a boundary condition has fixed it, or when evaluating a definite integral.

Frequently asked questions

What are the anti-derivatives of sine and cosine?

\(\displaystyle\int \sin x\,dx=-\cos x+C\) and \(\displaystyle\int \cos x\,dx=\sin x+C\). Anti-differentiating \(\sin\) introduces a minus sign; \(\cos\) does not.

Why does the integral of sine have a minus sign?

Because \(\dfrac{d}{dx}\cos x=-\sin x\). To get \(\sin x\) back when you differentiate, you must start from \(-\cos x\), so \(\displaystyle\int \sin x\,dx=-\cos x+C\).

How do you anti-differentiate sin(kx) or cos(ax+b)?

Anti-differentiate as if the inner were \(x\), then divide by the coefficient of \(x\). So \(\displaystyle\int \cos(ax+b)\,dx=\dfrac{1}{a}\sin(ax+b)+C\); the \(+b\) does not change the factor.

Why do you always add plus C?

Because the derivative of any constant is \(0\), so infinitely many functions share the same derivative. The \(+C\) captures that whole family; a boundary point pins down one member.

How do you find the particular anti-derivative through a point?

Integrate to the general form ending in \(+C\), then substitute the point and solve for \(C\). E.g. \(f'=4\cos 2x\) through \(\left(\dfrac{\pi}{4},3\right)\) gives \(C=1\), so \(f=2\sin 2x+1\).

How do you evaluate a definite integral of sine or cosine?

Anti-differentiate (no \(+C\)), then substitute the limits and subtract. E.g. \(\displaystyle\int_{0}^{\pi}\sin x\,dx=\big[-\cos x\big]_{0}^{\pi}=1-(-1)=2\).

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