AP Calculus AB and BC

Trig Identities You Actually Need for Calculus

You need three families for AP Calculus: the Pythagorean identities (sin^2 x + cos^2 x = 1 and its two divided-down forms), the double-angle formulas for cosine that power-reduce sin^2 x and cos^2 x, and the sum formulas. Each one exists to rewrite a trig expression you cannot antidifferentiate into one you can.

When you reach for a trig identity

You pull out a trig identity when an expression is trigonometric and no basic antiderivative or single u-substitution fits it as written. The identity's only job is to rewrite that trig into a shape one of your known rules can finish. This is a rewrite-then-recognize task, and in AP it lives mostly in Topic 6.9 (integrating using substitution) and the derivative rules of Topics 2.7 and 2.10 that those integrals reverse.

The one question that decides

Does a basic antiderivative or a clean u-substitution already stare back at you? If yes, you need no identity. If instead you see an even power of sine or cosine, a squared tangent or cotangent, or a product like $\sin x\cos x$ , the identity that unlocks it is on this page. There is no AP formula sheet, so you rebuild these from memory, but the tables below show you only need to memorize a handful and derive the rest.

When the integrand looks like	Rewrite it with	Because it becomes
$\int \sin^2 x\,dx$ or $\int \cos^2 x\,dx$	A power-reduction form of $\cos 2x$	A squared trig function has no basic antiderivative; the linear $\cos 2x$ that replaces it does.
$\int \tan^2 x\,dx$	$1+\tan^2 x=\sec^2 x$	$\tan^2 x=\sec^2 x-1$ , and $\int \sec^2 x\,dx=\tan x$ .
$\int \cot^2 x\,dx$	$1+\cot^2 x=\csc^2 x$	$\cot^2 x=\csc^2 x-1$ , and $\int \csc^2 x\,dx=-\cot x$ .
An odd power, like $\int \sin^3 x\,dx$	$\sin^2 x=1-\cos^2 x$	Peels off one $\sin x$ for $du$ and leaves cosines, so $u=\cos x$ finishes it (Topic 6.9).
$\int 2\sin x\cos x\,dx$	$\sin 2x=2\sin x\cos x$	Collapses the product into a single sine you integrate in one step.
$\sqrt{\sin^2\theta+\cos^2\theta}$ inside an integrand	$\sin^2 x+\cos^2 x=1$	The radical simplifies to a constant, common in BC arc-length and parametric-speed setups.

The three Pythagorean identities

All three are one identity seen from three angles. Start from $\sin^2 x+\cos^2 x=1$ . Divide every term by $\cos^2 x$ and you get the tangent-secant form; divide every term by $\sin^2 x$ and you get the cotangent-cosecant form. You do not memorize three separate facts, you memorize one and divide.

Identity	How you get it	Its calculus job
$\sin^2 x+\cos^2 x=1$	Memorize outright (the unit circle)	Swap $\sin^2 x$ for $1-\cos^2 x$ (or the reverse) to free one factor for u-substitution, and collapse radicals in arc-length and parametric-speed integrands.
$1+\tan^2 x=\sec^2 x$	Divide the first identity by $\cos^2 x$	Turns $\int \tan^2 x\,dx$ into $\int(\sec^2 x-1)\,dx$ , because $\sec^2 x$ is the member you can antidifferentiate.
$1+\cot^2 x=\csc^2 x$	Divide the first identity by $\sin^2 x$	The same move for cotangent: $\int \cot^2 x\,dx=\int(\csc^2 x-1)\,dx$ .

The tangent case is the model worked example. You cannot integrate $\tan^2 x$ directly, but the identity trades it for something you can:

\int \tan^2 x\,dx = \int (\sec^2 x - 1)\,dx = \int \sec^2 x\,dx - \int 1\,dx = \tan x - x + C

Why sec squared and csc squared are the targets

You always convert toward $\sec^2 x$ and $\csc^2 x$ , never away from them, because $\sec^2 x$ is exactly the derivative of $\tan x$ and $\csc^2 x$ is exactly the derivative of $-\cot x$ : in prime notation $(\tan x)' = \sec^2 x$ and $(\cot x)' = -\csc^2 x$ (Topic 2.10). They are the only members of this family with a one-step antiderivative, $\tan x$ and $-\cot x$ respectively, so the identity exists to move your integrand onto that landing spot.

Double-angle formulas and power reduction

The double-angle formulas matter in calculus for one reason above all others: they are how you integrate an even power of sine or cosine. A squared trig function is not on your basic antiderivative list, so you trade it for a first power of $\cos 2x$ , which is. Cosine has three equivalent forms, and the two rewritten forms are the ones that do the work.

Formula	Which form, and when
$\sin 2x = 2\sin x\cos x$	Read right-to-left to collapse a $2\sin x\cos x$ product into one term; read left-to-right when a $\sin 2x$ needs breaking apart.
$\cos 2x = \cos^2 x - \sin^2 x$	The base form. The next two rows rewrite it with $\sin^2 x+\cos^2 x=1$ and are the ones you actually integrate with.
$\cos 2x = 2\cos^2 x - 1$	Solve this for $\cos^2 x$ to get the power-reduction formula for cosine.
$\cos 2x = 1 - 2\sin^2 x$	Solve this for $\sin^2 x$ to get the power-reduction formula for sine.

Solving those last two forms for the squared term gives the payoff, the two power-reduction formulas you reach for whenever an even power of sine or cosine stands alone in an integral:

Power-reduction form	Comes from solving	Use it to integrate
$\cos^2 x = \frac{1+\cos 2x}{2}$	$\cos 2x = 2\cos^2 x - 1$	$\int \cos^2 x\,dx$ : the right side is a constant plus $\cos 2x$ , both one-step integrals.
$\sin^2 x = \frac{1-\cos 2x}{2}$	$\cos 2x = 1 - 2\sin^2 x$	$\int \sin^2 x\,dx$ : the same, carrying a minus sign.

Here is the sine case worked all the way through, using $\int \cos 2x\,dx=\frac{1}{2}\sin 2x$ for the linear inside:

\begin{aligned}\int \sin^2 x\,dx &= \int \frac{1-\cos 2x}{2}\,dx = \frac{1}{2}\int (1-\cos 2x)\,dx \\ &= \frac{1}{2}\left(x - \frac{1}{2}\sin 2x\right)+C = \frac{x}{2} - \frac{\sin 2x}{4} + C\end{aligned}

How to remember which sign

The power-reduction formula matches its own name: $\cos^2 x$ takes the plus sign, $\sin^2 x$ takes the minus. If you blank on it, rederive in seconds: $\cos^2 x$ comes from solving $\cos 2x = 2\cos^2 x - 1$ , while $\sin^2 x$ comes from solving $\cos 2x = 1 - 2\sin^2 x$ (or from $\sin^2 x = 1 - \cos^2 x$ once you have the cosine form), rather than memorizing both.

Sum formulas, the source of everything above

The sum and difference formulas are the parents of the double-angle formulas: set $A=B$ and $\sin(A+B)$ becomes $\sin 2A$ while $\cos(A+B)$ becomes $\cos 2A$ . On the AP exam you rarely apply them to an integral directly, but they let you rebuild any double-angle formula you forget, and one of them is the engine behind the derivative of sine from the definition.

Formula	Watch the sign	Calculus role
$\sin(A+B) = \sin A\cos B + \cos A\sin B$	Sine keeps the sign: a plus stays a plus.	Expanding $\sin(x+h)$ with this is the first step in deriving $(\sin x)' = \cos x$ from the limit definition.
$\sin(A-B) = \sin A\cos B - \cos A\sin B$	A difference turns the plus into a minus.	Set $A=B$ as a check: it correctly gives $0$ .
$\cos(A+B) = \cos A\cos B - \sin A\sin B$	Cosine flips the sign: an outside plus becomes an inside minus.	Set $A=B$ to recover $\cos 2A=\cos^2 A-\sin^2 A$ , the base double-angle form.
$\cos(A-B) = \cos A\cos B + \sin A\sin B$	A difference turns the minus into a plus.	The least-used of the four on the AP exam; know it for the sign pattern.

What you can skip

The product-to-sum formulas, like $\sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$ , belong to a full trig course and to integrals such as $\int \sin 5x\cos 3x\,dx$ that the AP exam does not ask. Leave them out. The four sum formulas here, plus the double-angle and Pythagorean rows above, are the complete working set for both AB and BC.