Trigonometric Composites with a Linear Inner Function

Why the Inner Slope Matters

The derivative of a trigonometric composite includes a factor from the derivative of its inner angle. Reversing differentiation to find an antiderivative therefore requires accounting for that factor. For a linear inner angle, the factor is constant, so the adjustment is especially direct.

Throughout this tutorial, trigonometric angles are measured in radians, as required for the standard derivative formulas for sine and cosine. The goal is to recognize an integrand such as \(\sin(3x-\pi/4)\) or \(\cos(5-2x)\), find an antiderivative, and verify that the inner slope has been handled correctly.

Definition: A linear inner angle is an expression of the form \(ax+b\), where \(a\) and \(b\) are constants. Its derivative with respect to \(x\) is the constant \(a\).

The Two Basic Formulas

For real constants \(a\ne 0\) and \(b\), the antiderivatives are

\int \sin(ax+b)\,dx = -\frac{1}{a}\cos(ax+b)+C

\int \cos(ax+b)\,dx = \frac{1}{a}\sin(ax+b)+C.

Here \(C\) is an arbitrary constant. The condition \(a\ne 0\) is essential to these formulas: the factor \(1/a\) is not defined when the inner angle is constant. If \(a=0\), the integrands are the constants \(\sin(b)\) and \(\cos(b)\), so their antiderivatives are \(x\sin(b)+C\) and \(x\cos(b)+C\), respectively.

The sign in the sine formula comes from the derivative of cosine, while the denominator comes from differentiating the inner angle. The sign of \(a\) must be retained: if \(a\) is negative, then \(1/a\) is negative as well.

Why the Formulas Work

The chain rule says that the derivative of a composite is the derivative of its outer function, evaluated at the inner function, multiplied by the derivative of the inner function. In particular,

\frac{d}{dx}\cos(ax+b) = -\sin(ax+b)\cdot a

and

\frac{d}{dx}\sin(ax+b) = \cos(ax+b)\cdot a.

To obtain \(\sin(ax+b)\) as a derivative, the first expression must be multiplied by \(-1/a\). To obtain \(\cos(ax+b)\), the second must be multiplied by \(1/a\). Since \(a\) is a nonzero constant, these factors are valid. This proves both formulas.

Theorem: If \(a\ne 0\), differentiating \(-\frac{1}{a}\cos(ax+b)\) gives \(\sin(ax+b)\), and differentiating \(\frac{1}{a}\sin(ax+b)\) gives \(\cos(ax+b)\). Thus these expressions are antiderivatives on any interval where the functions are defined.

A Reliable Calculation Pattern

For an integral involving a sine or cosine of a linear angle, identify the coefficient \(a\) of \(x\) before integrating. Apply the appropriate formula with that coefficient, including its sign. If the integrand has a constant multiplier, carry it through the calculation. For a definite integral, evaluate the antiderivative at the upper bound and subtract its value at the lower bound.

Identify the inner angle.
Write it as \(ax+b\), and record the slope \(a\).

Choose the matching formula.
Sine gives a negative cosine; cosine gives a positive sine. Divide by \(a\) in either case.

Check by differentiation.
Differentiate the proposed antiderivative and confirm that every factor and sign returns the original integrand.

Worked Examples

Worked Example: Sine with a Positive Inner Slope

Find an antiderivative of \(\sin(3x-\pi/4)\).

The inner angle is \(3x-\pi/4\), so \(a=3\) and \(b=-\pi/4\). The sine formula gives

\int \sin(3x-\pi/4)\,dx = -\frac{1}{3}\cos(3x-\pi/4)+C.

Check by differentiating the proposed antiderivative:

\begin{aligned} \frac{d}{dx}\left(-\frac{1}{3}\cos(3x-\pi/4)\right) &=-\frac{1}{3}\left[-\sin(3x-\pi/4)\cdot 3\right]\\ &=\sin(3x-\pi/4). \end{aligned}

The factor \(-1/3\) cancels the negative sign from differentiating cosine and the factor \(3\) from the inner angle. The derivative matches the integrand, so the antiderivative is correct.

Worked Example: Cosine with a Negative Inner Slope

Find an antiderivative of \(\cos(5-2x)\).

Rewrite the inner angle as \(-2x+5\). Its slope is \(a=-2\), so the cosine formula gives

\int \cos(5-2x)\,dx = \frac{1}{-2}\sin(5-2x)+C = -\frac{1}{2}\sin(5-2x)+C.

The negative sign in the result is required because the inner slope is negative. Differentiate to check:

\begin{aligned} \frac{d}{dx}\left(-\frac{1}{2}\sin(5-2x)\right) &=-\frac{1}{2}\left[\cos(5-2x)\cdot(-2)\right]\\ &=\cos(5-2x). \end{aligned}

The two negative factors multiply to a positive one, leaving exactly the original integrand.

Worked Example: A Definite Sine Integral

Evaluate \(\int_0^{\pi/6}\sin(3x+\pi/6)\,dx\).

The inner slope is \(3\), so an antiderivative is \(-\frac{1}{3}\cos(3x+\pi/6)\). By the Fundamental Theorem of Calculus, the definite integral is the upper-bound value of this antiderivative minus its lower-bound value:

\begin{aligned} \int_0^{\pi/6}\sin(3x+\pi/6)\,dx &=\left[-\frac{1}{3}\cos(3x+\pi/6)\right]_0^{\pi/6}\\ &=-\frac{1}{3}\cos\left(3\cdot\frac{\pi}{6}+\frac{\pi}{6}\right) +\frac{1}{3}\cos\left(3\cdot0+\frac{\pi}{6}\right)\\ &=-\frac{1}{3}\cos(2\pi/3)+\frac{1}{3}\cos(\pi/6)\\ &=-\frac{1}{3}\left(-\frac{1}{2}\right) +\frac{1}{3}\left(\frac{\sqrt{3}}{2}\right)\\ &=\frac{1}{6}+\frac{\sqrt{3}}{6}\\ &=\frac{1+\sqrt{3}}{6}. \end{aligned}

To check the antiderivative used here, differentiate it:

\frac{d}{dx}\left(-\frac{1}{3}\cos(3x+\pi/6)\right) = -\frac{1}{3}\left[-3\sin(3x+\pi/6)\right] = \sin(3x+\pi/6).

The endpoint calculation also uses the correct order: upper value minus lower value. The resulting exact value is \((1+\sqrt{3})/6\).

Worked Example: Integrating a Sum

Find an antiderivative of \(4\sin(2x)+3\cos(2x)\).

An antiderivative of a sum is the sum of antiderivatives, and constant factors can be kept outside each integral. The slope of each inner angle \(2x\) is \(2\):

\begin{aligned} \int \left(4\sin(2x)+3\cos(2x)\right)\,dx &=4\left(-\frac{1}{2}\cos(2x)\right) +3\left(\frac{1}{2}\sin(2x)\right)+C\\ &=-2\cos(2x)+\frac{3}{2}\sin(2x)+C. \end{aligned}

Differentiate each term:

\begin{aligned} \frac{d}{dx}\left(-2\cos(2x)+\frac{3}{2}\sin(2x)\right) &=-2\left[-2\sin(2x)\right] +\frac{3}{2}\left[2\cos(2x)\right]\\ &=4\sin(2x)+3\cos(2x). \end{aligned}

Both terms return their corresponding parts of the integrand, confirming the result.

A Common Mistake: Omitting the Inner Slope

A frequent error is to integrate \(\sin(4x)\) as if its inner angle were just \(x\). For example, the calculation

\int \sin(4x)\,dx \overset{\text{wrong}}{=} -\cos(4x)+C

is incorrect. Differentiating the proposed answer reveals the extra factor:

\frac{d}{dx}\left[-\cos(4x)\right] = 4\sin(4x),

which is four times the required integrand. The inner slope is \(4\), so the correct antiderivative includes a factor of \(1/4\):

\int \sin(4x)\,dx = -\frac{1}{4}\cos(4x)+C, \qquad \frac{d}{dx}\left[-\frac{1}{4}\cos(4x)\right] = \sin(4x).

The same check catches sign errors when the slope is negative. Do not replace \(a\) by its absolute value: the formulas use the signed coefficient \(a\).

When the Method Applies

These formulas apply directly when the entire angle is linear in \(x\), such as \(7x+2\) or \(3-5x\). A constant factor multiplying the sine or cosine causes no difficulty, because it can be carried through the integral. Sums of terms can be handled term by term when each term has this form.

The formulas do not directly cover an angle whose derivative depends on \(x\), such as \(x^2\), because there is no single constant inner slope to divide by. The chain-rule check is a useful way to decide whether a proposed expression works: its derivative must reproduce the integrand exactly, not merely a multiple or a sign-reversed version.

For definite integrals, the same antiderivative formulas apply, provided the integrand is continuous on the interval of integration; sine and cosine of linear angles are continuous everywhere. The Fundamental Theorem of Calculus then evaluates the integral by subtracting the antiderivative at the lower limit from its value at the upper limit.

Takeaway: For a nonzero linear slope \(a\), an antiderivative of sine or cosine of \(ax+b\) must account for the factor \(a\) introduced by the chain rule. Divide by \(a\), keep its sign, and differentiate the result to check it.

Check Your Understanding

Try each question, and check any proposed antiderivative by differentiating it.

Find an antiderivative of \(\sin(6x+1)\).
Find an antiderivative of \(\cos(7-3x)\). Why does the coefficient of the resulting sine term have a negative sign?
Evaluate \(\int_0^{\pi/4}\cos(2x)\,dx\) exactly.
Differentiate \(\frac{1}{5}\sin(5x-\pi)\). Which integrand does it give?
What is an antiderivative of \(\sin(2)\), and why is the nonzero-slope formula inapplicable?