Trigonometric Integrands and Moving Endpoints
An integral with a variable endpoint describes accumulated change up to a point that depends on \(x\). Differentiating this accumulation function asks how quickly the accumulated value changes as \(x\) changes.
When the integrand is trigonometric, the same basic principle applies as with other continuous integrands: evaluate the integrand at the moving endpoint. If that endpoint is itself a function of \(x\), the chain rule supplies an additional factor.
For example, if the upper endpoint is \(g(x)\), then the integrand is evaluated at \(g(x)\), and the result is multiplied by \(g'(x)\). If instead the lower endpoint moves, the contribution has a minus sign.
The Variable-Endpoint Rule
Let \(f\) be continuous on an interval containing the relevant endpoints, and let \(g\) be differentiable. The Fundamental Theorem of Calculus and the chain rule give the following formulas. The lower or upper endpoint not named as variable is held fixed.
These formulas apply in particular when \(f(t)=\sin t\), \(f(t)=\cos t\), or a continuous expression built from trigonometric functions. Sine and cosine are continuous for every real input, so there is no restriction on the endpoint values for these integrands.
The variable \(t\) indicates the input at which the integrand is evaluated inside the integral. The endpoint \(g(x)\) tells us which value of \(t\) to use after applying the rule. For a trigonometric integrand, that means, for example, writing \(\sin(g(x))\), not \(\sin(x)\) unless the endpoint actually is \(x\).
Why the Rule Works
For the upper-endpoint formula, define an accumulation function \(A\) by
Since \(f\) is continuous, the Fundamental Theorem of Calculus Part I states that \(A'(u)=f(u)\). The integral with moving endpoint is \(A(g(x))\). Differentiating this composition by the chain rule gives
This derives the upper-endpoint formula. For a moving lower endpoint, use the identity \[ \int_{g(x)}^b f(t)\,dt=-\int_b^{g(x)} f(t)\,dt. \] The upper-endpoint formula applied to the integral on the right gives derivative \(f(g(x))g'(x)\). The minus sign in the identity therefore gives derivative \(-f(g(x))g'(x)\), as claimed.
The derivation shows why two distinct operations appear in the answer: the Fundamental Theorem evaluates \(f\) at the endpoint, and the chain rule multiplies by the endpoint’s derivative.
Worked Examples
Worked Example: Sine With a Composite Upper Endpoint
Differentiate \[ F(x)=\int_0^{2x+\pi/6}\sin t\,dt. \]
The fixed lower endpoint is \(0\), the integrand is \(f(t)=\sin t\), and the upper endpoint is \(g(x)=2x+\pi/6\). Its derivative is \[ g'(x)=2. \] Evaluate the integrand at the upper endpoint, then multiply by the endpoint derivative:
To check, evaluate the integral using the antiderivative \(-\cos t\): \[ F(x)=[-\cos t]_0^{2x+\pi/6} =-\cos\!\left(2x+\frac{\pi}{6}\right)+\cos(0) =1-\cos\!\left(2x+\frac{\pi}{6}\right). \] Differentiating this expression gives \[ F'(x)=\sin\!\left(2x+\frac{\pi}{6}\right)\cdot 2, \] which agrees with the variable-endpoint rule. As an exact value check, at \(x=\pi/4\) the endpoint is \(2(\pi/4)+\pi/6=\pi/2+\pi/6=2\pi/3\), so the derivative is \(2\sin(2\pi/3)=\sqrt{3}\). The differentiated explicit expression gives the same value.
Worked Example: Cosine With a Composite Lower Endpoint
Differentiate \[ G(x)=\int_{x^2}^{\pi/3}\cos t\,dt. \]
Here the lower endpoint is \(g(x)=x^2\), so \(g'(x)=2x\). The integrand evaluated at the moving endpoint is \(\cos(x^2)\). Because the moving endpoint is the lower endpoint, include a minus sign:
Check by evaluating the integral with the antiderivative \(\sin t\): \[ G(x)=[\sin t]_{x^2}^{\pi/3} =\sin\!\left(\frac{\pi}{3}\right)-\sin(x^2) =\frac{\sqrt{3}}{2}-\sin(x^2). \] Differentiating gives \[ G'(x)=0-\cos(x^2)\cdot 2x=-2x\cos(x^2), \] as above. For an exact value check, take \(x=\sqrt{\pi/3}\). Then \(x^2=\pi/3\), and \[ G'(x)=-2\sqrt{\frac{\pi}{3}}\cos\!\left(\frac{\pi}{3}\right) =-2\sqrt{\frac{\pi}{3}}\cdot\frac12 =-\sqrt{\frac{\pi}{3}}. \] The explicit derivative gives the same result at this value of \(x\).
Worked Example: A Sine Endpoint With a Trigonometric Integrand
Differentiate \[ H(x)=\int_0^{\sin x}(1+\cos t)\,dt. \]
The integrand is \(f(t)=1+\cos t\), and the upper endpoint is \(g(x)=\sin x\), with \(g'(x)=\cos x\). Evaluate the entire integrand at \(t=\sin x\), then multiply by \(\cos x\):
To check, an antiderivative of \(1+\cos t\) is \(t+\sin t\). Thus \[ H(x)=[t+\sin t]_0^{\sin x} =\sin x+\sin(\sin x)-(0+\sin 0) =\sin x+\sin(\sin x). \] Differentiating term by term gives \[ H'(x)=\cos x+\cos(\sin x)\cos x =\bigl(1+\cos(\sin x)\bigr)\cos x. \] At \(x=0\), the rule gives \(\bigl(1+\cos 0\bigr)\cos 0=2\). The explicit derivative also gives \(\cos 0+\cos(\sin 0)\cos 0=1+1=2\).
A Reliable Way to Apply the Rule
The main work is identifying the moving endpoint and keeping its value separate from its rate of change. The following sequence makes those roles explicit.
Write it as \(g(x)\). Note whether it is the upper or lower limit.
Replace its input \(t\) with the complete endpoint \(g(x)\), giving \(f(g(x))\).
Find \(g'(x)\) and include it as a factor.
Use a positive sign for a moving upper endpoint and a negative sign for a moving lower endpoint.
When useful, evaluate the integral explicitly and differentiate the resulting expression to compare.
Common Mistake: Evaluating at the Wrong Input
Consider \[ K(x)=\int_0^{3x}\cos t\,dt. \] The endpoint is \(3x\), not \(x\), and its derivative is \(3\). A common incorrect calculation is \[ K'(x)=\cos(x), \] which both evaluates the integrand at the wrong input and omits the endpoint’s rate of change.
The correct calculation keeps these two roles distinct:
For a check, \(K(x)=\sin(3x)-\sin(0)=\sin(3x)\), so direct differentiation gives \(K'(x)=3\cos(3x)\). The incorrect expression \(\cos(x)\) does not agree in general.
A separate sign check is important for a moving lower endpoint. For instance, in \(\int_{x}^{b}\sin t\,dt\), the derivative is \(-\sin(x)\), not \(+\sin(x)\). The sign changes because increasing the lower limit removes area from the accumulated integral.
Why These Derivatives Matter
An accumulation function can represent total change collected up to a moving input. Its derivative describes the instantaneous rate at which that total changes. With a trigonometric integrand, the instantaneous rate depends on the sine or cosine value at the current endpoint, adjusted for how quickly that endpoint moves.
This is why a composite endpoint cannot be treated as if it were simply \(x\). If the endpoint moves twice as fast as \(x\), the accumulated quantity changes with the corresponding factor of two. If it moves at a variable rate, that rate appears as the chain factor. The evaluation and the endpoint rate are both essential parts of the derivative.
Check Your Understanding
For each question, identify the moving endpoint before applying the rule. Then check your result by differentiation when an explicit evaluation is straightforward.
- Find the derivative of \(\displaystyle \int_0^{4x-\pi/6}\sin t\,dt\).
- Find the derivative of \(\displaystyle \int_{\cos x}^{\pi/2}\cos t\,dt\).
- For \(\displaystyle A(x)=\int_1^{x^3}(2+\sin t)\,dt\), what is the integrand evaluated at the endpoint, and what is the endpoint chain factor?
- Explain why \(\displaystyle \frac{d}{dx}\int_0^{\sin x}\cos t\,dt\) includes a factor of \(\cos x\).
- State the sign and derivative for a moving lower endpoint in \(\displaystyle \int_{x^2}^{b}\sin t\,dt\), where \(b\) is fixed.
