Why the Endpoints Must Change
Substitution is useful when an integrand contains a trigonometric function of an inner expression, together with the derivative of that expression. In a definite integral, the substitution changes not only the integrand but also the values at which the integral is evaluated. Transforming both endpoints keeps the new variable and its limits consistent.
For example, if \(u=4x\), then an original endpoint \(x=\pi/6\) corresponds to the new endpoint \(u=4(\pi/6)=2\pi/3\). The upper limit is not still \(\pi/6\): that value belongs to the \(x\)-variable, not the \(u\)-variable.
The Definite Substitution Formula
Suppose \(g\) is continuously differentiable on \([a,b]\), and \(F\) is continuous on an interval containing the values \(g(x)\) for \(a\le x\le b\). The definite substitution formula is
The new limits are written in the order \(g(a)\) to \(g(b)\). If \(g\) decreases, this order may be decreasing as well; the formula already accounts for that. For the examples here, the inner functions are linear, but the same endpoint rule applies to other differentiable inner functions too.
Why the Formula Works
Let \(A\) be an antiderivative of \(F\) on the interval containing the values of \(g\), so \(A'(u)=F(u)\). The chain rule gives the derivative of \(A(g(x))\):
Thus \(A(g(x))\) is an antiderivative of the original integrand. By the Fundamental Theorem of Calculus,
Since \(A\) is an antiderivative of \(F\), the right-hand side is also
This proves the substitution formula and explains why the endpoints must be transformed: the new antiderivative is evaluated at \(g(b)\) and \(g(a)\), not at \(b\) and \(a\).
Common Trigonometric Patterns
The following patterns apply when the derivative of the inner function supplies the required factor. In the table, \(u\) denotes the inner expression. Each formula is valid on an interval where the functions involved are defined; in particular, tangent and secant require \(\cos u\ne 0\).
| Integrand in \(u\) | An antiderivative | Useful condition |
|---|---|---|
| \(\sin u\) | \(-\cos u\) | Defined for all real \(u\) |
| \(\cos u\) | \(\sin u\) | Defined for all real \(u\) |
| \(\sec^2 u\) | \(\tan u\) | \(\cos u\ne 0\) |
| \(\sec u\tan u\) | \(\sec u\) | \(\cos u\ne 0\) |
| \(\tan u\) | \(-\ln|\cos u|\) | \(\cos u\ne 0\) |
| \(\sec u\) | \(\ln|\sec u+\tan u|\) | \(\cos u\ne 0\) |
For instance, the tangent formula follows because the derivative of \(-\ln|\cos u|\) is \(\sin u/\cos u=\tan u\), wherever \(\cos u\ne 0\). These formulas are patterns, not permission to ignore the rest of the integrand: the factor \(g'(x)\) must also be present, including any constant multiplier needed to make the substitution exact.
Worked Examples
Worked Example: A Composite Sine Integral
Evaluate \(\displaystyle\int_0^{\pi/6}4\sin(4x)\,dx\).
Set \(u=4x\). Then \(du=4\,dx\), so the factor \(4\,dx\) in the integral is exactly \(du\). Transform the endpoints:
Therefore,
Check the result by differentiating the corresponding antiderivative in \(x\): \(\frac{d}{dx}[-\cos(4x)]=4\sin(4x)\). Its endpoint difference is \(-\cos(2\pi/3)-[-\cos(0)]=3/2\), as obtained above.
Worked Example: A Composite Cosine Integral
Evaluate \(\displaystyle\int_0^{\pi/3}2\cos(2x)\,dx\).
Let \(u=2x\), so \(du=2\,dx\). At the lower endpoint, \(u=2(0)=0\); at the upper endpoint, \(u=2(\pi/3)=2\pi/3\). Thus
Check: \(\frac{d}{dx}\sin(2x)=2\cos(2x)\), and \(\sin(2\pi/3)-\sin(0)=\sqrt{3}/2\). Both the derivative and endpoint evaluation match the original integral.
Worked Example: A Composite Tangent Integral
Evaluate \(\displaystyle\int_0^{\pi/8}2\tan(2x)\,dx\).
Use \(u=2x\), for which \(du=2\,dx\). The transformed limits are \(u=2(0)=0\) and \(u=2(\pi/8)=\pi/4\). On the full interval \(0\le u\le\pi/4\), \(\cos u\) is positive, so tangent is defined and the logarithm formula applies.
Check: differentiating \(-\ln|\cos(2x)|\) gives \(2\tan(2x)\). The endpoint difference is \(-\ln(\cos(\pi/4))+\ln(\cos 0)=\frac12\ln 2\), confirming the value.
Worked Example: A Composite Secant-Tangent Integral
Evaluate \(\displaystyle\int_0^{\pi/9}3\sec(3x)\tan(3x)\,dx\).
Set \(u=3x\), so \(du=3\,dx\). The limits become \(u=3(0)=0\) and \(u=3(\pi/9)=\pi/3\). The interval from \(0\) to \(\pi/3\) contains no point where \(\cos u=0\); hence secant and tangent are defined throughout the transformed interval.
Check: \(\frac{d}{dx}\sec(3x)=3\sec(3x)\tan(3x)\), and \(\sec(\pi/3)-\sec(0)=2-1=1\). This agrees with the evaluated integral.
Endpoint Values and Domain Checks
After changing variables, evaluate the trigonometric functions at the new endpoints. Exact values from the unit circle are usually preferable to decimal approximations. For example, \(\sin(2\pi/3)=\sqrt{3}/2\), \(\cos(2\pi/3)=-1/2\), and \(\sec(\pi/3)=2\). Keeping these exact values makes the endpoint subtraction transparent.
For tangent and secant patterns, also verify that the interval does not cross a point where cosine is zero. If it does, the integrand may be undefined at an interior point, so the ordinary substitution formula for a continuous integrand on the whole interval cannot be applied as written. The examples above avoid such points, and their transformed integrands are continuous across their full intervals.
A Common Mistake: Leaving the Old Limits in Place
Consider the sine example \(\int_0^{\pi/6}4\sin(4x)\,dx\). If \(u=4x\), the integrand becomes \(\sin u\,du\), but the limits must become \(0\) and \(2\pi/3\). A common incorrect calculation keeps the old limits and writes
This is wrong because \(u\) is not being evaluated over the original \(x\)-interval; its endpoint value corresponding to \(x=\pi/6\) is \(u=2\pi/3\). The correct calculation is
The error is not a trigonometric simplification; it is a mismatch between the integration variable and its limits. Either transform the limits and finish in \(u\), or return to the original variable before using the original limits. Do not mix the two choices.
A Reliable Evaluation Sequence
Write \(u=g(x)\) and check whether the remaining factor includes \(g'(x)\), with the correct constant.
Calculate \(g(a)\) and \(g(b)\) from the original lower and upper limits.
Use a sine, cosine, tangent, or secant antiderivative, checking that it is defined over the transformed interval.
Substitute the transformed endpoints, simplify the trigonometric values, and verify by differentiating the corresponding antiderivative.
Why These Integrals Matter
Definite substitution lets an integral be evaluated in the variable that makes its structure simplest. Composite trigonometric patterns occur whenever a sine, cosine, tangent, or secant is applied to a scaled or otherwise transformed input. Recognizing the derivative factor and carrying the endpoints through the change of variable turns those expressions into standard trigonometric integrals.
The essential discipline is to keep three things aligned: the transformed integrand, the new differential, and the transformed limits. When those pieces agree, the Fundamental Theorem of Calculus gives the definite value directly. When they do not, even a familiar antiderivative can produce an incorrect answer.
Check Your Understanding
For each question, state any domain condition needed and use exact values where possible.
- Evaluate \(\displaystyle\int_0^{\pi/4}2\sin(2x)\,dx\), showing the transformed endpoints.
- For \(\displaystyle\int_0^{\pi/6}3\cos(3x)\,dx\), what are the new limits after setting \(u=3x\), and what is the integral's value?
- Evaluate \(\displaystyle\int_0^{\pi/8}2\sec^2(2x)\,dx\). Verify the result by differentiating an antiderivative.
- Evaluate \(\displaystyle\int_0^{\pi/9}3\tan(3x)\,dx\), and explain why the tangent antiderivative is valid on the transformed interval.
- In a definite substitution, why is it incorrect to change the integrand to a new variable but retain the old numerical limits?
