Why the Sine Substitution Fits
Some trigonometric integrals become algebraic once their sine and cosine factors are expressed using a single variable. The substitution \(t=\sin x\) is especially useful when the integrand contains an odd power of cosine: one factor of \(\cos x\) pairs with \(dx\), while every remaining cosine factor occurs in an even power and can be replaced using \(\cos^2 x=1-\sin^2 x\).
The sign of cosine still matters when changing limits on a definite integral. On an interval where \(\cos x<0\), the quantity \(t=\sin x\) decreases as \(x\) increases. The correct transformed limits preserve that direction; they should not be reversed merely to make them appear in increasing order.
Setting Up the Substitution
Suppose the integrand has the form
where \(m\) is a nonnegative integer and \(R\) is a rational function: a quotient of polynomials wherever its denominator is nonzero. Separate the odd power of cosine into an even power and one remaining factor:
Now set \(t=\sin x\), so \(dt=\cos x\,dx\). The integral becomes
This is a rational integral in \(t\), which can then be evaluated using algebraic integration methods. When translating an answer back to \(x\), replace \(t\) with \(\sin x\). The substitution is valid wherever the original integrand is defined; in particular, avoid intervals containing points where a denominator in \(R(\sin x)\) is zero.
A Derivation Using the Chain Rule
The substitution formula can also be checked by differentiating. Let \(H\) be an antiderivative of \(R(t)(1-t^2)^m\) on an interval where this function is defined. By the chain rule,
Thus \(H(\sin x)\) is an antiderivative of the original integrand. This verifies the method directly and also explains why the same expression can be used on either side of a point where cosine changes sign, as long as the expression is defined there.
Worked Examples
Worked Example: Simplifying a Denominator After Substitution
Evaluate \(\int \frac{\cos^3 x}{1+\sin x}\,dx\) on any interval where the integrand is defined.
Write \(\cos^3 x=\cos^2 x\cos x\), and let \(t=\sin x\), so \(dt=\cos x\,dx\). Then
Since \(1-t^2=(1-t)(1+t)\), the quotient simplifies to \(1-t\), provided \(t\ne -1\). This restriction matches the original denominator, which is zero when \(\sin x=-1\). Therefore,
To check, differentiate the result:
The original integrand is also \(\cos x(1-\sin x)\), because \[ \frac{\cos^3 x}{1+\sin x} = \frac{\cos x(1-\sin^2 x)}{1+\sin x} = \cos x(1-\sin x) \] where \(1+\sin x\ne0\). The derivative check agrees.
Worked Example: A Higher Odd Power of Cosine
Evaluate \(\int \frac{\cos^5 x}{1+\sin^2 x}\,dx\).
Set \(t=\sin x\), so \(dt=\cos x\,dx\). The remaining fourth power of cosine becomes \((1-t^2)^2\):
Divide the polynomials. The quotient is \(t^2-3\), since \[ (t^2+1)(t^2-3)=t^4-2t^2-3. \] The remainder is \(4\), so
Integrating each term gives
Substituting \(t=\sin x\), the answer is \[ \frac{\sin^3 x}{3}-3\sin x+4\arctan(\sin x)+C. \] For a check, the derivative of the expression as a function of \(t\) is \[ t^2-3+\frac{4}{1+t^2} = \frac{(t^2-3)(1+t^2)+4}{1+t^2} = \frac{t^4-2t^2+1}{1+t^2} = \frac{(1-t^2)^2}{1+t^2}. \] Multiplying by \(dt/dx=\cos x\) gives \(\cos x(1-\sin^2 x)^2/(1+\sin^2 x)\), which equals the original integrand.
Worked Example: Definite Limits When Cosine Is Negative
Evaluate \[ \int_{2\pi/3}^{\pi}\frac{\cos^3 x}{1+\sin x}\,dx. \] On this interval, \(\cos x<0\), so \(\sin x\) decreases as \(x\) increases. With \(t=\sin x\), the lower limit becomes \(\sin(2\pi/3)=\sqrt{3}/2\), and the upper limit becomes \(\sin(\pi)=0\). Thus
Evaluating the bounds,
This value is negative, as expected: \(\cos^3 x<0\) on the interval and \(1+\sin x>0\). The reversed order of the transformed limits carries the sign correctly.
Cosine's Sign and the Limits of Integration
On an interval where \(\cos x>0\), \(t=\sin x\) increases with \(x\). Where \(\cos x<0\), it decreases. The change-of-variables formula uses the actual transformed endpoints in their original order, and therefore accounts for this orientation automatically:
provided the integrands are defined and the substitution is applied over an interval on which \(t=\sin x\) is one-to-one. In particular, an interval on which cosine has a fixed sign is monotone for this substitution. If the original interval crosses a point where \(\cos x=0\), split it there and apply the formula to each monotone piece.
There is no need to write \(\cos x=\sqrt{1-\sin^2 x}\) or \(\cos x=-\sqrt{1-\sin^2 x}\) in this method. The even power is replaced by \(1-t^2\), and the remaining factor \(\cos x\,dx\) is replaced by \(dt\). Introducing a square root can obscure the sign that is already handled by the transformed limits.
A Common Sign Mistake
For the definite integral in the preceding example, a tempting but incorrect step is to reorder the limits from \(0\) to \(\sqrt{3}/2\) without also changing the sign. That would give
which is positive. But \(t\) actually moves from \(\sqrt{3}/2\) down to \(0\), not from \(0\) up to \(\sqrt{3}/2\). Reversing the limits requires a minus sign:
The corrected sign agrees with the sign of the original integrand. A reliable check for definite integrals is to compare the answer's sign with the sign of the integrand over the interval.
When This Method Is Useful
The sine substitution is a direct choice when the sine-dependent factors form a rational function of \(\sin x\) and the differential includes a positive odd power of cosine. It reduces the trigonometric integral to a rational one without needing to solve for \(x\) in terms of \(t\). The essential procedure is to reserve one \(\cos x\,dx\) for \(dt\), then convert every remaining even power of cosine using \(1-\sin^2 x\).
Before applying the method, check two details. First, make sure the denominator is nonzero on the interval of integration, or split the integral at any singularity and treat each defined interval appropriately. Second, for definite integrals, check whether \(\sin x\) is increasing or decreasing. If cosine changes sign inside the interval, split at the zero of cosine so that the substitution is one-to-one on each piece.
Check Your Understanding
For each question, track both the algebra and the orientation of the substitution.
- For an integrand containing \(\cos^7 x\,dx\) and a rational function of \(\sin x\), how many cosine factors become \(dt\), and how can the others be rewritten?
- Use \(t=\sin x\) to set up, but do not necessarily evaluate, \(\int \frac{\cos^3 x}{2+\sin x}\,dx\).
- Why does \(\cos^2 x=1-\sin^2 x\) avoid the need to choose a sign for a square root?
- If a definite integral is taken over an interval where cosine is negative, what happens to the transformed limits?
- What should be done before applying the substitution to an interval that crosses a point where \(\cos x=0\)?
