When the Cosine Substitution Fits
An integral involving both sine and cosine can sometimes be simplified by treating cosine as the new variable. The key pattern is an odd power of sine multiplied by a factor that is rational in cosine. The odd power leaves one factor of \(\sin x\,dx\) to become a differential, while the remaining even power can be written using \(\cos x\).
This method does not apply to every trigonometric integral. It is useful when the factors left after extracting one sine factor can all be expressed rationally in \(\cos x\). The resulting integral is then rational in the new variable and can be handled using standard methods for rational functions.
Deriving the Transformation
Set \(t=\cos x\). Differentiating gives \(dt=-\sin x\,dx\), or \(\sin x\,dx=-dt\). Also, from \(\sin^2 x+\cos^2 x=1\), we have \(\sin^2 x=1-t^2\). Therefore,
The minus sign is part of the substitution: it comes from the derivative of cosine. It should not be dropped, even when the transformed integral is later rewritten with its bounds in increasing order.
Because \(k\) is an integer, \((1-t^2)^k\) is a polynomial. Multiplying it by the rational function \(R(t)\) still gives a rational function. Thus the substitution reduces the trigonometric integral to a rational integral.
Why the Substitution Works
The transformation can be justified directly by the chain rule. Suppose \(R\) is defined for the values of \(\cos x\) under consideration, and let \(G\) be an antiderivative of \(-R(t)(1-t^2)^k\) on the corresponding \(t\)-interval. Then
Apply the chain rule to \(G(\cos x)\). Using \(t=\cos x\) and \(1-\cos^2x=\sin^2x\) gives
So \(G(\cos x)\) is an antiderivative of the original integrand wherever the expressions are defined. This argument also shows why the substitution does not require choosing an inverse cosine for an indefinite integral.
Worked Examples
Worked Example: An Odd Sine Power Over a Linear Cosine Factor
Evaluate \(\int \frac{\sin^3 x}{2+\cos x}\,dx\). The denominator is rational in \(\cos x\), and \(\sin^3x\) contains one sine factor for the differential and one even power.
Let \(t=\cos x\). Then \(dt=-\sin x\,dx\), and \(\sin^2x=1-t^2\). Hence
Divide the numerator by the denominator. Since \((-t+2)(t+2)=-t^2+4\), the remainder needed to obtain \(1-t^2\) is \(-3\). Therefore,
Including the minus sign from the substitution,
Integrate each term and substitute \(t=\cos x\):
To check, differentiating the expression in \(t\) gives \[ t-2+\frac{3}{t+2} = \frac{t^2-1}{t+2}. \] Multiplying by \(dt/dx=-\sin x\) gives \[ -\sin x\,\frac{\cos^2x-1}{\cos x+2} = \frac{\sin x(1-\cos^2x)}{2+\cos x} = \frac{\sin^3x}{2+\cos x}, \] as required.
Worked Example: A Fifth Power and a Quadratic Denominator
Evaluate \(\int \frac{\sin^5x}{1+\cos^2x}\,dx\). Use \(t=\cos x\), so that \(\sin^4x=(1-t^2)^2\) and \(\sin x\,dx=-dt\):
Expand the numerator and divide:
Thus the transformed integrand, including the minus sign, is
Integrating and returning to \(x\) gives
For a check, the derivative of the expression in \(t\) is \(-t^2+3-\frac{4}{1+t^2}\). Multiplying by \(dt/dx=-\sin x\) gives \[ \sin x\left(t^2-3+\frac{4}{1+t^2}\right) = \sin x\,\frac{t^4-2t^2+1}{1+t^2} = \frac{\sin x(1-t^2)^2}{1+t^2} = \frac{\sin^5x}{1+\cos^2x}. \]
Worked Example: Definite Bounds and Orientation
Evaluate \(\int_0^{\pi/2}\frac{\sin^5x}{1+\cos x}\,dx\). The substitution sends the lower bound \(x=0\) to \(t=1\), and the upper bound \(x=\pi/2\) to \(t=0\). Keeping the minus sign and the transformed bounds together gives
Factor \(1-t^2=(1-t)(1+t)\). Since \(1+t\) is nonzero on this interval,
Therefore,
The result is \(\frac{5}{12}\). The bounds are reversed because cosine decreases from \(1\) to \(0\) on \([0,\pi/2]\); the minus sign in the differential compensates for that reversal.
Definite Integrals and the Interval
For a definite integral, the bounds must be transformed along with the differential. If \(x\) runs from \(a\) to \(b\), then \(t=\cos x\) runs from \(\cos a\) to \(\cos b\), and the transformed integral is
A particularly straightforward change of variables is when cosine is monotone on \([a,b]\), so each \(t\)-value in the resulting interval corresponds to at most one \(x\)-value. If cosine is not monotone, split the \(x\)-interval into pieces on which it is monotone before applying the ordinary one-to-one change-of-variables rule.
There is also a useful chain-rule perspective. If the transformed integrand has a primitive throughout the range of \(\cos x\), the composed primitive gives the endpoint formula directly, even if the path of \(t=\cos x\) turns around. For example, on \([\pi/2,3\pi/2]\), cosine goes from \(0\) down to \(-1\) and then back to \(0\). Splitting at \(x=\pi\) makes the cancellation explicit:
This applies to the original integral \(\int_{\pi/2}^{3\pi/2}\frac{\sin^3x}{2+\cos x}\,dx\). The denominator does not vanish on the relevant range \(t\in[-1,0]\), and the two pieces cancel. If a rational factor has a pole in the range, the integral must instead be considered on intervals where the integrand is defined; endpoint substitution alone does not remove a singularity.
A Common Sign Error
A frequent mistake is to write \(dt=\sin x\,dx\). That is the derivative of \(-\cos x\), not of \(\cos x\). For example, the incorrect calculation
loses the minus sign. In fact, the derivative with respect to \(x\) of an antiderivative of the right-hand side is the negative of the original integrand, because \(dt/dx=-\sin x\). The correction is
For definite integrals, another common error is to reverse the bounds but forget to account for the minus sign—or to keep the minus sign and reverse the bounds a second time. Transform the differential and both bounds first, then simplify the signs.
When to Use the Method
Before substituting, check the structure rather than relying on the appearance of the integrand. Extract one factor of \(\sin x\); express every remaining factor using \(\cos x\); then use \(\sin^2x=1-\cos^2x\). If this leaves a rational function of \(\cos x\), the substitution is a good fit. If other factors cannot be written rationally in cosine, this substitution may not simplify the integral.
- An odd sine power is needed to supply the factor \(\sin x\,dx\).
- The other factors must be expressible as a rational function of \(\cos x\).
- The transformed denominator must remain nonzero on the interval being integrated.
- For definite integrals, map both bounds and preserve the orientation sign.
The method is valuable because it turns a trigonometric integral into an algebraic problem without approximations. Its reliability depends on two small but essential details: the identity \(\sin^2x=1-\cos^2x\) and the negative sign in \(d(\cos x)=-\sin x\,dx\).
Check Your Understanding
For each question, identify the transformed integrand and pay particular attention to the sign and the interval.
- Under \(t=\cos x\), what does \(\int \frac{\sin^7x}{3+\cos^2x}\,dx\) become in the variable \(t\)?
- Why does an odd power of sine leave a factor suitable for \(dt\), while the remaining sine power can be replaced by \(1-t^2\)?
- What are the transformed bounds when \(x\) runs from \(0\) to \(\pi/3\)?
- On what interval of \(x\) is cosine monotone on \([0,\pi]\), and why can splitting at \(\pi/2\) help with a direct change of variables?
- What sign error results from using \(dt=\sin x\,dx\) when \(t=\cos x\)?
