Why Simplify Before Substituting?
In a mixed sine-cosine integral, a factor that looks suitable for substitution may not be enough by itself. The remaining powers must also be rewritten in terms of the proposed new variable. A useful order is to simplify the trigonometric powers first, identify the differential factor second, and then substitute.
For odd powers, this often means separating one sine or cosine factor and using \(\sin^2 x+\cos^2 x=1\) to rewrite the rest. When both powers are even, a power-reduction identity can turn the product into a sum of simpler trigonometric terms. The examples below show both patterns.
Two Ways to Reduce Powers
The Pythagorean identity gives a direct way to replace the square of one trigonometric function with the other:
For example, if an integrand contains \(\sin^3 x\cos^2 x\), separate \(\sin^3 x=\sin^2 x\sin x\). Rewrite the squared sine in terms of cosine, leaving a factor \(\sin x\,dx\) that matches the derivative of \(-\cos x\).
When both powers are even, the half-angle power-reduction identities are often more effective:
These identities can reduce products of even powers to sums of cosines, which can then be integrated term by term. A linear substitution such as \(u=4x\) may help organize the integration of a multiple-angle term; its differential must be adjusted as well.
Worked Example: An Odd Sine Power and an Even Cosine Power
Worked Example: Reducing with \(u=\cos x\)
Evaluate \(\int \sin^3 x\cos^2 x\,dx\).
First separate one factor of sine, then use \(\sin^2 x=1-\cos^2 x\):
Set \(u=\cos x\). Then \(du=-\sin x\,dx\), so \(\sin x\,dx=-du\). The remaining factors become \(1-u^2\) and \(u^2\):
Check by differentiating the result:
The derivative matches the integrand. The minus sign from \(du=-\sin x\,dx\) is essential.
Worked Example: An Odd Cosine Power
Worked Example: Reducing with \(u=\sin x\)
Evaluate \(\int \sin^2 x\cos^3 x\,dx\).
Separate one factor of cosine and rewrite the remaining square using \(\cos^2 x=1-\sin^2 x\):
Set \(u=\sin x\), so \(du=\cos x\,dx\). The integral becomes a polynomial in \(u\):
Check the answer by differentiating:
Here the cosine factor supplies the differential directly. Rewriting the other cosine factors as \(1-\sin^2 x\) ensures that the whole remaining expression is in \(u\).
Worked Example: Both Powers Even
Worked Example: Power Reduction Before a Linear Substitution
Evaluate \(\int \sin^2 x\cos^2 x\,dx\).
Use the half-angle identities. Multiplying them gives:
The last equality follows from \(\cos^2 2x=(1+\cos 4x)/2\). Now integrate the reduced expression. For the multiple-angle term, let \(u=4x\), so \(du=4\,dx\) and \(dx=du/4\):
Check by differentiating:
The substitution contributes a factor of \(1/4\), which combines with the outside factor \(1/8\) to give \(1/32\). Keeping these constants visible prevents a common coefficient error.
Why the Combined Method Works
The simplification step and the substitution step serve different purposes. A trigonometric identity rewrites powers without changing the value of the integrand. A substitution changes the variable of integration, including the differential. When the identity has expressed all remaining trigonometric factors in the chosen variable, the substitution converts the integral into a form that can be integrated directly.
Look for an odd sine or cosine power, or recognize that both powers are even.
For an odd power, separate one factor and rewrite the leftover square. For even powers, consider power reduction.
Choose a substitution whose derivative matches the separated factor, including its sign and any constant.
Integrate in the new variable and substitute the original trigonometric expression back.
Differentiate the proposed antiderivative and confirm that it recovers the original integrand.
The reason substitution works is the chain rule in reverse. If \(u=g(x)\), then \(du=g'(x)\,dx\). An integral containing the factor \(g'(x)\,dx\) and all other factors expressible in \(g(x)\) can therefore be rewritten as an integral in \(u\). The preliminary reduction is what makes that last condition hold.
A Common Mistake: Substituting Before Reducing
In the first example, setting \(u=\cos x\) gives \(du=-\sin x\,dx\). A tempting but incorrect step is to replace \(\sin x\,dx\) and then treat \(\sin^2 x\) as if it were another factor of \(u\):
This is wrong because \(u=\cos x\), not \(\sin x\), and one factor of \(\sin^2 x\) remains after using \(du\). Correctly reduce that square first:
A second frequent error is dropping a constant in a linear substitution. If \(u=4x\), then \(du=4\,dx\), not \(du=dx\). Solving for the differential gives \(dx=du/4\), so that factor must remain in the integral.
When Each Route Is Useful
| Pattern in the integrand | Useful first move |
|---|---|
| Odd power of sine | Separate one sine factor; rewrite the remaining even sine power in cosine. |
| Odd power of cosine | Separate one cosine factor; rewrite the remaining even cosine power in sine. |
| Both powers even | Apply power-reduction identities to express the integrand using multiple angles. |
These are practical patterns, not rules that every integral must follow. The aim is to find a simplification that matches a derivative factor and makes the remaining expression manageable. Once an antiderivative is found, differentiation is a reliable check for both algebra and signs.
Check Your Understanding
For each question, identify the reduction or substitution that makes the integral easier before carrying out the calculation.
- In \(\int \sin^5 x\cos^2 x\,dx\), which factor would you separate, and what identity would you use for the remaining sine power?
- For \(\int \sin^2 x\cos^3 x\,dx\), why is \(u=\sin x\) a suitable substitution after reducing the cosine power?
- Which type of identity is a natural starting point for \(\int \sin^2 x\cos^2 x\,dx\), and why?
- If \(u=6x\), what is \(dx\) in terms of \(du\)?
- What must be checked about the remaining factors before a proposed substitution can be applied?
