Polynomial Factors Over √(a²−x²)

Why the Power of $x$ Matters

An integral with a square root in the denominator can look complicated even when its numerator is a simple power of $x$. For integrals of the form $\int \frac{x^n}{\sqrt{a^2-x^2}}\,dx$, the substitution $x=a\sin\theta$ removes the radical. What remains is an integral of a power of sine.

The parity of $n$ then guides the calculation. If $n$ is odd, one sine factor can be set aside for a direct substitution. If $n$ is even, a reduction formula lowers the power by two until the integral of $1$ remains. In both cases, the answer must eventually be written in terms of $x$.

Setup: Assume $a>0$ and let $n$ be a nonnegative integer. We consider the real-valued integrand on $-a<x<a$, where $\sqrt{a^2-x^2}>0$. Choose $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$, so that $\cos\theta>0$.

Transforming the Integral

Set $x=a\sin\theta$. Then $dx=a\cos\theta\,d\theta$, and, because $\cos\theta>0$ on the chosen range, $\sqrt{a^2-x^2}=a\cos\theta$. Thus the differential and radical cancel, while the numerator becomes a power of $a\sin\theta$:

\int \frac{x^n}{\sqrt{a^2-x^2}}\,dx = a^n\int \sin^n\theta\,d\theta.

The factor $a^n$ is important: it comes from the numerator, since $x^n=(a\sin\theta)^n$. The factors $a\cos\theta$ in $dx$ and in the denominator cancel one another, but they do not cancel the $a^n$.

Odd Powers: Save One Sine Factor

Let $n=2m+1$, where $m$ is a nonnegative integer. Write $\sin^{2m+1}\theta$ as $\sin^{2m}\theta\sin\theta$, and use $\sin^2\theta=1-\cos^2\theta$. With $u=\cos\theta$, $du=-\sin\theta\,d\theta$, so the remaining integral is a polynomial integral in $u$:

a^{2m+1}\int \sin^{2m+1}\theta\,d\theta = -a^{2m+1}\int (1-u^2)^m\,du.

Expanding $(1-u^2)^m$ by the binomial theorem gives a finite sum. Since $\cos\theta=\sqrt{a^2-x^2}/a$ on the chosen range, this yields the following expression directly in $x$:

Odd-power formula: For $n=2m+1$, $$ \int \frac{x^{2m+1}}{\sqrt{a^2-x^2}}\,dx = -\sum_{k=0}^{m} \frac{(-1)^k\binom{m}{k}}{2k+1} a^{2m-2k}(a^2-x^2)^{k+\frac12}+C. $$ The formula follows by expanding $(1-u^2)^m$, integrating each power of $u$, and substituting $u=\sqrt{a^2-x^2}/a$.

Worked Example: An Odd Power with $a=2$

Evaluate $\int \frac{x^3}{\sqrt{4-x^2}}\,dx$. Under $x=2\sin\theta$, we have $dx=2\cos\theta\,d\theta$, $\sqrt{4-x^2}=2\cos\theta$, and $x^3=8\sin^3\theta$. Therefore:

\int \frac{x^3}{\sqrt{4-x^2}}\,dx = 8\int \sin^3\theta\,d\theta = 8\int (1-\cos^2\theta)\sin\theta\,d\theta.

Set $u=\cos\theta$, so $du=-\sin\theta\,d\theta$. Then:

8\int (1-\cos^2\theta)\sin\theta\,d\theta = -8\int(1-u^2)\,du = -8\left(u-\frac{u^3}{3}\right)+C.

Since $u=\sqrt{4-x^2}/2$, substitute back:

-8\left(\frac{\sqrt{4-x^2}}{2} -\frac{1}{3}\left(\frac{\sqrt{4-x^2}}{2}\right)^3\right)+C = -4\sqrt{4-x^2} +\frac{1}{3}(4-x^2)^{3/2}+C.

To check, differentiate. Writing $R=4-x^2$, the derivative is:

\frac{d}{dx}\left(-4R^{1/2}+\frac13R^{3/2}\right) = \frac{4x}{\sqrt R}-x\sqrt R = \frac{4x-xR}{\sqrt R} = \frac{x(4-(4-x^2))}{\sqrt R} = \frac{x^3}{\sqrt{4-x^2}}.

This is the original integrand.

Even Powers: Reduce the Power

When $n$ is even, splitting off one sine factor does not leave a polynomial in $\cos\theta$, because the remaining power of sine is still odd only after a different decomposition. A systematic approach is to reduce the sine power by two repeatedly.

For a nonnegative integer $n\geq 2$, let $I_n=\int\sin^n\theta\,d\theta$. Integration by parts with $u=\sin^{n-1}\theta$ and $dv=\sin\theta\,d\theta$ gives $du=(n-1)\sin^{n-2}\theta\cos\theta\,d\theta$ and $v=-\cos\theta$. Therefore:

I_n = -\sin^{n-1}\theta\cos\theta +(n-1)\int\sin^{n-2}\theta\cos^2\theta\,d\theta.

Now use $\cos^2\theta=1-\sin^2\theta$. The integral on the right becomes $I_{n-2}-I_n$, so rearranging gives the recurrence:

nI_n = -\sin^{n-1}\theta\cos\theta+(n-1)I_{n-2}, \qquad I_n = -\frac{\sin^{n-1}\theta\cos\theta}{n} +\frac{n-1}{n}I_{n-2}.

This derivation explains why the method works: the identity for $\cos^2\theta$ turns the integration-by-parts result into an equation involving $I_n$ and $I_{n-2}$, which can be solved for $I_n$. In the original variable, define $F_n$ as an antiderivative of $\frac{x^n}{\sqrt{a^2-x^2}}$. The recurrence becomes:

Even-power reduction: For $n\geq2$, $$ F_n(x) = -\frac{x^{n-1}\sqrt{a^2-x^2}}{n} +\frac{(n-1)a^2}{n}F_{n-2}(x), $$ with base cases $F_0(x)=\arcsin(x/a)$ and $F_1(x)=-\sqrt{a^2-x^2}$. Each recurrence step reduces the power by two.

Applying the Reduction Formula

Worked Example: A Quadratic Numerator

Evaluate $\int \frac{x^2}{\sqrt{9-x^2}}\,dx$. Here $a=3$ and $n=2$. The transformed integral is $9\int\sin^2\theta\,d\theta$. Apply the recurrence with $n=2$:

I_2 = -\frac{\sin\theta\cos\theta}{2} +\frac12 I_0 = -\frac{\sin\theta\cos\theta}{2} +\frac{\theta}{2}.

Multiplying by $9$ and using $\sin\theta=x/3$, $\cos\theta=\sqrt{9-x^2}/3$, and $\theta=\arcsin(x/3)$, we obtain:

9I_2 = -\frac{x\sqrt{9-x^2}}{2} +\frac92\arcsin\left(\frac{x}{3}\right)+C.

To check, write $R=9-x^2$. Differentiating the first term gives $-\frac12\sqrt R+\frac{x^2}{2\sqrt R}$, and differentiating the second gives $\frac{9}{2\sqrt R}$. Their sum is:

-\frac{\sqrt R}{2} +\frac{x^2+9}{2\sqrt R} = \frac{-R+x^2+9}{2\sqrt R} = \frac{x^2}{\sqrt R} = \frac{x^2}{\sqrt{9-x^2}}.

The derivative agrees with the integrand.

Worked Example: A Fourth Power Numerator

Evaluate $\int \frac{x^4}{\sqrt{4-x^2}}\,dx$. Use the recurrence first at $n=4$, then at $n=2$:

F_4 = -\frac{x^3\sqrt{4-x^2}}{4} +\frac{3\cdot4}{4}F_2 = -\frac{x^3\sqrt{4-x^2}}{4}+3F_2, $$ $$ F_2 = -\frac{x\sqrt{4-x^2}}{2} +\frac{4}{2}F_0 = -\frac{x\sqrt{4-x^2}}{2} +2\arcsin\left(\frac{x}{2}\right).

Substitute the expression for $F_2$ into $F_4$, multiplying each term by $3$:

F_4 = -\frac{x^3\sqrt{4-x^2}}{4} -\frac{3x\sqrt{4-x^2}}{2} +6\arcsin\left(\frac{x}{2}\right)+C.

For a derivative check, put $R=4-x^2$. The derivative of the first term is $-\frac{3x^2\sqrt R}{4}+\frac{x^4}{4\sqrt R}$. The derivative of the second is $-\frac{3\sqrt R}{2}+\frac{3x^2}{2\sqrt R}$, and that of the third is $6/\sqrt R$. Combining over the common denominator $4\sqrt R$ gives:

\frac{-3x^2R+x^4-6R+6x^2+24}{4\sqrt R} = \frac{4x^4}{4\sqrt R} = \frac{x^4}{\sqrt{4-x^2}}.

In the simplification, $-3x^2R=-12x^2+3x^4$ and $-6R=-24+6x^2$; the remaining terms combine to $4x^4$. Thus the antiderivative checks.

Choosing the Efficient Route

Both parity cases begin with the same substitution, but they simplify differently. The useful distinction is not merely whether the transformed integral looks trigonometric; it is whether a factor can be isolated to make a direct substitution, or whether reducing the power is more efficient.

Power of $x$	Useful step after substitution	Typical endpoint
Odd: $n=2m+1$	Save $\sin\theta\,d\theta$ and set $u=\cos\theta$.	A finite sum of powers of $a^2-x^2$.
Even: $n=2m$	Apply the sine-power reduction formula until $I_0$.	Radical terms and $\arcsin(x/a)$.

The inverse sine appears in the even case because the reduction eventually reaches $I_0=\int 1\,d\theta=\theta$. Returning to $x$ gives $\theta=\arcsin(x/a)$ on the chosen range. For odd powers, the direct substitution integrates powers of $\cos\theta$ and produces only powers of the radical.

A Common Scaling Error

A frequent mistake is to cancel the differential against the radical correctly but then forget the scale factor in $x^n$. For example, with $a=3$ and $n=2$, writing $x=3\sin\theta$ gives $dx=3\cos\theta\,d\theta$, $\sqrt{9-x^2}=3\cos\theta$, and $x^2=9\sin^2\theta$. The correct transformed integral is:

\int \frac{x^2}{\sqrt{9-x^2}}\,dx = \int\frac{9\sin^2\theta}{3\cos\theta}\,3\cos\theta\,d\theta = 9\int\sin^2\theta\,d\theta.

A wrong calculation might cancel the factors $3\cos\theta$ and then write just $\int\sin^2\theta\,d\theta$, dropping the $9$. That answer is too small by a factor of $9$. Keeping $x^2=(3\sin\theta)^2=9\sin^2\theta$ visible during the substitution prevents the error.

Summary

For $\int \frac{x^n}{\sqrt{a^2-x^2}}\,dx$, the sine substitution removes the radical and gives $a^n\int\sin^n\theta\,d\theta$. For odd $n$, save one sine factor and substitute $u=\cos\theta$. For even $n$, use the recurrence to lower the power by two, ending at $I_0=\theta$. In either method, respect the chosen angle range when replacing $\cos\theta$ by the positive square root, retain every factor of $a$, and differentiate the final expression to check it.

Check Your Understanding

Use the parity of the power to decide which method is most efficient, and check each result by differentiation.

Under $x=a\sin\theta$, what do $dx$ and $\sqrt{a^2-x^2}$ become when $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$?
Why does an odd power $x^{2m+1}$ allow a direct substitution $u=\cos\theta$ after rewriting the remaining sine power?
Use the recurrence to express $F_6$ in terms of $F_4$, and then in terms of $F_2$.
Why does the even-power method eventually produce an inverse-sine term?
For $a=4$ and $n=3$, what factor multiplies $\int\sin^3\theta\,d\theta$ after the substitution?

Power of \(x\)	Useful step after substitution	Typical endpoint
Odd: \(n=2m+1\)	Save \(\sin\theta\,d\theta\) and set \(u=\cos\theta\).	A finite sum of powers of \(a^2-x^2\).
Even: \(n=2m\)	Apply the sine-power reduction formula until \(I_0\).	Radical terms and \(\arcsin(x/a)\).

Polynomial Factors Over √(a²−x²)

What You'll Learn

Why the Power of \(x\) Matters

Transforming the Integral

Odd Powers: Save One Sine Factor

Worked Example: An Odd Power with \(a=2\)

Even Powers: Reduce the Power