Trigonometric Substitution and Branches

Turn a radical into an identity

Three recurring shapes suggest substitutions, with $a>0$ :

$\sqrt{a^2-x^2}$ : use $x=a\sin\theta$ .
$\sqrt{a^2+x^2}$ : use $x=a\tan\theta$ .
$\sqrt{x^2-a^2}$ : use a suitable secant substitution on one domain branch.

The substitution is only part of the work. Replace $dx$ , choose an angle interval, simplify the radical using its nonnegative square root, then return to the original variable. In general $\sqrt{u^2}=|u|$ , not $u$ .

Visual guide

VISUAL GUIDEA triangle explains the radical

Right triangle

For x = a sin θ with −π/2 ≤ θ ≤ π/2, √(a² − x²) = a cos θ. The pictured first-quadrant triangle uses a = 5 and x = 3, giving the adjacent side 4. The angle restriction controls the sign of cosine.

Worked example: a circular radical

For $\int dx/\sqrt{9-x^2}$ on $-3<x<3$ , let $x=3\sin\theta$ with $-\pi/2<\theta<\pi/2$ . Cosine is positive on this interval, so $\sqrt{9-x^2}=3\cos\theta$ and $dx=3\cos\theta\,d\theta$ .

The factors cancel and the integral is $\theta+C=\arcsin(x/3)+C$ . The branch choice makes the cancellation valid. The antiderivative is finite at some endpoint values, but the original integrand is singular at $x=\pm3$ .

Worked example: a less immediate simplification

Consider $\int x^2/\sqrt{4-x^2}\,dx$ on $(-2,2)$ . Set $x=2\sin\theta$ on the same principal sine branch. Then the integrand becomes $4\sin^2\theta\,d\theta$ . Using a half-angle identity gives $2\theta-\sin2\theta+C$ .

Now $\theta=\arcsin(x/2)$ , $\sin\theta=x/2$ , and $\cos\theta=\sqrt{4-x^2}/2$ . Thus

\int\frac{x^2}{\sqrt{4-x^2}}dx=2\arcsin(x/2)-\frac{x\sqrt{4-x^2}}2+C.

The triangle or inverse relation is a conversion tool, not a reason to assume every original input is positive.

Completing the square first

A radical such as $\sqrt{x^2+4x+8}$ becomes $\sqrt{(x+2)^2+4}$ . First shift to $u=x+2$ , then use the positive-sum pattern. This separates an algebraic simplification from the trigonometric change of variable.

For $\sqrt{x^2-a^2}$ , the domain has two disconnected branches. A secant substitution must be chosen to represent the branch being studied, and the absolute value in $\sqrt{\tan^2\theta}$ must be respected. Hyperbolic substitutions can be convenient alternatives but have their own domain bookkeeping.

Practice

Evaluate $\int_0^1 dx/\sqrt{4-x^2}$ .
Which substitution matches $\sqrt{16+x^2}$ ?
Why can $\sqrt{9\cos^2\theta}$ not always be replaced by $3\cos\theta$ ?

Show worked solutions

The antiderivative is $\arcsin(x/2)$ , giving $\pi/6$ .
Use $x=4\tan\theta$ with $-\pi/2<\theta<\pi/2$ , where secant is positive.
The square root is $3|\cos\theta|$ . Removing the absolute value requires an angle interval where cosine is nonnegative.