Trigonometric Substitution

A square root of a quadratic — \sqrt{a^2 - x^2}, \sqrt{a^2 + x^2}, \sqrt{x^2 - a^2} — is poison to ordinary substitution: there is no inner derivative lying around to absorb. The cure is a clever trigonometric substitution that uses a Pythagorean identity to kill the root outright.

\sqrt{a^2 - x^2} → x = a\sin\theta, using 1 - \sin^2\theta = \cos^2\theta.
\sqrt{a^2 + x^2} → x = a\tan\theta, using 1 + \tan^2\theta = \sec^2\theta.
\sqrt{x^2 - a^2} → x = a\sec\theta, using \sec^2\theta - 1 = \tan^2\theta.

Each substitution is engineered so the quadratic becomes a perfect square and the root dissolves into a single trig function.

Worked example: \displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}}

The root is \sqrt{a^2 - x^2}, so we reach for x = a\sin\theta.

Step 1 — substitute x = a\sin\theta and differentiate. Then dx = a\cos\theta\, d\theta:

x = a\sin\theta, \qquad dx = a\cos\theta\, d\theta.

Step 2 — simplify the root. Factor out a^2 and apply 1 - \sin^2\theta = \cos^2\theta:

\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = \sqrt{a^2(1 - \sin^2\theta)} = \sqrt{a^2\cos^2\theta} = a\cos\theta.

The root is gone — that is the whole point. (We take \theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}], where \cos\theta \ge 0, so \sqrt{\cos^2\theta} = \cos\theta with no absolute value.)

Step 3 — assemble the new integral. Put the dx from Step 1 over the simplified root from Step 2:

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

Step 4 — cancel. The a\cos\theta top and bottom annihilate, leaving the easiest integral there is:

\int \frac{a\cos\theta\, d\theta}{a\cos\theta} = \int 1\, d\theta = \theta + C.

Step 5 — back-substitute to x. From x = a\sin\theta we get \sin\theta = x/a, hence \theta = \arcsin(x/a):

\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\!\left(\frac{x}{a}\right) + C.

The area-of-a-circle integral

The same substitution, applied to \int \sqrt{a^2 - x^2}\, dx (no longer in the denominator), gives — after a power-reduction step from trig integrals —

\int \sqrt{a^2 - x^2}\, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\!\left(\frac{x}{a}\right) + C.

As a sanity check, run it across the full semicircle. The curve y = \sqrt{a^2 - x^2} is the upper half of the circle of radius a, so its area from -a to a should be half of \pi a^2:

\int_{-a}^{a} \sqrt{a^2 - x^2}\, dx = \frac{a^2}{2}\Big[\arcsin(1) - \arcsin(-1)\Big] = \frac{a^2}{2}\left(\frac{\pi}{2} + \frac{\pi}{2}\right) = \frac{\pi a^2}{2}.

Exactly the area of a semicircle — calculus agreeing with geometry.

For an integrand containing a root of a quadratic (here a > 0):

\sqrt{a^2 - x^2}: substitute x = a\sin\theta, giving \sqrt{a^2 - x^2} = a\cos\theta and dx = a\cos\theta\, d\theta.
\sqrt{a^2 + x^2}: substitute x = a\tan\theta, giving \sqrt{a^2 + x^2} = a\sec\theta and dx = a\sec^2\theta\, d\theta.
\sqrt{x^2 - a^2}: substitute x = a\sec\theta, giving \sqrt{x^2 - a^2} = a\tan\theta and dx = a\sec\theta\tan\theta\, d\theta.

After integrating in \theta you must return to x, and that is where a reference right triangle earns its keep. For x = a\sin\theta, read \sin\theta = \tfrac{x}{a} = \tfrac{\text{opposite}}{\text{hypotenuse}}: draw a right triangle with hypotenuse a and the side opposite \theta equal to x. Pythagoras then fills in the adjacent side as \sqrt{a^2 - x^2} — and every trig function of \theta can be read straight off the triangle. For example \cos\theta = \tfrac{\sqrt{a^2 - x^2}}{a}, which is exactly how the \tfrac{x}{2}\sqrt{a^2-x^2} term re-materialises in x. The diagram below is that triangle.

And whenever the integrand is the top half of a circle, the semicircle area \tfrac12 \pi a^2 is a free correctness check on the whole computation — as we just saw.

The reference triangle, step by step

Step through the construction of the right triangle that turns \theta back into x. Hypotenuse a, opposite side x, and — by Pythagoras — adjacent side \sqrt{a^2 - x^2}.