Recipe

Integrate using a trigonometric substitution

Use when: the integrand contains \sqrt{a^2-x^2}, \sqrt{a^2+x^2}, or \sqrt{x^2-a^2} and no algebraic substitution removes the radical

Goal: the radical replaced by a single trig factor, integrated in theta, and converted back to a function of x

Taught in Trigonometric substitution

1. 1

   Compare the expression under the radical to the three standard shapes and pick the matching substitution.

   Example

   \displaystyle\int\frac{dx}{x^2\sqrt{x^2+4}} has the shape \sqrt{a^2+x^2} with a=2, so substitute x=2\tan\theta, dx=2\sec^2\theta\,d\theta.

   • shape is \sqrt{a^2-x^2} — substitute x=a\sin\theta
   • shape is \sqrt{a^2+x^2} — substitute x=a\tan\theta
   • shape is \sqrt{x^2-a^2} — substitute x=a\sec\theta

   Why: each substitution is chosen specifically so the matching Pythagorean identity turns the radical into a perfect square, and only the substitution that matches the sign pattern under the root does that.

2. 2

   Replace x and dx with the chosen substitution, then use the matching Pythagorean identity to collapse the radical into a single trig factor.

   Example

   \sqrt{x^2+4}=\sqrt{4\tan^2\theta+4}=2\sqrt{\tan^2\theta+1}=2\sec\theta, so the integral becomes \int\frac{2\sec^2\theta\,d\theta}{(2\tan\theta)^2(2\sec\theta)} = \int\frac{2\sec^2\theta}{8\tan^2\theta\sec\theta}\,d\theta = \frac14\int\frac{\sec\theta}{\tan^2\theta}\,d\theta

   Why: this is the entire point of the substitution; without it, the radical has no elementary antiderivative in x directly.

3. 3

   Rewrite the trig expression in terms of \sin\theta and \cos\theta if needed, then integrate.

   Example

   \dfrac{\sec\theta}{\tan^2\theta}=\dfrac{\cos\theta}{\sin^2\theta}. With u=\sin\theta, du=\cos\theta\,d\theta: \frac14\int\frac{\cos\theta}{\sin^2\theta}\,d\theta = \frac14\int\frac{du}{u^2} = -\frac{1}{4u}+C = -\frac{1}{4\sin\theta}+C

   Why: after Step 2 the integral is an ordinary trig integral, solvable by the same techniques as any other trig integrand.

4. 4

   Draw a right triangle labeled from the substitution equation, read off whatever trig ratio the answer needs, then substitute it back in terms of x.

   Example

   From \tan\theta=x/2: opposite =x, adjacent =2, hypotenuse =\sqrt{x^2+4}. So \sin\theta=\dfrac{x}{\sqrt{x^2+4}}, and \int\frac{dx}{x^2\sqrt{x^2+4}} = -\frac{1}{4\sin\theta}+C = -\frac{\sqrt{x^2+4}}{4x}+C

   • the antiderivative in theta is theta itself (no other trig function of theta remains) — convert directly with the substitution equation solved for theta (like \theta=\arcsin(x/a)); no triangle is needed.
   • the antiderivative in theta involves \sin\theta, \cos\theta, \tan\theta, or another ratio — build the triangle and read that ratio off it directly, as above.

   Why: the answer has to be a function of x, the original variable, not of theta, which was only ever a bookkeeping device.