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Three radicals cleared by three identities, with the range of $\theta$ as a hypothesis and a right triangle to convert back.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to recognise which of the three quadratic radicals you are looking at, choose the substitution whose Pythagorean identity matches that pattern of signs, fix the range of $\theta$ and say what dropping the absolute value depends on, replace $x$, the radical and $dx$ together, convert an indefinite answer back with a reference triangle or change the limits of a definite one, and recognise the cases where an ordinary substitution is far shorter.
Every substitution so far has simplified the integrand: you spotted an inner function whose derivative was present and replaced it with a single letter. The integral got shorter.
This one goes the other way. It replaces $x$ — a perfectly simple thing — with $a\sin\theta$, which is not simple at all, and the integrand gets longer before it gets shorter. The reason is that a square root of a quadratic cannot be removed by algebra, and the Pythagorean identities are the only tools that turn a sum or difference of squares into a perfect square.
After the integral is evaluated the answer is in $\theta$, and the question was in $x$. The reference triangle is how you convert back: draw a right triangle in which the substitution is true, read the other ratios off it, and substitute them.
For $x = a\sin\theta$ the triangle has opposite $x$, hypotenuse $a$, and therefore adjacent $\sqrt{a^2 - x^2}$; so $\cos\theta = \dfrac{\sqrt{a^2 - x^2}}{a}$ and $\tan\theta = \dfrac{x}{\sqrt{a^2 - x^2}}$. A definite integral needs none of this: change the limits with the variable and the answer is a number in either language.
A square root of a quadratic can always be brought, by completing the square, to one of exactly three forms, and each is cleared by exactly one substitution.
| Radical | Substitute | Identity used | Becomes |
|---|---|---|---|
| $\sqrt{a^2 - x^2}$ | $x = a\sin\theta$ | $1 - \sin^2 = \cos^2$ | $a\cos\theta$ |
| $\sqrt{a^2 + x^2}$ | $x = a\tan\theta$ | $1 + \tan^2 = \sec^2$ | $a\sec\theta$ |
| $\sqrt{x^2 - a^2}$ | $x = a\sec\theta$ | $\sec^2 - 1 = \tan^2$ | $a\tan\theta$ |
The range of $\theta$ is part of the substitution. Take $\theta \in [-\pi/2, \pi/2]$ for sine and tangent, and $\theta \in [0, \pi/2)$ for secant with $x \ge a$. Those choices make the resulting $\cos\theta$, $\sec\theta$ or $\tan\theta$ non-negative, which is what permits dropping the absolute value when the square root is taken. Without them the third line of the table is false half the time.
What replaces what. Substituting $x = a\sin\theta$ also replaces $dx$ by $a\cos\theta\,d\theta$; forgetting that is the most common way the calculation fails silently.
When it is not needed. If the derivative of the inside is present — $\int x\sqrt{a^2 - x^2}\,dx$ has an $x$ sitting outside — then an ordinary substitution $u = a^2 - x^2$ is far shorter. Trigonometric substitution is for the radicals that are alone.
Another way: picture
Draw the right triangle with hypotenuse $a$ and one leg $x$. The angle $\theta$ opposite $x$ satisfies $\sin\theta = x/a$, and the remaining leg is $\sqrt{a^2 - x^2}$ by Pythagoras. The substitution is nothing more than the decision to describe the point by its angle instead of by its coordinate — and the triangle is how you get back.
Another way: steps
$\sqrt{a^2 - x^2}$ with $x = a\sin\theta$ is $\sqrt{a^2\cos^2\theta} = a|\cos\theta|$, not $a\cos\theta$. The two differ wherever $\cos\theta < 0$, which is most of the circle.
Restricting $\theta$ to $[-\pi/2, \pi/2]$ is what removes the absolute value, and it costs nothing because $\sin$ already covers the whole of $[-1, 1]$ on that interval — the substitution reaches every $x$ the integrand is defined at. That is the sense in which the restriction is free: it is not a loss of generality, it is a choice of which of two equally good $\theta$ to use.
For the secant substitution it is not free. $\sec\theta$ on $[0, \pi/2)$ reaches only $x \ge a$, so $x \le -a$ is a separate case needing its own range, and an integral crossing between them is improper anyway because the integrand is undefined on $(-a, a)$. Handling only one branch and quoting the answer for both is a real error, and it is the one this substitution is most often used to commit. A test is its hypotheses. Every convergence test in this course is a theorem with conditions, and a test quoted where its conditions fail has not been applied — it has been guessed with. Naming the hypothesis you checked is part of the answer, not a flourish on top of it.
Forgetting to replace $dx$. $x = a\tan\theta$ means $dx = a\sec^2\theta\,d\theta$. An integral with the right integrand and the wrong differential is simply a different integral.
Dropping the absolute value without the range. See above; it is a hypothesis, not a convention.
Reaching for it when an ordinary substitution works. A factor of $x$ outside the root means $u = a^2 \pm x^2$ finishes in two lines. Trigonometric substitution would also work, in about fifteen.
Converting back with algebra instead of the triangle. $\theta = \arcsin(x/a)$ is correct but useless: the answer usually contains $\sin 2\theta$ or $\tan\theta$, and writing $\sin(2\arcsin(x/a))$ is not an answer anybody can read. Draw the triangle.
Keeping the old limits. If you substitute in a definite integral and do not change the limits, you have written $\int_{-a}^{a}$ of a function of $\theta$, which means nothing.
Not completing the square first. $\sqrt{x^2 + 6x + 13}$ is none of the three forms until it is written $\sqrt{(x+3)^2 + 4}$ — which is the next lesson.
Every substitution up to now was justified by making things simpler, so a learner meeting $x = a\sin\theta$ reasonably objects: the integrand had one square root, and now it has sines, cosines, a $d\theta$ and a trigonometric integral to do. This looks like a step backwards, and for about two lines it is.
What has been bought is the removal of the root. $\sqrt{a^2 - x^2}$ cannot be simplified by any algebraic operation — there is no rearrangement of a difference of squares under a root that gets rid of it. But $\sqrt{a^2 - a^2\sin^2\theta}$ is $\sqrt{a^2\cos^2\theta}$, and a square root of a square is not a difficulty at all. The identity is the only tool available that turns the quadratic into a perfect square, and that is why the method exists.
So the test of whether you have chosen correctly is not "does this look shorter?" but "has the radical gone?" If it has, the rest is the parity rules from the previous lesson and a triangle. If it has not — which happens when the wrong one of the three is chosen — the substitution has cost you everything and bought nothing, and the fix is to look at the signs again.
$\displaystyle\int \dfrac{dx}{\sqrt{4 + x^2}}$: a constant plus a square, so $x = 2\tan\theta$ and $dx = 2\sec^2\theta\,d\theta$.
Read the signs; replace $x$ and $dx$ together.
The radical becomes $2\sec\theta$, so the integral is $\displaystyle\int \dfrac{2\sec^2\theta}{2\sec\theta}d\theta = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C$.
Almost everything cancels, which is the sign of the right choice.
The triangle has opposite $x$, adjacent $2$, hypotenuse $\sqrt{4 + x^2}$, so $\sec\theta = \dfrac{\sqrt{4+x^2}}{2}$ and $\tan\theta = \dfrac{x}{2}$: the answer is $\ln\left|\dfrac{\sqrt{4 + x^2} + x}{2}\right| + C$.
Read the ratios off the triangle.
$\displaystyle\int_0^{3}\dfrac{dx}{(9 + x^2)^{3/2}}$: take $x = 3\tan\theta$, $dx = 3\sec^2\theta\,d\theta$, and $(9 + x^2)^{3/2} = 27\sec^3\theta$.
The power of the radical is no obstacle; the identity clears it whole.
Limits: $x = 0$ gives $\theta = 0$; $x = 3$ gives $\tan\theta = 1$, so $\theta = \pi/4$.
Change the limits and the triangle is never needed.
$\displaystyle\int_0^{\pi/4}\dfrac{3\sec^2\theta}{27\sec^3\theta}d\theta = \dfrac{1}{9}\int_0^{\pi/4}\cos\theta\,d\theta = \dfrac{1}{9}\cdot\dfrac{\sqrt2}{2} = \dfrac{\sqrt2}{18}$.
A number, in either variable.
A constant minus a square, so $x = 2\sin\theta$ with $\theta \in [-\pi/2, \pi/2]$, and $dx = 2\cos\theta\,d\theta$.
The range is what lets the root be $2\cos\theta$.
Limits: $x = 0$ gives $\theta = 0$, $x = 2$ gives $\theta = \pi/2$. The integral is $4\displaystyle\int_0^{\pi/2}\cos^2\theta\,d\theta$.
Change the limits with the variable.
An even power: $\displaystyle\int_0^{\pi/2}\cos^2 = \dfrac{\pi}{4}$, so the answer is $\pi$ — a quarter of the circle of radius $2$, as the picture says.
Match each radical to the substitution that clears it.
| $x = 9\sin\theta$ | $x = 9\tan\theta$ | $x = 9\sec\theta$ | |
|---|---|---|---|
| $\sqrt{81 - x^{2}}$ | |||
| $\sqrt{81 + x^{2}}$ | |||
| $\sqrt{x^{2} - 81}$ |
Which substitution clears $\sqrt{a^2 - x^2}$?
For which $x$ does the substitution $x = 3\sin\theta$, with $\theta$ in $[-\pi/2, \pi/2]$, reach? Give the interval.
This task has no paper form; do it on a device.
Complete the table of the three trigonometric substitutions, with $a = 4$.
| Substitute | The root becomes | |
|---|---|---|
| $\sqrt{16 - x^{2}}$ | ||
| $\sqrt{16 + x^{2}}$ | ||
| $\sqrt{x^{2} - 16}$ |
$\displaystyle\int_{-3}^{3} \sqrt{9 - x^{2}}\,dx = k\pi$. Find $k$.
The coefficient k is answer.
The ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ encloses an area $k\pi$. Find $k$.
The coefficient k is answer.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each radical to the substitution that clears it.
| $x = 2\sin\theta$ | $x = 2\tan\theta$ | $x = 2\sec\theta$ | |
|---|---|---|---|
| $\sqrt{4 - x^{2}}$ | |||
| $\sqrt{4 + x^{2}}$ | |||
| $\sqrt{x^{2} - 4}$ |
You can clear a quadratic radical and say what the restriction on $\theta$ is doing. Say in your own words why making the integrand longer is the method rather than a mistake. Next: the algebra that puts a quadratic into one of these three forms in the first place.
10. Your turn: $\displaystyle\int_0^{2}\sqrt{4 - x^{2}}\,dx$, step 3
And the geometry agrees.