Corrected exercises: trigonometric substitution (MATH 141)

a) Under $\sqrt{25 - x^2}$ a constant MINUS $x^2$: the identity $1 - \sin^2\theta = \cos^2\theta$ collapses it, so $x = 5\sin\theta$ with $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$, $dx = 5\cos\theta\,d\theta$ and $\sqrt{25 - 25\sin^2\theta} = 5|\cos\theta| = 5\cos\theta$, because $\cos\theta \ge 0$ on that interval. Under $\sqrt{x^2 + 9}$ a SUM: $1 + \tan^2\theta = \sec^2\theta$, so $x = 3\tan\theta$ with $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, $dx = 3\sec^2\theta\,d\theta$ and $\sqrt{9\tan^2\theta + 9} = 3|\sec\theta| = 3\sec\theta$, since $\sec\theta > 0$ there. Under $\sqrt{x^2 - 16}$ $x^2$ MINUS a constant: $\sec^2\theta - 1 = \tan^2\theta$, so $x = 4\sec\theta$, $dx = 4\sec\theta\tan\theta\,d\theta$ and $\sqrt{16\sec^2\theta - 16} = 4|\tan\theta|$, equal to $4\tan\theta$ on $0 \le \theta < \frac{\pi}{2}$ (for $x \ge 4$) and on $\pi \le \theta < \frac{3\pi}{2}$ (for $x \le -4$), the two intervals of Stewart's convention, where $\tan\theta \ge 0$. The rule to memorise is the sign pattern, not the letters: which term is subtracted decides the identity.

b) The form is $a^2 - x^2$ with $a = 5$, and there is no odd power of $x$ outside the root to hand to a plain substitution. So $x = 5\sin\theta$, $dx = 5\cos\theta\,d\theta$, $\sqrt{25 - x^2} = 5\cos\theta$, and $\int \frac{5\cos\theta\,d\theta}{25\sin^2\theta \cdot 5\cos\theta} = \frac{1}{25}\int \csc^2\theta\,d\theta = -\frac{1}{25}\cot\theta + C$. To return to $x$, draw the triangle that $\sin\theta = \frac{x}{5}$ describes: opposite side $x$, hypotenuse $5$, adjacent side $\sqrt{25 - x^2}$ by Pythagoras (figure of the solution, left). Then $\cot\theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{\sqrt{25 - x^2}}{x}$, and $\int \frac{dx}{x^2\sqrt{25 - x^2}} = -\frac{\sqrt{25 - x^2}}{25x} + C$. Leaving $\cot\theta$, or writing $\cot(\arcsin\frac{x}{5})$, costs the return mark: the question is in $x$.

c) Write $F(x) = -\frac{1}{25}\cdot\frac{\sqrt{25 - x^2}}{x}$. By the quotient rule, $\frac{d}{dx}\frac{\sqrt{25 - x^2}}{x} = \frac{\frac{-x}{\sqrt{25 - x^2}}\cdot x - \sqrt{25 - x^2}}{x^2} = \frac{-x^2 - (25 - x^2)}{x^2\sqrt{25 - x^2}} = \frac{-25}{x^2\sqrt{25 - x^2}}$. Multiplying by $-\frac{1}{25}$ gives $F'(x) = \frac{1}{x^2\sqrt{25 - x^2}}$, the integrand. On an exam without a calculator this check costs one minute and is the only proof you have that the triangle was read correctly.

d) Yes. The identity $1 - \cos^2\theta = \sin^2\theta$ works just as well, and on $0 \le \theta \le \pi$ the sine is non-negative, so $\sqrt{25 - x^2} = 5\sin\theta$. Now $dx = -5\sin\theta\,d\theta$, note the MINUS, and the integral becomes $\int \frac{-5\sin\theta\,d\theta}{25\cos^2\theta\cdot 5\sin\theta} = -\frac{1}{25}\int\sec^2\theta\,d\theta = -\frac{1}{25}\tan\theta + C$. The triangle of $\cos\theta = \frac{x}{5}$ has $x$ ADJACENT and $\sqrt{25 - x^2}$ opposite, so $\tan\theta = \frac{\sqrt{25 - x^2}}{x}$ and the answer is again $-\frac{\sqrt{25 - x^2}}{25x} + C$. The cosine choice is legal but carries two extra traps, the sign of $dx$ and a triangle drawn the other way round; the sine is the default for a reason.

e) The domain of $\sqrt{25 - x^2}$ is $[-5, 5]$, swept once by $5\sin\theta$ on $[-\frac{\pi}{2}, \frac{\pi}{2}]$; that of $\sqrt{x^2 + 9}$ is every real, swept once by $3\tan\theta$ on $(-\frac{\pi}{2}, \frac{\pi}{2})$. But $\sqrt{x^2 - 16}$ lives on TWO pieces, $x \ge 4$ and $x \le -4$, and $x = 4\sec\theta$ needs one interval of $\theta$ for each. For a definite integral on the negative piece, the new limits must be read in the interval chosen for that piece, and the sign of $\tan\theta$ must be checked there: with the principal choice $\frac{\pi}{2} < \theta \le \pi$ the tangent is NEGATIVE and the root is $-4\tan\theta$. Exercise 4 shows what forgetting this costs.

a) The form is $a^2 - x^2$ with $a = 3$: $x = 3\sin\theta$, $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$, $dx = 3\cos\theta\,d\theta$, and $\sqrt{9 - 9\sin^2\theta} = 3|\cos\theta| = 3\cos\theta$ because the cosine is non-negative on that interval. The integral becomes $\int 3\cos\theta\cdot 3\cos\theta\,d\theta = 9\int\cos^2\theta\,d\theta$. An even power of the cosine alone calls for the half-angle formula of section 7.2, $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$: $9\int\frac{1 + \cos 2\theta}{2}\,d\theta = \frac{9}{2}\theta + \frac{9}{4}\sin 2\theta + C$. Forgetting $dx$ here, and integrating $3\cos\theta\,d\theta$, gives $3\sin\theta = x$: an answer so simple it should raise an alarm.

b) $\sin\theta = \frac{x}{3}$ describes a right triangle with $x$ opposite $\theta$ and hypotenuse $3$, so the adjacent side is $\sqrt{9 - x^2}$ and $\cos\theta = \frac{\sqrt{9 - x^2}}{3}$, with the $3$ in the denominator. The term $\sin 2\theta$ cannot be read on the triangle as it stands: expand it first, $\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\frac{x}{3}\cdot\frac{\sqrt{9 - x^2}}{3} = \frac{2x\sqrt{9 - x^2}}{9}$. Then $\frac{9}{4}\sin 2\theta = \frac{x}{2}\sqrt{9 - x^2}$, and $\theta = \arcsin\frac{x}{3}$, which is legitimate because $\theta$ was taken in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the range of arcsine. Final answer: $\int\sqrt{9 - x^2}\,dx = \frac{9}{2}\arcsin\frac{x}{3} + \frac{x}{2}\sqrt{9 - x^2} + C$. Writing $\sin 2\theta = \frac{2x}{3}$ is the classic slip: it doubles the angle's sine as if it were linear.

c) $\frac{d}{dx}\left[\frac{9}{2}\arcsin\frac{x}{3}\right] = \frac{9}{2}\cdot\frac{1/3}{\sqrt{1 - x^2/9}} = \frac{9}{2}\cdot\frac{1}{\sqrt{9 - x^2}}$, and by the product rule $\frac{d}{dx}\left[\frac{x}{2}\sqrt{9 - x^2}\right] = \frac{1}{2}\sqrt{9 - x^2} - \frac{x^2}{2\sqrt{9 - x^2}}$. Over the common denominator $\sqrt{9 - x^2}$ the numerator is $\frac{9}{2} + \frac{9 - x^2}{2} - \frac{x^2}{2} = 9 - x^2$, and $\frac{9 - x^2}{\sqrt{9 - x^2}} = \sqrt{9 - x^2}$. The two terms of the antiderivative are not independent: neither one alone differentiates to anything useful, only their sum does.

d) $\int_0^2 \sqrt{9 - x^2}\,dx = \left[\frac{9}{2}\arcsin\frac{x}{3} + \frac{x}{2}\sqrt{9 - x^2}\right]_0^2 = \frac{9}{2}\arcsin\frac{2}{3} + \sqrt 5$. Now the geometry. Join the origin $O$ to $P(2, \sqrt 5)$, which is on the circle since $4 + 5 = 9$. The segment $OP$ cuts the shaded region into the right triangle $O$, $(2, 0)$, $P$, of area $\frac{1}{2}\cdot 2\cdot\sqrt 5 = \sqrt 5$, and the sector between the positive $y$-axis and $OP$. That sector has radius $3$ and an angle $\alpha$ measured from the vertical with $\sin\alpha = \frac{2}{3}$, so its area is $\frac{1}{2}\cdot 9\cdot\alpha = \frac{9}{2}\arcsin\frac{2}{3}$. The antiderivative IS triangle plus sector: $\theta$ is the angle at the centre. To bracket: $\frac{1}{2} < \frac{2}{3} < \frac{\sqrt 2}{2}$, so $\frac{\pi}{6} < \arcsin\frac{2}{3} < \frac{\pi}{4}$, and the value lies between $\frac{3\pi}{4} + \sqrt 5 \approx 4.6$ and $\frac{9\pi}{8} + \sqrt 5 \approx 5.8$. That is consistent with the region itself: it contains the rectangle of height $\sqrt 5$, area $2\sqrt 5 \approx 4.5$, and fits in the rectangle of height $3$, area $6$.

e) $\int_{-3}^{3}\sqrt{9 - x^2}\,dx = \frac{9}{2}\left(\arcsin 1 - \arcsin(-1)\right) + 0 = \frac{9}{2}\cdot\pi = \frac{9\pi}{2}$: the area of the upper half disc, $\frac{1}{2}\pi\cdot 3^2$. And $\int_0^3\sqrt{9 - x^2}\,dx = \frac{9}{2}\cdot\frac{\pi}{2} = \frac{9\pi}{4}$, the quarter disc. The product term vanishes at both ends because the root does. These two values are the standard sanity check of this antiderivative: if your formula does not give $\frac{\pi a^2}{2}$ between $-a$ and $a$, a coefficient is wrong, usually the $\frac{1}{2}$ of the half-angle formula.

a) The form $x^2 + a^2$ with $a = 2$ gives $x = 2\tan\theta$, $dx = 2\sec^2\theta\,d\theta$, $\sqrt{x^2 + 4} = 2\sec\theta$. So $\int\frac{2\sec^2\theta\,d\theta}{4\tan^2\theta\cdot 2\sec\theta} = \frac{1}{4}\int\frac{\sec\theta}{\tan^2\theta}\,d\theta$. Rewrite in sines and cosines: $\frac{\sec\theta}{\tan^2\theta} = \frac{1}{\cos\theta}\cdot\frac{\cos^2\theta}{\sin^2\theta} = \frac{\cos\theta}{\sin^2\theta}$, and with $w = \sin\theta$ the integral is $\frac{1}{4}\int w^{-2}\,dw = -\frac{1}{4\sin\theta} + C$. The triangle of $\tan\theta = \frac{x}{2}$ has $x$ opposite, $2$ adjacent and hypotenuse $\sqrt{x^2 + 4}$, so $\sin\theta = \frac{x}{\sqrt{x^2 + 4}}$ and the answer is $-\frac{\sqrt{x^2 + 4}}{4x} + C$. Converting to sines and cosines is the reflex when a quotient of $\sec$ and $\tan$ does not match a formula of the table.

b) $x = 2\tan\theta$, $dx = 2\sec^2\theta\,d\theta$ and $(x^2 + 4)^{3/2} = (4\sec^2\theta)^{3/2} = 8\sec^3\theta$. The integrand becomes $\frac{2\sec^2\theta}{8\sec^3\theta} = \frac{1}{4}\cos\theta$. The limits move with the variable: $x = 0$ gives $\tan\theta = 0$, so $\theta = 0$; $x = 2$ gives $\tan\theta = 1$, so $\theta = \frac{\pi}{4}$. Then $\int_0^2\frac{dx}{(x^2 + 4)^{3/2}} = \frac{1}{4}\int_0^{\pi/4}\cos\theta\,d\theta = \frac{1}{4}\sin\frac{\pi}{4} = \frac{\sqrt 2}{8}$. No triangle was needed: once the limits are in $\theta$, the computation ends in $\theta$. Writing the new limits next to the substitution, on the same line, is what earns the mark and prevents the error of part d).

c) With $G(x) = \frac{x}{4}(x^2 + 4)^{-1/2}$, the product rule gives $G'(x) = \frac{1}{4}(x^2 + 4)^{-1/2} - \frac{x^2}{4}(x^2 + 4)^{-3/2} = \frac{(x^2 + 4) - x^2}{4(x^2 + 4)^{3/2}} = \frac{1}{(x^2 + 4)^{3/2}}$. This is exactly what the triangle would have produced from $\frac{1}{4}\sin\theta$, since $\sin\theta = \frac{x}{\sqrt{x^2 + 4}}$. Then $G(2) - G(0) = \frac{2}{4\sqrt 8} = \frac{2}{8\sqrt 2} = \frac{1}{4\sqrt 2} = \frac{\sqrt 2}{8}$. The two routes agree, and on an exam the route in $\theta$ is the one to take: fewer steps, fewer places to slip.

d) Keeping the limits gives $\frac{1}{4}\left[\sin\theta\right]_0^2 = \frac{1}{4}\sin 2$, a number with no reason to be right. The deeper problem: the substitution was made with $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, and $2 > \frac{\pi}{2} \approx 1.57$. The value $\theta = 2$ is not in that interval, and it corresponds to $x = 2\tan 2$, which is NEGATIVE since the tangent is negative in the second quadrant: the student is integrating over a stretch of $x$ that has nothing to do with $[0, 2]$. The limits of a definite integral always belong to the variable of integration; when the variable changes, they change with it.

e) $(4\sec^2\theta)^{3/2} = 8\left(\sqrt{\sec^2\theta}\right)^3 = 8|\sec\theta|^3$, and on $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ the cosine is positive, so $\sec\theta > 0$ and the absolute value drops. That is the reason for the interval, not a formality. For the size: on $[0, 2]$, $x^2 + 4 \ge 4$, so $(x^2 + 4)^{3/2} \ge 8$ and the integrand is at most $\frac{1}{8}$; over an interval of length $2$ the integral is at most $\frac{1}{4}$. And $\frac{\sqrt 2}{8} \approx \frac{1.41}{8} \approx 0.18$ satisfies it. A bound like this takes twenty seconds and catches a lost factor of $2$ or $4$.

a) $x = \sec\theta$, $dx = \sec\theta\tan\theta\,d\theta$, and on $0 \le \theta < \frac{\pi}{2}$ the tangent is non-negative, so $\sqrt{\sec^2\theta - 1} = \tan\theta$. The limits: $x = 1$ gives $\cos\theta = 1$, $\theta = 0$; $x = 2$ gives $\cos\theta = \frac{1}{2}$, $\theta = \frac{\pi}{3}$. The integrand becomes $\frac{\tan\theta}{\sec\theta}\cdot\sec\theta\tan\theta\,d\theta = \tan^2\theta\,d\theta = (\sec^2\theta - 1)\,d\theta$. So the integral is $\left[\tan\theta - \theta\right]_0^{\pi/3} = \sqrt 3 - \frac{\pi}{3}$. Positive, as it must be: $\sqrt 3 \approx 1.73$ and $\frac{\pi}{3} \approx 1.05$, and the integrand is positive on $(1, 2]$. The identity $\tan^2\theta = \sec^2\theta - 1$ is the only tool needed, and it comes straight from the table of section 7.2.

b) On $[-2, -1]$ the numerator $\sqrt{x^2 - 1}$ is non-negative and the denominator $x$ is negative: the curve is BELOW the axis and the integral is negative, as the shaded region shows. With Stewart's convention, $x = -1$ gives $\sec\theta = -1$, so $\theta = \pi$, and $x = -2$ gives $\cos\theta = -\frac{1}{2}$ with $\pi \le \theta < \frac{3\pi}{2}$, so $\theta = \frac{4\pi}{3}$. On that interval the tangent is non-negative (third quadrant), so again $\sqrt{x^2 - 1} = \tan\theta$ and the integrand is $\tan^2\theta\,d\theta$. The limits run from $\frac{4\pi}{3}$ (for $x = -2$) to $\pi$ (for $x = -1$), in that order: $\int_{4\pi/3}^{\pi}\tan^2\theta\,d\theta = \left[\tan\theta - \theta\right]_{4\pi/3}^{\pi} = (0 - \pi) - \left(\sqrt 3 - \frac{4\pi}{3}\right) = \frac{\pi}{3} - \sqrt 3$. Negative. The upper limit is smaller than the lower one, and that is fine: the limits are wherever the substitution sends them, never reordered by hand.

c) With $\frac{\pi}{2} < \theta \le \pi$: $x = -2$ gives $\theta = \frac{2\pi}{3}$ and $x = -1$ gives $\theta = \pi$. In the second quadrant $\tan\theta \le 0$, so $\sqrt{\tan^2\theta} = |\tan\theta| = -\tan\theta$. The integrand becomes $\frac{-\tan\theta}{\sec\theta}\cdot\sec\theta\tan\theta\,d\theta = -\tan^2\theta\,d\theta$, and $\int_{2\pi/3}^{\pi}(-\tan^2\theta)\,d\theta = -\left[\tan\theta - \theta\right]_{2\pi/3}^{\pi} = -\left[(0 - \pi) - \left(-\sqrt 3 - \frac{2\pi}{3}\right)\right] = -\left(\sqrt 3 - \frac{\pi}{3}\right) = \frac{\pi}{3} - \sqrt 3$. The same number as in b). Both conventions are correct; what is never correct is writing $\tan\theta$ for the root without looking at the quadrant.

d) Let $f(x) = \frac{\sqrt{x^2 - 1}}{x}$. Then $f(-x) = \frac{\sqrt{x^2 - 1}}{-x} = -f(x)$: $f$ is odd, and the figure shows it, the left branch is the right branch turned half a turn about the origin. The substitution $x = -u$ turns $\int_{-2}^{-1}f(x)\,dx$ into $-\int_1^2 f(u)\,du = -\left(\sqrt 3 - \frac{\pi}{3}\right) = \frac{\pi}{3} - \sqrt 3$. On an exam this is the fastest route of all, and the right one to use as a CHECK of b) when the secant has been used.

e) He took $\frac{\pi}{2} < \theta \le \pi$, the principal range of arcsecant, and wrote $\sqrt{x^2 - 1} = \tan\theta$ as on the positive branch. There $\tan\theta < 0$, so the root he used is negative, and every value of the integrand has the wrong sign: he obtained exactly the opposite of the answer. The five-second check is the SIGN of the integrand: on $[-2, -1]$ it is negative, so the integral must be negative, and $\sqrt 3 - \frac{\pi}{3} > 0$ is impossible. It costs the whole question, and it is the error the figure was drawn to prevent.

a) The ellipse is symmetric about both axes: replacing $x$ by $-x$ or $y$ by $-y$ leaves its equation unchanged. So its area is four times the area of the part in the first quadrant, which is the region under $y = \frac{3}{4}\sqrt{16 - x^2}$ for $0 \le x \le 4$. The area is $4\int_0^4\frac{3}{4}\sqrt{16 - x^2}\,dx = 3\int_0^4\sqrt{16 - x^2}\,dx$. Stating the symmetry is part of the answer: a formula like $4\int$ without a reason is a formula the marker cannot check.

b) Form $a^2 - x^2$ with $a = 4$: $x = 4\sin\theta$, $dx = 4\cos\theta\,d\theta$, and $\sqrt{16 - x^2} = 4\cos\theta$ since $\cos\theta \ge 0$ on $[0, \frac{\pi}{2}]$. The limits: $x = 0$ gives $\theta = 0$, and $x = 4$ gives $\sin\theta = 1$, $\theta = \frac{\pi}{2}$. So $3\int_0^4\sqrt{16 - x^2}\,dx = 3\int_0^{\pi/2}16\cos^2\theta\,d\theta = 48\int_0^{\pi/2}\frac{1 + \cos 2\theta}{2}\,d\theta = 48\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_0^{\pi/2} = 48\cdot\frac{\pi}{4} = 12\pi$. The term $\sin 2\theta$ vanishes at both limits, which is typical when the limits are $0$ and $\frac{\pi}{2}$: no triangle, no return to $x$.

c) The strip is symmetric about the $x$-axis, so its area is $2\int_0^2\frac{3}{4}\sqrt{16 - x^2}\,dx = \frac{3}{2}\int_0^2\sqrt{16 - x^2}\,dx$. With $x = 4\sin\theta$, $x = 2$ gives $\sin\theta = \frac{1}{2}$, so $\theta = \frac{\pi}{6}$. The area is $\frac{3}{2}\int_0^{\pi/6}16\cos^2\theta\,d\theta = 24\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_0^{\pi/6} = 24\left(\frac{\pi}{12} + \frac{\sqrt 3}{8}\right) = 2\pi + 3\sqrt 3$. Here $\sin 2\theta = \sin\frac{\pi}{3} = \frac{\sqrt 3}{2}$ does NOT vanish, and dropping it is the usual loss. Size: the height of the strip decreases from $6$ at $x = 0$ to $2\cdot\frac{3}{4}\sqrt{12} = 3\sqrt 3$ at $x = 2$, so the area lies between $2\cdot 3\sqrt 3 = 6\sqrt 3 \approx 10.4$ and $2\cdot 6 = 12$. And $2\pi + 3\sqrt 3 \approx 6.28 + 5.20 = 11.48$ fits.

d) By the same symmetry, the area is $4\int_0^a\frac{b}{a}\sqrt{a^2 - x^2}\,dx$. With $x = a\sin\theta$, $dx = a\cos\theta\,d\theta$, $\sqrt{a^2 - x^2} = a\cos\theta$ and limits $0$ and $\frac{\pi}{2}$: $4\cdot\frac{b}{a}\int_0^{\pi/2}a^2\cos^2\theta\,d\theta = 4ab\cdot\frac{\pi}{4} = \pi ab$. For $a = 4$, $b = 3$ this is $12\pi$, as in b). For a circle, $a = b = r$ gives $\pi r^2$. A general formula that does not reduce to the circle is wrong, and this check takes one line.