Corrected exercises: strategy for integration (MATH 141)

Simplification: EXPAND the square and divide term by term. $(1+\sqrt{x})^2 = 1 + 2\sqrt{x} + x$, so the integrand is $x^{-1} + 2x^{-1/2} + 1$. Then $\int (x^{-1} + 2x^{-1/2} + 1)\,dx = \ln x + 4\sqrt{x} + x + C$. The absolute value in $\ln|x|$ can be dropped because $\sqrt{x}$ already forces $x > 0$. Check by differentiating: $\frac{1}{x} + \frac{2}{\sqrt{x}} + 1$, which is the expanded integrand. The substitution $u = 1 + \sqrt{x}$ also works, but it gives $x = (u-1)^2$, $dx = 2(u-1)\,du$ and the new integrand $\frac{2u^2}{u-1}$, which then needs a long division: three extra lines for the same answer, and three extra places to drop a sign.

Simplification: rewrite EVERYTHING in sines and cosines. $\frac{\sin x}{\tan x} = \sin x \cdot \frac{\cos x}{\sin x} = \cos x$ and $\frac{\sec x}{\tan x} = \frac{1}{\cos x}\cdot\frac{\cos x}{\sin x} = \frac{1}{\sin x} = \csc x$. So the integral is $\int (\cos x + \csc x)\,dx = \sin x + \ln|\csc x - \cot x| + C$, where $\int \csc x\,dx = \ln|\csc x - \cot x| + C$ is the companion of $\int \sec x\,dx$. Check: $\frac{d}{dx}\ln|\csc x - \cot x| = \frac{-\csc x\cot x + \csc^2 x}{\csc x - \cot x} = \csc x$. The trap is to see $\tan x$ and reach for $u = \tan x$: its derivative $\sec^2 x$ is nowhere in the integrand, and the attempt goes in circles.

Simplification: split the numerator along the denominator. $(x+1)^2 = x^2 + 2x + 1 = (x^2 + 1) + 2x$, so $\frac{(x+1)^2}{x^2+1} = 1 + \frac{2x}{x^2+1}$. The second piece is of the form $\frac{g'(x)}{g(x)}$ with $g(x) = x^2 + 1$. Hence $\int \frac{(x+1)^2}{x^2+1}\,dx = x + \ln(x^2+1) + C$, with no absolute value since $x^2 + 1 > 0$. The tempting substitution $u = x + 1$ turns the denominator into $u^2 - 2u + 2$, which is worse. Partial fractions would start with a long division, since numerator and denominator have the same degree, and that division is exactly the one line written here.

Simplification: a DIFFERENCE OF SQUARES. $e^{2x} - 1 = (e^x)^2 - 1 = (e^x - 1)(e^x + 1)$, so the fraction reduces to $e^x - 1$ and $\int (e^x - 1)\,dx = e^x - x + C$. The route $u = e^x$, $dx = \frac{du}{u}$ gives $\int \frac{u^2 - 1}{u(u+1)}\,du = \int \frac{u - 1}{u}\,du = u - \ln u + C$, the same answer after the same cancellation, reached later. Whenever $e^{2x}$ sits next to $e^x$, ask first whether it factors.

Simplification: the double-angle identity $\cos 2x = 2\cos^2 x - 1$ gives $1 + \cos 2x = 2\cos^2 x$. So $\int \frac{dx}{1 + \cos 2x} = \frac{1}{2}\int \sec^2 x\,dx = \frac{1}{2}\tan x + C$, on any interval where $\cos x \neq 0$. The substitution $u = 1 + \cos 2x$ fails because it needs $du = -2\sin 2x\,dx$ in the numerator, and the numerator is $1$: a substitution is not a change of name, it demands that the derivative of the new variable be present, up to a constant factor. That test, is the derivative there, is the one to run before every substitution in this chapter.

The pair: $u = \ln x$, $du = \frac{dx}{x}$, and the factor $\frac{1}{x}$ is sitting in the denominator. Bounds: $x = 1$ gives $u = 0$, $x = e$ gives $u = 1$. Then $\int_0^1 \frac{du}{1+u^2} = \arctan 1 - \arctan 0 = \frac{\pi}{4}$. Changing the bounds at the moment of the substitution means never returning to $x$. The trap is to read $1 + (\ln x)^2$ as the thing to substitute: $u = 1 + (\ln x)^2$ needs $\frac{2\ln x}{x}$ in the numerator, and there is no $\ln x$ upstairs.

The pair: $u = \tan x$, $du = \sec^2 x\,dx$, which is exactly the numerator. Seeing powers of $\tan x$ and a $\sec^2 x$ in the same integrand is the signal, and it has nothing to do with the trigonometric-integral recipes. The integral becomes $\int \frac{du}{u^2 + 4u + 5}$. The quadratic has discriminant $16 - 20 = -4 < 0$, so it does not factor: COMPLETE THE SQUARE, $u^2 + 4u + 5 = (u+2)^2 + 1$. Then $\int \frac{du}{(u+2)^2 + 1} = \arctan(u + 2) + C = \arctan(\tan x + 2) + C$. The classic wrong answer is $\ln(\tan^2 x + 4\tan x + 5)$: its derivative is $\frac{(2\tan x + 4)\sec^2 x}{\tan^2 x + 4\tan x + 5}$, and the numerator here is only $\sec^2 x$. A logarithm needs the derivative of the WHOLE denominator on top.

The pair: $u = \sqrt{x}$, $du = \frac{dx}{2\sqrt{x}}$, so $\frac{dx}{\sqrt{x}} = 2\,du$, and $x = u^2$. Bounds: $1 \mapsto 1$ and $3 \mapsto \sqrt{3}$. Then $\int_1^{\sqrt{3}} \frac{2\,du}{1 + u^2} = 2\left(\arctan\sqrt{3} - \arctan 1\right) = 2\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \frac{\pi}{6}$. Partial fractions cannot even start on $\frac{1}{\sqrt{x}(1+x)}$, which is not a rational function; the substitution is what makes it one, and here the rational function it produces is already an arctangent.

The pair: $u = \sin x$, $du = \cos x\,dx$. The integral becomes $\int \frac{du}{u^2 - 3u + 2} = \int \frac{du}{(u-1)(u-2)}$, a rational function with two distinct linear factors: partial fractions. Write $\frac{1}{(u-1)(u-2)} = \frac{A}{u-1} + \frac{B}{u-2}$. Covering up: $u = 1$ gives $A = \frac{1}{1-2} = -1$, $u = 2$ gives $B = \frac{1}{2-1} = 1$. So $\int = -\ln|u-1| + \ln|u-2| + C = \ln\left|\frac{\sin x - 2}{\sin x - 1}\right| + C$. Since $\sin x - 2 < 0$ and $\sin x - 1 < 0$ wherever the integrand is defined, the quotient is positive and the answer reads $\ln\left(\frac{2 - \sin x}{1 - \sin x}\right) + C$. Two techniques, chained: the substitution produced the rational function, partial fractions finished it.

With $u = \sin^2 x$: $du = 2\sin x\cos x\,dx$ and $\sin^4 x = u^2$. Bounds: $0 \mapsto 0$, $\frac{\pi}{2} \mapsto 1$. So the integral is $\frac{1}{2}\int_0^1 \frac{du}{1+u^2} = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8}$. With $u = \sin x$ instead, one gets $\int_0^1 \frac{u\,du}{1 + u^4}$, which still needs a second substitution $w = u^2$ to reach the same arctangent. Both are correct; the better choice reads $\sin^4 x$ as $(\sin^2 x)^2$ and spots that $\sin x \cos x$ is half the derivative of $\sin^2 x$. Plausibility: the integrand lies between $0$ and $\frac{1}{2}$ on an interval of length $\frac{\pi}{2}$, and $\frac{\pi}{8} \approx 0.39$ is well inside.

Substitution: $t = \sqrt{x}$, so $x = t^2$ and $dx = 2t\,dt$; bounds $0 \mapsto 0$, $1 \mapsto 1$. The integral becomes $\int_0^1 2t e^t\,dt$. Parts on this product: $u = t$ (it simplifies when differentiated), $dv = e^t\,dt$, so $du = dt$, $v = e^t$, and $\int t e^t\,dt = te^t - e^t = (t-1)e^t$. Hence $\int_0^1 e^{\sqrt{x}}\,dx = 2\left[(t-1)e^t\right]_0^1 = 2\left(0 - (-1)\right) = 2$. In terms of $x$, the antiderivative is $2(\sqrt{x} - 1)e^{\sqrt{x}} + C$. Check: on $[0, 1]$ the integrand runs from $e^0 = 1$ to $e^1 = e$ and the interval has length $1$, so the value must lie between $1$ and $e \approx 2.72$. It does.

As written, nothing pairs up: the exponent is $\sin x$, whose derivative $\cos x$ does not appear. The double-angle identity $\sin 2x = 2\sin x\cos x$ makes it appear: the integrand is $2\sin x \cdot e^{\sin x}\cos x$. Now $t = \sin x$, $dt = \cos x\,dx$, bounds $0 \mapsto 0$ and $\frac{\pi}{6} \mapsto \frac{1}{2}$, and the integral becomes $\int_0^{1/2} 2te^t\,dt$, the same product as in a). So the value is $2\left[(t - 1)e^t\right]_0^{1/2} = 2\left(-\frac{1}{2}e^{1/2} + 1\right) = 2 - \sqrt{e}$. Three moves in a row, an identity, a substitution and integration by parts, and none of them is visible on the original integrand. Plausibility: $\sqrt{e} \approx 1.65$ gives about $0.35$, and the integrand stays below $\sin\frac{\pi}{3}\,e^{1/2} \approx 1.43$ on an interval of length $\frac{\pi}{6} \approx 0.52$, so the value is below $0.75$.

Write $x^3\,dx = x^2 \cdot x\,dx$: the factor $x$ is half the derivative of $x^2$. So $t = x^2$, $dt = 2x\,dx$, bounds $0 \mapsto 0$ and $\sqrt{\pi} \mapsto \pi$, and the integral becomes $\frac{1}{2}\int_0^{\pi} t\cos t\,dt$. Parts with $u = t$, $dv = \cos t\,dt$, $v = \sin t$: $\int t\cos t\,dt = t\sin t + \cos t$. So the value is $\frac{1}{2}\left[t\sin t + \cos t\right]_0^{\pi} = \frac{1}{2}\left((0 - 1) - (0 + 1)\right) = -1$. Sign: $\cos(x^2) > 0$ only while $x^2 < \frac{\pi}{2}$, and on the rest of the interval, where $x^3$ is largest, it is negative; a negative total is expected. Integrating by parts first with $u = x^3$ would need an antiderivative of $\cos(x^2)$, which does not exist in elementary form: the substitution is not a convenience here, it is the only way in.

$\frac{d}{dx}e^{\sqrt{x}} = e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}$ by the chain rule, which is NOT $e^{\sqrt{x}}$. The error: the student renamed $\sqrt{x}$ as $t$ but left $dx$ as if it were $dt$. A substitution replaces three things, the inner function, the differential and (for a definite integral) the bounds. Here $dx = 2t\,dt$, and that factor $2t$ is precisely what makes part a) an integration by parts. Differentiating the final answer takes ten seconds and catches this every time.

With $u = e^{\sqrt{x}}$ and $dv = dx$: $du = \frac{e^{\sqrt{x}}}{2\sqrt{x}}\,dx$ and $v = x$, so $\int e^{\sqrt{x}}\,dx = xe^{\sqrt{x}} - \int x\cdot\frac{e^{\sqrt{x}}}{2\sqrt{x}}\,dx = xe^{\sqrt{x}} - \frac{1}{2}\int \sqrt{x}\,e^{\sqrt{x}}\,dx$. The new integral is harder than the one we started with, and a second round of parts makes it worse again. The rule that comes out: when the difficulty is a composition, attack the inner function first; parts is for PRODUCTS, and the substitution is what turns the composition into one.

Rationalizing substitution: $t = \sqrt{x}$, $x = t^2$, $dx = 2t\,dt$; bounds $9 \mapsto 3$, $16 \mapsto 4$. The integral becomes $\int_3^4 \frac{t^2}{t^2 - t - 2}\cdot 2t\,dt = \int_3^4 \frac{2t^3}{(t-2)(t+1)}\,dt$, and the denominator does not vanish on $[3, 4]$. Degrees: $3$ over $2$, so DIVIDE first: $2t^3 = (t^2 - t - 2)(2t + 2) + (6t + 4)$, that is $\frac{2t^3}{t^2 - t - 2} = 2t + 2 + \frac{6t + 4}{(t-2)(t+1)}$. Then split: $\frac{6t + 4}{(t-2)(t+1)} = \frac{A}{t-2} + \frac{B}{t+1}$ with $A = \frac{16}{3}$ (cover up at $t = 2$) and $B = \frac{-2}{-3} = \frac{2}{3}$ (at $t = -1$). So the value is $\left[t^2 + 2t + \frac{16}{3}\ln(t - 2) + \frac{2}{3}\ln(t + 1)\right]_3^4 = (24 - 15) + \frac{16}{3}\ln 2 + \frac{2}{3}\ln\frac{5}{4} = 9 + 4\ln 2 + \frac{2}{3}\ln 5$, using $\frac{2}{3}\ln\frac{5}{4} = \frac{2}{3}\ln 5 - \frac{4}{3}\ln 2$. Three techniques in a row, and skipping the division is the classic slip: the right side of a decomposition never produces the polynomial part $2t + 2$.

The roots have indices $2$ and $3$, and their least common multiple is $6$: set $t = x^{1/6}$, so $x = t^6$, $\sqrt{x} = t^3$, $\sqrt[3]{x} = t^2$ and $dx = 6t^5\,dt$. Bounds: $1 \mapsto 1$ and $64 \mapsto 2$, since $2^6 = 64$. The integral becomes $\int_1^2 \frac{6t^5}{t^3 + t^2}\,dt = 6\int_1^2 \frac{t^3}{t+1}\,dt$ after cancelling $t^2$. Degree $3$ over degree $1$: long division gives $\frac{t^3}{t+1} = t^2 - t + 1 - \frac{1}{t+1}$. So the value is $\left[2t^3 - 3t^2 + 6t - 6\ln(t+1)\right]_1^2 = (16 - 12 + 12 - 6\ln 3) - (2 - 3 + 6 - 6\ln 2) = 11 - 6\ln 3 + 6\ln 2 = 11 + 6\ln\frac{2}{3}$. Plausibility: $6\ln\frac{2}{3} \approx 6(0.69 - 1.10) \approx -2.4$, so about $8.6$; the integrand falls from $\frac{1}{2}$ to $\frac{1}{12}$ over an interval of length $63$, so the value lies between $5.25$ and $31.5$. It does.

Substitution $t = e^x$, $dx = \frac{dt}{t}$, bounds $0 \mapsto 1$ and $\ln 2 \mapsto 2$. The integral becomes $\int_1^2 \frac{t^3}{t^2 + 5t + 6}\cdot\frac{dt}{t} = \int_1^2 \frac{t^2}{(t+2)(t+3)}\,dt$. Degrees $2$ and $2$: divide, $\frac{t^2}{t^2 + 5t + 6} = 1 - \frac{5t + 6}{(t+2)(t+3)}$. Split: $\frac{5t + 6}{(t+2)(t+3)} = \frac{A}{t+2} + \frac{B}{t+3}$ with $A = \frac{-4}{1} = -4$ and $B = \frac{-9}{-1} = 9$. So the integrand is $1 + \frac{4}{t+2} - \frac{9}{t+3}$ and the value is $\left[t + 4\ln(t+2) - 9\ln(t+3)\right]_1^2 = 1 + 4\ln\frac{4}{3} - 9\ln\frac{5}{4}$. Plausibility: the integrand grows from $\frac{1}{12}$ at $x = 0$ to $\frac{8}{20} = \frac{2}{5}$ at $x = \ln 2$, so the value lies between $\frac{\ln 2}{12} \approx 0.06$ and $\frac{2}{5}\ln 2 \approx 0.28$. The exact value is a small difference of larger terms, about $0.14$, and two-decimal logarithms are too coarse to confirm it; the check that protects the answer here is on the decomposition itself, for instance at $t = 0$: $1 + \frac{4}{2} - \frac{9}{3} = 0$, which is indeed $\frac{0^2}{2 \cdot 3}$.

With $t = \sqrt{x}$ in b), $\sqrt[3]{x} = t^{2/3}$: the integrand is still not rational, and the substitution has bought nothing. The LCM of ALL the indices is the one that clears every root. In c), the factor $\frac{1}{t}$ coming from $dx = \frac{dt}{t}$ was dropped: the student is integrating a different function, of degree $3$ over $2$, which needs a longer division and gives a wrong number. Writing $dx$ in terms of $dt$ on its own line before substituting is the gesture that prevents both.

On $[9, 16]$ the integrand falls from $\frac{9}{9 - 3 - 2} = \frac{9}{4}$ at $x = 9$ to $\frac{16}{16 - 4 - 2} = \frac{8}{5}$ at $x = 16$. The interval has length $7$, so the integral lies between $7 \cdot \frac{8}{5} = 11.2$ and $7 \cdot \frac{9}{4} = 15.75$. And $9 + 4\ln 2 + \frac{2}{3}\ln 5 \approx 9 + 2.77 + 1.07 = 12.84$, inside the bracket. Dropping the polynomial part $2t + 2$ of the division, a very common slip, would remove the $9$ and give about $3.8$, far outside.

First: $u = x^2$, because $du = 2x\,dx$ and the factor $x$ is present. $\int_0^1 xe^{x^2}\,dx = \frac{1}{2}\int_0^1 e^u\,du = \frac{e - 1}{2}$. Second: the same $u = x^2$, writing $x^3\,dx = x^2 \cdot x\,dx = u\cdot\frac{du}{2}$; what is left, $\frac{1}{2}\int_0^1 ue^u\,du$, is a product, so integrate by parts, differentiating the polynomial factor $u$ and integrating $e^u$: $\int ue^u\,du = ue^u - \int e^u\,du = (u-1)e^u$, and the value is $\frac{1}{2}\left[(u-1)e^u\right]_0^1 = \frac{1}{2}(0 + 1) = \frac{1}{2}$. Third: no factor $x$, so no substitution, and parts only trades it for something worse. The antiderivative is not elementary, yet the integral EXISTS, since $e^{x^2}$ is continuous on $[0, 1]$. The integrand increases from $1$ to $e$ there, so $1 \le \int_0^1 e^{x^2}\,dx \le e$. Exercise 7 sharpens this bracket.

First: $u = 4 - x^2$, $du = -2x\,dx$, and the factor $x$ is present: $\int_0^1 \frac{x\,dx}{\sqrt{4-x^2}} = \left[-\sqrt{4 - x^2}\right]_0^1 = -\sqrt{3} + 2 = 2 - \sqrt{3}$. Second: no factor $x$, but the form $\frac{1}{\sqrt{a^2 - x^2}}$ is in the table: $\left[\arcsin\frac{x}{2}\right]_0^1 = \arcsin\frac{1}{2} = \frac{\pi}{6}$. Third: the root is now in the NUMERATOR, neither trick applies, and the tool is the trigonometric substitution $x = 2\sin\theta$. Faster still is the GEOMETRY of the figure: the region under the circle $x^2 + y^2 = 4$ between $x = 0$ and $x = 1$ splits into a triangle $T$ with vertices $(0,0)$, $(1,0)$, $(1,\sqrt{3})$, of area $\frac{\sqrt{3}}{2}$, and a sector $S$ of radius $2$ between the directions of $60^\circ$ and $90^\circ$, of angle $\frac{\pi}{6}$ and area $\frac{1}{2}\cdot 4\cdot\frac{\pi}{6} = \frac{\pi}{3}$. Total $\frac{\pi}{3} + \frac{\sqrt{3}}{2}$, which the substitution confirms: $\left[\frac{x}{2}\sqrt{4-x^2} + 2\arcsin\frac{x}{2}\right]_0^1 = \frac{\sqrt{3}}{2} + \frac{\pi}{3}$.

First: $u = \ln x$, $du = \frac{dx}{x}$: $\int_0^1 u\,du = \frac{1}{2}$. Second: the $\frac{1}{x}$ is there, but a second $\frac{1}{x}$ is left over, and it is not a function of $\ln x$ unless one writes it $e^{-u}$. The direct route is parts, with $u = \ln x$ (the factor that simplifies when differentiated) and $dv = \frac{dx}{x^2}$, $v = -\frac{1}{x}$: $\left[-\frac{\ln x}{x}\right]_1^e + \int_1^e \frac{dx}{x^2} = -\frac{1}{e} + \left(1 - \frac{1}{e}\right) = 1 - \frac{2}{e}$. Third: $u = \ln x$ again, bounds $e \mapsto 1$ and $e^2 \mapsto 2$: $\int_1^2 \frac{du}{u} = \ln 2$. The first and third are both substitutions, but the first gives a power of $u$ and the third a logarithm: once substituted, the form decides again.