Corrected exercises: trigonometric integrals (MATH 141)

a) The power of $\sin x$ is $3$, odd, so detach one $\sin x$ for the differential: $u = \cos x$, $du = -\sin x\,dx$. The two factors left, $\sin^2 x = 1 - \cos^2 x = 1 - u^2$, convert cleanly. Then $\int \sin^3 x \sqrt{\cos x}\,dx = \int \sin^2 x \sqrt{\cos x}\,\sin x\,dx = -\int (1 - u^2)u^{1/2}\,du = -\int (u^{1/2} - u^{5/2})\,du = -\frac{2}{3}u^{3/2} + \frac{2}{7}u^{7/2} + C = -\frac{2}{3}\cos^{3/2} x + \frac{2}{7}\cos^{7/2} x + C$. The square root is harmless because it sits on the function we SUBSTITUTE: $\sqrt{\cos x}$ becomes $u^{1/2}$, a power like any other. The method never needed the other power to be an integer, only the saved power to be odd. Check by differentiating: $\sin x\sqrt{\cos x} - \sin x\cos^{5/2} x = \sin x\sqrt{\cos x}(1 - \cos^2 x) = \sin^3 x\sqrt{\cos x}$.

b) Here the ODD power is on the cosine, $5$, so the roles swap: detach $\cos x\,dx = du$ with $u = \sin x$, and write $\cos^4 x = (\cos^2 x)^2 = (1 - u^2)^2$. Then $\int \cos^5 x \sin^4 x\,dx = \int (1 - u^2)^2 u^4\,du = \int (u^4 - 2u^6 + u^8)\,du = \frac{u^5}{5} - \frac{2u^7}{7} + \frac{u^9}{9} + C = \frac{\sin^5 x}{5} - \frac{2\sin^7 x}{7} + \frac{\sin^9 x}{9} + C$. The even power $4$ of the sine is simply carried along as $u^4$. Trying to save a factor of $\sin x$ instead would leave $\sin^3 x$, an odd power of the function you are not substituting, and the method stalls.

c) Both powers are odd, so both routes work. With $u = \sin x$ (save $\cos x\,dx$, convert $\cos^2 x = 1 - u^2$): $\int u^3(1 - u^2)\,du = \frac{\sin^4 x}{4} - \frac{\sin^6 x}{6} + C_1$. With $u = \cos x$ (save $\sin x\,dx = -du$, convert $\sin^2 x = 1 - u^2$): $-\int (1 - u^2)u^3\,du = -\frac{\cos^4 x}{4} + \frac{\cos^6 x}{6} + C_2$. They LOOK different, and a marker accepts either. To compare, put $s = \sin^2 x$, so $\cos^2 x = 1 - s$: the difference is $\frac{s^2}{4} - \frac{s^3}{6} + \frac{(1 - s)^2}{4} - \frac{(1 - s)^3}{6}$. Expanding, every term in $s$ cancels and what remains is $\frac{1}{4} - \frac{1}{6} = \frac{1}{12}$. Two antiderivatives of the same function on an interval differ by a constant, here $\frac{1}{12}$: neither student is wrong. When the two powers are odd, choose the SMALLER one to save, the polynomial is shorter.

d) The power of $\cos x$ is $3$, odd: save $\cos x\,dx = du$ with $u = \sin x$, and convert $\cos^2 x = 1 - u^2$. Bounds: $x = 0$ gives $u = 0$ and $x = \frac{\pi}{2}$ gives $u = 1$. So $\int_0^{\pi/2} \sin^2 x \cos^3 x\,dx = \int_0^1 u^2(1 - u^2)\,du = \frac{1}{3} - \frac{1}{5} = \frac{2}{15}$. Once the bounds are changed, $x$ never comes back. Sanity check against the figure: the curve stays below $0.19$ on an interval of length $\frac{\pi}{2} \approx 1.57$, so the area is below $0.3$, and $\frac{2}{15} \approx 0.13$ fits.

e) He writes $\int \sin^2 x \cos^3 x\,dx = \int \sin x\cos^3 x\,(\sin x\,dx)$ and sets $u = \cos x$, $\sin x\,dx = -du$. What is left to convert is ONE factor $\sin x$, and $\sin x = \pm\sqrt{1 - u^2}$, with a sign that depends on the interval: $-\int u^3\sqrt{1 - u^2}\,du$ is a harder integral than the one he started with. That is the whole rule of the chapter: an EVEN number of factors converts through $\sin^2 x = 1 - \cos^2 x$, an odd number leaves a square root. So the factor you detach must come from the odd power, and the even power is the one you convert.

a) The curve oscillates between $0$ and $1$ around the dashed line $y = \frac{1}{2}$, and the part above the line is the mirror image of the gap below it: the region has the same area as the rectangle of height $\frac{1}{2}$ on $[0, \pi]$, that is $\frac{\pi}{2}$. Computation: $\int_0^\pi \sin^2 x\,dx = \int_0^\pi \frac{1 - \cos 2x}{2}\,dx = \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_0^\pi = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. The figure IS the half-angle formula: $\sin^2 x = \frac{1}{2} - \frac{\cos 2x}{2}$ is a constant $\frac{1}{2}$ plus a wave that integrates to zero over whole periods.

b) $\cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1}{4}(1 + 2\cos 2x + \cos^2 2x)$. The square $\cos^2 2x$ is again an even power, so use the formula a SECOND time, with angle $2x$: $\cos^2 2x = \frac{1 + \cos 4x}{2}$. Then $\cos^4 x = \frac{1}{4} + \frac{\cos 2x}{2} + \frac{1 + \cos 4x}{8} = \frac{3}{8} + \frac{\cos 2x}{2} + \frac{\cos 4x}{8}$, and $\int \cos^4 x\,dx = \frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$. Two uses of the formula for a fourth power, and each one doubles the frequency. The usual slip is forgetting the inner chain rule factor: $\int \cos 4x\,dx = \frac{\sin 4x}{4}$, not $\sin 4x$.

c) Pairing is the shortcut: $\sin x\cos x = \frac{\sin 2x}{2}$, so $(\sin x\cos x)^2 = \frac{\sin^2 2x}{4}$, and $\sin^2 x = \frac{1 - \cos 2x}{2}$. The integrand becomes $\frac{1 - \cos 2x}{2}\cdot\frac{\sin^2 2x}{4} = \frac{1}{8}\left(\sin^2 2x - \sin^2 2x\cos 2x\right)$. The first piece is an even power again: $\int \sin^2 2x\,dx = \int \frac{1 - \cos 4x}{2}\,dx = \frac{x}{2} - \frac{\sin 4x}{8}$. The second piece has an ODD power of $\cos 2x$, so it is a substitution: $w = \sin 2x$, $dw = 2\cos 2x\,dx$, and $\int \sin^2 2x\cos 2x\,dx = \frac{\sin^3 2x}{6}$. Altogether $\int \sin^4 x\cos^2 x\,dx = \frac{1}{8}\left(\frac{x}{2} - \frac{\sin 4x}{8} - \frac{\sin^3 2x}{6}\right) + C = \frac{x}{16} - \frac{\sin 4x}{64} - \frac{\sin^3 2x}{48} + C$. Note that halving the degree produced a term with an odd power: the two gestures of the chapter often alternate inside one integral.

d) $\int_0^{\pi/4} \cos^2 x\,dx = \int_0^{\pi/4} \frac{1 + \cos 2x}{2}\,dx = \left[\frac{x}{2} + \frac{\sin 2x}{4}\right]_0^{\pi/4} = \frac{\pi}{8} + \frac{\sin(\pi/2)}{4} = \frac{\pi}{8} + \frac{1}{4}$. Check the sign in the formula before using it: at $x = 0$, $\cos^2 0 = 1$ and $\frac{1 + \cos 0}{2} = 1$, while the wrong version $\frac{1 - \cos 2x}{2}$ gives $0$. Order of magnitude: $\cos^2 x$ runs from $1$ down to $\frac{1}{2}$ on an interval of length $\frac{\pi}{4} \approx 0.79$, so the answer lies between $0.39$ and $0.79$; $\frac{\pi}{8} + \frac{1}{4} \approx 0.64$ fits.

e) First, differentiate: $\frac{d}{dx}\frac{\sin^3 x}{3} = \sin^2 x\cos x$, which is not $\sin^2 x$. The power rule $\int u^2\,du = \frac{u^3}{3}$ needs $du = \cos x\,dx$ in the integrand, and there is no $\cos x$ to play that role. Second, use part a): $\left[\frac{\sin^3 x}{3}\right]_0^\pi = 0$, whereas $\sin^2 x \ge 0$ and is not identically zero, so its integral on $[0, \pi]$ must be positive, and it is $\frac{\pi}{2}$. An antiderivative guessed by analogy is always checked by one differentiation before it is used.

a) The power of $\sec x$ is $4$, even: save $\sec^2 x\,dx = du$ with $u = \tan x$, and convert the remaining $\sec^2 x = 1 + \tan^2 x = 1 + u^2$. Then $\int \tan^6 x\sec^4 x\,dx = \int \tan^6 x\,\sec^2 x\,(\sec^2 x\,dx) = \int u^6(1 + u^2)\,du = \frac{u^7}{7} + \frac{u^9}{9} + C = \frac{\tan^7 x}{7} + \frac{\tan^9 x}{9} + C$. The power of the tangent, even here, did not matter at all: it is carried along as $u^6$.

b) Now the power of $\sec x$ is $3$, odd, so saving $\sec^2 x$ would leave one $\sec x = \sqrt{1 + u^2}$: the route of a) is closed. But the power of $\tan x$ is $5$, odd: save $\sec x\tan x\,dx = du$ with $u = \sec x$, and convert the four remaining tangents, $\tan^4 x = (\sec^2 x - 1)^2 = (u^2 - 1)^2$. What is left of the secants is $\sec^2 x = u^2$. So $\int \tan^5 x\sec^3 x\,dx = \int \tan^4 x\sec^2 x\,(\sec x\tan x\,dx) = \int (u^2 - 1)^2u^2\,du = \int (u^6 - 2u^4 + u^2)\,du = \frac{\sec^7 x}{7} - \frac{2\sec^5 x}{5} + \frac{\sec^3 x}{3} + C$. Count the factors before writing anything: one $\sec x$ and one $\tan x$ are spent on $du$, so the powers left are $\tan^4$ and $\sec^2$, not $\tan^5$ and $\sec^3$.

c) With no secant, save nothing yet: split one $\tan^2 x$ and convert it, $\tan^3 x = \tan x(\sec^2 x - 1) = \tan x\sec^2 x - \tan x$. The first piece is the pattern of a) with $u = \tan x$: $\int \tan x\sec^2 x\,dx = \frac{\tan^2 x}{2}$. The second is the known $\int \tan x\,dx = \ln|\sec x|$. So $\int \tan^3 x\,dx = \frac{\tan^2 x}{2} - \ln|\sec x| + C$. A classmate who used $u = \sec x$ on the first piece gets $\frac{\sec^2 x}{2} - \ln|\sec x| + C$: the same family, since $\sec^2 x = 1 + \tan^2 x$ shifts the constant by $\frac{1}{2}$.

d) The power of $\tan x$ is $1$, odd: save $\sec x\tan x\,dx = du$ with $u = \sec x$, and the rest is $\sec^2 x = u^2$. Bounds: $\sec 0 = 1$ and $\sec\frac{\pi}{3} = \frac{1}{\cos(\pi/3)} = 2$. So $\int_0^{\pi/3} \tan x\sec^3 x\,dx = \int_1^2 u^2\,du = \frac{8 - 1}{3} = \frac{7}{3}$. The new lower bound is $1$, not $0$: $\sec 0 = 1$ is the most common slip on this kind of question, and it would give $\frac{8}{3}$.

e) The power of $\sec x$ is $1$, odd, so saving $\sec^2 x$ is impossible without a square root; the power of $\tan x$ is $2$, even, so saving $\sec x\tan x$ leaves one $\tan x = \sqrt{u^2 - 1}$. Neither substitution closes. What remains is the identity used as a pure REWRITING: $\tan^2 x\sec x = (\sec^2 x - 1)\sec x = \sec^3 x - \sec x$, so $\int \tan^2 x\sec x\,dx = \int \sec^3 x\,dx - \int \sec x\,dx$. Both integrals on the right are the two antiderivatives nobody guesses, and they are the subject of the next exercise.

a) By the chain rule, $F'(x) = \frac{\sec x\tan x + \sec^2 x}{\sec x + \tan x} = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} = \sec x$, on every interval where $\cos x \neq 0$ and $\sec x + \tan x \neq 0$. The absolute value is there because $\frac{d}{dx}\ln|g| = \frac{g'}{g}$ holds for $g$ of either sign. So $\int \sec x\,dx = \ln|\sec x + \tan x| + C$. The formula is found historically by multiplying $\sec x$ by $\frac{\sec x + \tan x}{\sec x + \tan x}$, which makes the numerator the derivative of the denominator; on an exam it is quoted, and checked by this differentiation if in doubt.

b) With $u = \sec x$, $dv = \sec^2 x\,dx$: $du = \sec x\tan x\,dx$ and $v = \tan x$. Then $I = \int \sec^3 x\,dx = \sec x\tan x - \int \sec x\tan^2 x\,dx$. Convert with $\tan^2 x = \sec^2 x - 1$: $\int \sec x\tan^2 x\,dx = \int \sec^3 x\,dx - \int \sec x\,dx = I - \ln|\sec x + \tan x|$. So $I = \sec x\tan x - I + \ln|\sec x + \tan x|$. The unknown $I$ reappeared with a MINUS sign, which is exactly what makes the equation solvable: $2I = \sec x\tan x + \ln|\sec x + \tan x|$, hence $\int \sec^3 x\,dx = \frac{1}{2}\left(\sec x\tan x + \ln|\sec x + \tan x|\right) + C$. The constant is added at the end only, once $I$ has been isolated. Had it reappeared with a plus sign, the equation would read $0 = \dots$ and the choice of $u$ would have to change.

c) From Exercise 3 e), $\int \tan^2 x\sec x\,dx = \int \sec^3 x\,dx - \int \sec x\,dx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln|\sec x + \tan x| - \ln|\sec x + \tan x| + C = \frac{1}{2}\left(\sec x\tan x - \ln|\sec x + \tan x|\right) + C$. The logarithms combine to $-\frac{1}{2}\ln|\dots|$: dropping the $\frac{1}{2}$ of part b) at this step is the classic way to lose the mark.

d) At $x = \frac{\pi}{4}$: $\sec\frac{\pi}{4} = \sqrt 2$ and $\tan\frac{\pi}{4} = 1$. At $x = 0$: $\sec 0 = 1$, $\tan 0 = 0$, and $\ln 1 = 0$. So $\int_0^{\pi/4} \sec^3 x\,dx = \frac{1}{2}\left(\sqrt 2\cdot 1 + \ln(\sqrt 2 + 1)\right) - \frac{1}{2}(0 + \ln 1) = \frac{\sqrt 2 + \ln(1 + \sqrt 2)}{2}$. This is the exact answer expected without a calculator; no decimal is required.

e) The figure gives two bounds. The curve stays above height $1$, so the area exceeds the rectangle $\frac{\pi}{4}\cdot 1 \approx 0.79$. The curve bends upward (it is convex), so it stays below its chord, and the area is less than the trapezoid $\frac{\pi}{4}\cdot\frac{1 + 2\sqrt 2}{2} \approx 0.79 \times 1.91 \approx 1.50$, using $\sec^3\frac{\pi}{4} = (\sqrt 2)^3 = 2\sqrt 2$. Now estimate the answer: $1 + \sqrt 2 \approx 2.41$ lies between $2$ and $e \approx 2.72$, so $\ln(1 + \sqrt 2)$ lies between $\ln 2 \approx 0.69$ and $1$, and the answer lies between $\frac{1.41 + 0.69}{2} = 1.05$ and $\frac{1.41 + 1}{2} \approx 1.21$. It sits well inside $(0.79,\ 1.50)$: plausible. A sign error in part b), giving $\frac{\sqrt 2 - \ln(1 + \sqrt 2)}{2} < 0.37$, would fall below the rectangle and be caught at once.

a) With $A = 5x$, $B = 3x$: $\sin 5x\cos 3x = \frac{1}{2}\left[\sin 2x + \sin 8x\right]$. Then $\int \sin 5x\cos 3x\,dx = \frac{1}{2}\left(-\frac{\cos 2x}{2} - \frac{\cos 8x}{8}\right) + C = -\frac{\cos 2x}{4} - \frac{\cos 8x}{16} + C$. Why not substitute? $u = \sin 5x$ gives $du = 5\cos 5x\,dx$, and there is a $\cos 3x$, not a $\cos 5x$: the frequencies do not match, so no factor is the differential of anything. And $\int \sin 5x\,dx\cdot\int \cos 3x\,dx$ is not an option: an integral of a product is never the product of the integrals.

b) $\sin 2x\sin 6x = \frac{1}{2}\left[\cos(2x - 6x) - \cos 8x\right] = \frac{1}{2}\left[\cos 4x - \cos 8x\right]$, since cosine is even, $\cos(-4x) = \cos 4x$. Then $\int \sin 2x\sin 6x\,dx = \frac{\sin 4x}{8} - \frac{\sin 8x}{16} + C$. The minus sign belongs to the SUM angle: $\sin A\sin B$ is half of $\cos(A - B)$ minus $\cos(A + B)$. Swapping the two is caught by a check at $x = \frac{\pi}{4}$, not at $x = 0$ where both versions give $0$: $\sin\frac{\pi}{2}\sin\frac{3\pi}{2} = -1$, and $\frac{1}{2}[\cos\pi - \cos 2\pi] = -1$. Correct.

c) Here $A = 2x < B = 5x$: $\sin 2x\cos 5x = \frac{1}{2}\left[\sin(-3x) + \sin 7x\right] = \frac{1}{2}\left[-\sin 3x + \sin 7x\right]$. Sine is ODD, so the negative angle brings out a sign; it is the step students skip, writing $\sin 3x$ for $\sin(-3x)$. Then $\int \sin 2x\cos 5x\,dx = \frac{1}{2}\left(\frac{\cos 3x}{3} - \frac{\cos 7x}{7}\right) + C = \frac{\cos 3x}{6} - \frac{\cos 7x}{14} + C$. With the sign lost, the answer would be $-\frac{\cos 3x}{6} - \frac{\cos 7x}{14}$, whose derivative is $\frac{\sin 3x + \sin 7x}{2} = \sin 5x\cos 2x$: the integral of a different product.