Integrating sin & cos Products - Calculus 2

1. Which of the following describes the first step in the most efficient method to evaluate $\int \sin^4(x)\cos^4(x) dx$?
   Answer: Use the double-angle identity $\sin(2x) = 2\sin(x)\cos(x)$ to simplify the integrand to $\frac{1}{16}\sin^4(2x)$. The integrand can be written as $(\sin(x)\cos(x))^4$. Using the identity $\sin(x)\cos(x) = \frac{1}{2}\sin(2x)$, the expression becomes $(\frac{1}{2}\sin(2x))^4 = \frac{1}{16}\sin^4(2x)$. This simplifies the problem to integrating $\frac{1}{16}\int \sin^4(2x) dx$, which can then be solved using half-angle identities. This is more efficient than directly applying half-angle identities to $\sin^4(x)$ and $\cos^4(x)$ and multiplying the resulting complex expressions.
2. Evaluate the integral $\int \frac{\sin^3(x)}{\cos^4(x)} \, dx$.
   Answer: $\frac{\sec^3(x)}{3} - \sec(x) + C$ This can be written as $\int \tan^3(x) \sec(x) \, dx$. The standard method for this form is to factor out $\sec(x)\tan(x)$. $\int \tan^2(x) (\sec(x)\tan(x)) \, dx$. Then use the identity $\tan^2(x) = \sec^2(x) - 1$. The integral becomes $\int (\sec^2(x) - 1) \sec(x)\tan(x) \, dx$. Let $u = \sec(x)$, so $du = \sec(x)\tan(x) \, dx$. The integral in $u$ is $\int (u^2 - 1) \, du = \frac{u^3}{3} - u + C$. Substituting back gives $\frac{\sec^3(x)}{3} - \sec(x) + C$.
3. To evaluate $\int \sin^6 x \cos^4 x \, dx$, the first step should be to apply which identity?
   Answer: $\sin^2 x = \frac{1-\cos(2x)}{2}$ and $\cos^2 x = \frac{1+\cos(2x)}{2}$ Since both sine and cosine have even powers, we cannot use the substitution method (which requires at least one odd power). We must use power reduction formulas. We apply $\sin^2 x = \frac{1-\cos(2x)}{2}$ and $\cos^2 x = \frac{1+\cos(2x)}{2}$ to get $\sin^6 x \cos^4 x = \left(\frac{1-\cos(2x)}{2}\right)^3 \left(\frac{1+\cos(2x)}{2}\right)^2$. Choices A and B suggest substitution, which doesn't work when both powers are even. Choice D uses an incorrect identity.
4. Which integral requires the use of integration by parts rather than standard sine-cosine techniques?
   Answer: $\int x \sin^2 x \, dx$ Choice B contains the factor $x$, which cannot be handled by standard sine-cosine power techniques (odd power substitution or even power reduction). It requires integration by parts with $u = x$ and $dv = \sin^2 x \, dx$. All other choices involve only powers of sine and cosine: Choice A uses power reduction since both powers are even, Choice C uses $u = \cos x$ substitution since sine has odd power, and Choice D uses power reduction since both powers are even.
5. Let $f(x) = \sin^n(x)$ for a positive even integer $n$. Which statement accurately describes the integral $I = \int_0^{2\pi} f(x) dx$?
   Answer: $I = 2 \int_0^{\pi} f(x) dx$ and $I > 0$. When evaluating integrals of even powers of trigonometric functions over symmetric intervals, you need to analyze both the symmetry properties and the sign of the function. For $f(x) = \sin^n(x)$ where $n$ is a positive even integer, the key insight is that $\sin^n(x) \geq 0$ for all $x$ since any real number raised to an even power is non-negative. This immediately tells us the integral $I = \int_0^{2\pi} \sin^n(x) dx$ must be positive, since we're integrating a non-negative function over a positive interval. The function $\sin^n(x)$ has period $\pi$ when $n$ is even (since $\sin^n(x + \pi) = (-\sin(x))^n = \sin^n(x)$). Therefore, $\int_0^{2\pi} \sin^n(x) dx = 2\int_0^{\pi} \sin^n(x) dx$, confirming that statement D is correct. Option A incorrectly assumes the integral equals zero due to symmetry. While $\sin(x)$ itself has canceling positive and negative regions, $\sin^n(x)$ for even $n$ is always non-negative, so there's no cancellation. Option B makes a false equivalence between integrals over different intervals and incorrectly concludes the result is zero. Option C correctly identifies a symmetry relationship but wrongly claims $I < 0$, which contradicts the fact that we're integrating a non-negative function. Study tip: When dealing with even powers of sine or cosine, remember that the result is always non-negative, eliminating any answer choices claiming the integral is zero or negative.
6. Which statement about $\int \sin^m x \cos^n x \, dx$ is always true when both $m$ and $n$ are positive even integers?
   Answer: The integral requires repeated application of power reduction formulas When both $m$ and $n$ are even, neither substitution method works (since we need at least one odd power for substitution). We must use power reduction formulas $\sin^2 x = \frac{1-\cos(2x)}{2}$ and $\cos^2 x = \frac{1+\cos(2x)}{2}$, and may need to apply them repeatedly if the powers are large. Choices A and C suggest substitution methods that don't work when both powers are even. Choice D is false; these integrands are not generally odd functions.
7. Given the strategies in the passage, which integral would require the most applications of half-angle identities to evaluate?
   Answer: $\int \cos^6(x) dx$ We only need half-angle identities when both powers are even. This eliminates option B. For option A, $\int \sin^2(x)\cos^4(x) dx = \int (\frac{1-\cos(2x)}{2})(\frac{1+\cos(2x)}{2})^2 dx$. This will expand into terms with $\cos(2x), \cos^2(2x), \cos^3(2x)$. The $\cos^2(2x)$ term needs one half-angle identity. The $\cos^3(2x)$ term can be solved with substitution. For option D, $\int \sin^4(x) dx = \int (\frac{1-\cos(2x)}{2})^2 dx = \frac{1}{4}\int(1-2\cos(2x)+\cos^2(2x))dx$. This requires one more half-angle identity for the $\cos^2(2x)$ term. For option C, $\int \cos^6(x) dx = \int (\frac{1+\cos(2x)}{2})^3 dx = \frac{1}{8}\int(1+3\cos(2x)+3\cos^2(2x)+\cos^3(2x))dx$. The $\cos^2(2x)$ term requires a half-angle identity, leading to a $\cos(4x)$ term. The $\cos^3(2x)$ term can be solved with substitution. However, if one consistently applies half-angle identities, $\int\cos^6(x)dx$ will require more steps. For instance, after the first step, you get $\cos^2(2x)$ which becomes $\cos(4x)$. And another term. Then $\int\cos^8(x)dx$ would need more. Comparing the options, $\cos^6(x)$ has the highest even power, requiring expansion to the third power, which creates more terms needing identities than squaring as in the other options.
8. Evaluate $\int_{\pi}^{3\pi/2} \cos^3(x) \, dx$.
   Answer: $-\frac{2}{3}$ When you encounter an integral of $\cos^3(x)$, you need a strategy to handle the odd power. The key technique is to separate one cosine factor and use the Pythagorean identity to convert the remaining even power. Start by rewriting: $\cos^3(x) = \cos^2(x) \cdot \cos(x) = (1 - \sin^2(x)) \cos(x)$ This sets up a perfect substitution. Let $u = \sin(x)$, so $du = \cos(x) \, dx$. The integral becomes: $\int (1 - u^2) \, du = u - \frac{u^3}{3} = \sin(x) - \frac{\sin^3(x)}{3}$ Now evaluate from $\pi$ to $\frac{3\pi}{2}$: $\left[\sin(x) - \frac{\sin^3(x)}{3}\right]_{\pi}^{3\pi/2}$ At $x = \frac{3\pi}{2}$: $\sin\left(\frac{3\pi}{2}\right) = -1$, so the expression equals $-1 - \frac{(-1)^3}{3} = -1 + \frac{1}{3} = -\frac{2}{3}$ At $x = \pi$: $\sin(\pi) = 0$, so the expression equals $0$ The definite integral is $-\frac{2}{3} - 0 = -\frac{2}{3}$, which is answer choice D. Choice A ($\frac{2}{3}$) results from a sign error when evaluating the bounds. Choice B ($-\frac{4}{3}$) likely comes from incorrectly handling the $\sin^3$ term. Choice C ($0$) might result from incorrectly assuming symmetry properties that don't apply here. Strategy tip: For odd powers of cosine or sine, always separate one factor and use substitution. Remember to be extra careful with signs when evaluating trigonometric functions at specific angles.
9. Evaluate $\int_{-\pi/2}^{\pi/2} \sin^3(x)\cos^2(x) \, dx$.
   Answer: $0$ The integrand is an odd function, and the integral is over a symmetric interval. Let $f(x) = \sin^3(x)\cos^2(x)$. Then $f(-x) = \sin^3(-x)\cos^2(-x) = (-\sin(x))^3(\cos(x))^2 = -\sin^3(x)\cos^2(x) = -f(x)$. The integral of an odd function over an interval $[-a, a]$ is always zero. Therefore, the value of the integral is 0.
10. Which of the following is equivalent to $\int \frac{1}{\sin^2(x)\cos^2(x)} dx$?
    Answer: $-2\cot(2x) + C$ One method is to use the identity $\sin(2x) = 2\sin(x)\cos(x)$, so $\sin(x)\cos(x) = \frac{1}{2}\sin(2x)$. The integrand is $\frac{1}{(\frac{1}{2}\sin(2x))^2} = \frac{4}{\sin^2(2x)} = 4\csc^2(2x)$. The integral is $\int 4\csc^2(2x) dx$. Using the substitution $u=2x$, $du=2dx$, we get $\int 4\csc^2(u) \frac{du}{2} = 2 \int \csc^2(u) du = -2\cot(u) + C = -2\cot(2x) + C$. Another method is to write $1 = \sin^2(x) + \cos^2(x)$ in the numerator, giving $\int \frac{\sin^2(x) + \cos^2(x)}{\sin^2(x)\cos^2(x)} dx = \int (\frac{1}{\cos^2(x)} + \frac{1}{\sin^2(x)}) dx = \int (\sec^2(x) + \csc^2(x)) dx = \tan(x) - \cot(x) + C$. The two answers, $-2\cot(2x)$ and $\tan(x) - \cot(x)$, are equivalent via double angle identities.
11. What is the average value of the function $f(x) = \cos^3(x)$ on the interval $[0, \pi/2]$?
    Answer: $\frac{4}{3\pi}$ The average value is $\frac{1}{b-a}\int_a^b f(x) dx$. Here, it is $\frac{1}{\pi/2 - 0} \int_0^{\pi/2} \cos^3(x) dx = \frac{2}{\pi} \int_0^{\pi/2} \cos^3(x) dx$. To evaluate the integral, write $\int \cos^2(x)\cos(x) dx = \int(1-\sin^2(x))\cos(x)dx$. Let $u=\sin(x)$, so $du=\cos(x)dx$. The bounds change from $0, \pi/2$ to $0, 1$. The integral is $\int_0^1 (1-u^2)du = [u-\frac{u^3}{3}]_0^1 = 1-\frac{1}{3} = \frac{2}{3}$. The average value is $\frac{2}{\pi} \cdot \frac{2}{3} = \frac{4}{3\pi}$.
12. To evaluate $\int \sin^3(x) \cos^5(x) \, dx$, one could use the substitution $u=\cos(x)$ or $u=\sin(x)$. Which statement correctly describes the resulting polynomial integrals in $u$?
    Answer: Both substitutions lead to integrals of seventh-degree polynomials. Case 1: Let $u = \cos(x)$, $du = -\sin(x)dx$. Integral is $\int \sin^2(x)\cos^5(x)\sin(x)dx = \int (1-\cos^2 x)\cos^5 x \sin x dx$. This becomes $-\int (1-u^2)u^5 du = \int (u^7-u^5)du$. The integrand is a 7th degree polynomial. Case 2: Let $u = \sin(x)$, $du = \cos(x)dx$. Integral is $\int \sin^3(x)\cos^4(x)\cos(x)dx = \int \sin^3 x (1-\sin^2 x)^2 \cos x dx$. This becomes $\int u^3(1-u^2)^2 du = \int u^3(1-2u^2+u^4)du = \int (u^3-2u^5+u^7)du$. The integrand is also a 7th degree polynomial. Therefore, both substitutions lead to integrals of seventh-degree polynomials.
13. Evaluate $\int_{0}^{2\pi} \sin^2(x)\cos^3(x) \, dx$.
    Answer: $0$ We can analyze the symmetry of the integrand $f(x) = \sin^2(x)\cos^3(x)$ around $x=\pi$. Let's consider the integral from $\pi$ to $2\pi$. Let $u = x - \pi$, so $x=u+\pi$ and $dx=du$. The bounds become $0$ to $\pi$. The integral becomes $\int_0^\pi \sin^2(u+\pi)\cos^3(u+\pi) du = \int_0^\pi (-\sin u)^2(-\cos u)^3 du = -\int_0^\pi \sin^2(u)\cos^3(u) du$. Thus, $\int_0^{2\pi} f(x) dx = \int_0^\pi f(x) dx + \int_\pi^{2\pi} f(x) dx = \int_0^\pi f(x) dx - \int_0^\pi f(x) dx = 0$. Alternatively, evaluating the indefinite integral with $u=\sin x$ gives $\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$. Evaluating from 0 to $2\pi$ gives $(\frac{0}{3}-\frac{0}{5}) - (\frac{0}{3}-\frac{0}{5}) = 0$.
14. The evaluation of $\int \sin^m(x)\cos^n(x)dx$ using the substitution $u = \sin(x)$ leads to the integral $\int (u^6 - 2u^4 + u^2)du$. What were the original integer powers $m$ and $n$?
    Answer: $m=2, n=5$ The substitution $u = \sin(x)$ implies $du = \cos(x)dx$. This strategy is used when the power of cosine, $n$, is odd. The integrand in $u$ is $\int (u^6 - 2u^4 + u^2)du = \int u^2(u^4 - 2u^2 + 1)du = \int u^2(u^2-1)^2 du$. The general form after substitution is $\int u^m (1-u^2)^{(n-1)/2} du$. Comparing this with $\int u^2(1-u^2)^2 du$, we can identify $m=2$ and $(n-1)/2 = 2$. Solving for $n$ gives $n-1=4$, so $n=5$. The original integral was $\int \sin^2(x)\cos^5(x)dx$.
15. Evaluate the integral $\int \cos(5x)\cos(3x) \, dx$.
    Answer: $\frac{\sin(2x)}{4} + \frac{\sin(8x)}{16} + C$ This integral is solved using the product-to-sum identity $\cos(A)\cos(B) = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$. With $A=5x$ and $B=3x$, the integrand becomes $\frac{1}{2}[\cos(2x) + \cos(8x)]$. Integrating this expression gives $\int \frac{1}{2}(\cos(2x) + \cos(8x)) \, dx = \frac{1}{2} [\frac{\sin(2x)}{2} + \frac{\sin(8x)}{8}] + C = \frac{\sin(2x)}{4} + \frac{\sin(8x)}{16} + C$.
16. If $\int \sin^5 x \, dx = -\cos x + \frac{2}{3}\cos^3 x - \frac{1}{5}\cos^5 x + C$, what is $\int \sin^5 x \cos^2 x \, dx$?
    Answer: $-\frac{1}{3}\cos^3 x + \frac{2}{5}\cos^5 x - \frac{1}{7}\cos^7 x + C$ We have $\sin^5 x \cos^2 x = \sin^4 x \cdot \sin x \cdot \cos^2 x = (1-\cos^2 x)^2 \cos^2 x \sin x$. With $u = \cos x$, $du = -\sin x \, dx$, this becomes $-\int (1-u^2)^2 u^2 \, du = -\int (1-2u^2+u^4)u^2 \, du = -\int (u^2-2u^4+u^6) \, du = -\frac{u^3}{3} + \frac{2u^5}{5} - \frac{u^7}{7} + C$. Choice B has wrong coefficient for $\cos^5 x$. Choice C has wrong signs. Choice D incorrectly includes lower powers.
17. For the integral $\int \sin^7 x \, dx$, after the first substitution step, which integral must be evaluated?
    Answer: $-\int (1-u^2)^3 \, du$ where $u = \cos x$ When you encounter integrals of odd powers of sine or cosine, the key strategy is to separate one factor and use substitution with the Pythagorean identity. For $\int \sin^7 x \, dx$, rewrite this as $\int \sin^6 x \cdot \sin x \, dx$. Since the power of sine is odd, you can express $\sin^6 x = (\sin^2 x)^3 = (1 - \cos^2 x)^3$ using the identity $\sin^2 x + \cos^2 x = 1$. The integral becomes $\int (1 - \cos^2 x)^3 \sin x \, dx$. Now substitute $u = \cos x$, so $du = -\sin x \, dx$, which means $\sin x \, dx = -du$. Substituting: $\int (1 - u^2)^3 \sin x \, dx = \int (1 - u^2)^3 (-du) = -\int (1 - u^2)^3 \, du$ This confirms answer D is correct. Looking at the wrong answers: A has the right integrand but wrong sign and substitution variable. B uses $u = \sin x$ but keeps the original power structure $u^6(1-u^2)$ without properly applying the identity—this would come from a cosine integral approach. C has the right sign and substitution variable but the wrong integrand form. Study tip: For odd powers of trig functions, always separate one factor for substitution and convert the remaining even power using $\sin^2 x + \cos^2 x = 1$. Watch your signs carefully when substituting—$du = -\sin x \, dx$ when $u = \cos x$.
18. Find an antiderivative for $f(x) = \frac{\sin^3(x)}{\cos^2(x)}$.
    Answer: $\sec(x) + \cos(x) + C$ The power of sine is odd. Rewrite the integral as $\int \frac{\sin^2(x)\sin(x)}{\cos^2(x)} dx = \int \frac{(1-\cos^2(x))\sin(x)}{\cos^2(x)} dx$. Let $u = \cos(x)$, so $du = -\sin(x)dx$. The integral becomes $\int \frac{1-u^2}{u^2}(-du) = \int \frac{u^2-1}{u^2} du = \int (1 - u^{-2}) du$. Integrating with respect to $u$ gives $u - \frac{u^{-1}}{-1} + C = u + u^{-1} + C$. Substituting back $u = \cos(x)$ yields $\cos(x) + \frac{1}{\cos(x)} + C = \cos(x) + \sec(x) + C$.
19. The integral $\int \sin^4(x) \cos^3(x) \, dx$ is evaluated using the substitution $u = \sin(x)$. Which of the following is the resulting integral in terms of $u$?
    Answer: $\int (u^4 - u^6) \, du$ To use the substitution $u = \sin(x)$, we need a factor of $\cos(x) dx$ for $du$. We split $\cos^3(x)$ into $\cos^2(x) \cos(x)$. The integral becomes $\int \sin^4(x) \cos^2(x) \cos(x) \, dx$. We convert the remaining even power of cosine to sine using $\cos^2(x) = 1 - \sin^2(x)$. This gives $\int \sin^4(x) (1 - \sin^2(x)) \cos(x) \, dx$. Now, substituting $u = \sin(x)$ and $du = \cos(x) dx$, we get $\int u^4 (1 - u^2) \, du = \int (u^4 - u^6) \, du$.
20. The integral $\int \sin^2(x) \cos(2x) \, dx$ can be solved by first applying a trigonometric identity to one of the terms. Which approach is most effective?
    Answer: Replace $\sin^2(x)$ with $\frac{1-\cos(2x)}{2}$ to get an integral in terms of powers of $\cos(2x)$. By replacing $\sin^2(x)$ with the half-angle identity $\frac{1-\cos(2x)}{2}$, the integral becomes $\int \frac{1-\cos(2x)}{2} \cos(2x) \, dx = \frac{1}{2} \int (\cos(2x) - \cos^2(2x)) \, dx$. The first term is easily integrated. The second term, $\cos^2(2x)$, requires one more application of the half-angle identity. This approach systematically reduces the complexity. Approach A results in $\int \sin^2(x)(\cos^2(x) - \sin^2(x)) dx = \int (\sin^2(x)\cos^2(x) - \sin^4(x)) dx$, which is more complicated to solve.

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