Calculus 2 Quiz: Integrating Sin And Cos Products
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Integrating Sin And Cos ProductsQuestion 1 of 20

Evaluate the definite integral ∫0π/2sin⁡5(x)cos⁡2(x) dx\int_{0}^{\pi/2} \sin^5(x) \cos^2(x) \, dx

235\frac{2}{35}
8105\frac{8}{105}
−8105-\frac{8}{105}
421\frac{4}{21}
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Calculus 2 Quiz

Calculus 2 Quiz: Integrating Sin And Cos Products

Practice Integrating Sin And Cos Products in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Integrating Sin And Cos Products, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Evaluate the definite integral ∫0π/2sin⁡5(x)cos⁡2(x) dx\int_{0}^{\pi/2} \sin^5(x) \cos^2(x) \, dx

  1. 235\frac{2}{35}
  2. 8105\frac{8}{105} (correct answer)
  3. −8105-\frac{8}{105}
  4. 421\frac{4}{21}
Explanation: The power of sine is odd. We save one sine factor and convert the rest to cosine. ∫0π/2sin⁡4(x)cos⁡2(x)sin⁡(x) dx=∫0π/2(1−cos⁡2(x))2cos⁡2(x)sin⁡(x) dx\int_{0}^{\pi/2} \sin^4(x) \cos^2(x) \sin(x) \, dx = \int_{0}^{\pi/2} (1-\cos^2(x))^2 \cos^2(x) \sin(x) \, dx. Let u=cos⁡(x)u = \cos(x), so du=−sin⁡(x)dxdu = -\sin(x) dx. The bounds change: x=0⇒u=1x=0 \Rightarrow u=1 and x=π/2⇒u=0x=\pi/2 \Rightarrow u=0. The integral becomes −∫10(1−u2)2u2 du=∫01(1−2u2+u4)u2 du=∫01(u2−2u4+u6) du-\int_{1}^{0} (1-u^2)^2 u^2 \, du = \int_{0}^{1} (1-2u^2+u^4)u^2 \, du = \int_{0}^{1} (u^2 - 2u^4 + u^6) \, du. Evaluating this gives [u33−2u55+u77]01=13−25+17=35−42+15105=8105[\frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7}]_0^1 = \frac{1}{3} - \frac{2}{5} + \frac{1}{7} = \frac{35 - 42 + 15}{105} = \frac{8}{105}.

Question 2

Which of the following is an antiderivative of f(x)=4sin⁡2(x)cos⁡2(x)f(x) = 4\sin^2(x)\cos^2(x)?

  1. x2+sin⁡(4x)8+C\frac{x}{2} + \frac{\sin(4x)}{8} + C
  2. x2−sin⁡(4x)8+C\frac{x}{2} - \frac{\sin(4x)}{8} + C (correct answer)
  3. x−sin⁡(2x)+Cx - \sin(2x) + C
  4. 49sin⁡3(x)cos⁡3(x)+C\frac{4}{9} \sin^3(x) \cos^3(x) + C
Explanation: First, simplify the integrand using the double angle identity sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x). So, f(x)=(2sin⁡(x)cos⁡(x))2=sin⁡2(2x)f(x) = (2\sin(x)\cos(x))^2 = \sin^2(2x). To integrate sin⁡2(2x)\sin^2(2x), we use the half-angle identity sin⁡2(θ)=1−cos⁡(2θ)2\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}. Here, θ=2x\theta = 2x, so ∫sin⁡2(2x) dx=∫1−cos⁡(4x)2 dx=12∫(1−cos⁡(4x)) dx=12(x−sin⁡(4x)4)+C=x2−sin⁡(4x)8+C\int \sin^2(2x) \, dx = \int \frac{1 - \cos(4x)}{2} \, dx = \frac{1}{2} \int (1 - \cos(4x)) \, dx = \frac{1}{2} (x - \frac{\sin(4x)}{4}) + C = \frac{x}{2} - \frac{\sin(4x)}{8} + C.

Question 3

The integral ∫sin⁡4(x)cos⁡3(x) dx\int \sin^4(x) \cos^3(x) \, dx is evaluated using the substitution u=sin⁡(x)u = \sin(x). Which of the following is the resulting integral in terms of uu?

  1. ∫(u4−u6) du\int (u^4 - u^6) \, du (correct answer)
  2. ∫(u4−u7) du\int (u^4 - u^7) \, du
  3. −∫(u4−u6) du-\int (u^4 - u^6) \, du
  4. ∫(u5−u7) du\int (u^5 - u^7) \, du
Explanation: To use the substitution u=sin⁡(x)u = \sin(x), we need a factor of cos⁡(x)dx\cos(x) dx for dudu. We split cos⁡3(x)\cos^3(x) into cos⁡2(x)cos⁡(x)\cos^2(x) \cos(x). The integral becomes ∫sin⁡4(x)cos⁡2(x)cos⁡(x) dx\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)\cos^2(x) = 1 - \sin^2(x). This gives ∫sin⁡4(x)(1−sin⁡2(x))cos⁡(x) dx\int \sin^4(x) (1 - \sin^2(x)) \cos(x) \, dx. Now, substituting u=sin⁡(x)u = \sin(x) and du=cos⁡(x)dxdu = \cos(x) dx, we get ∫u4(1−u2) du=∫(u4−u6) du\int u^4 (1 - u^2) \, du = \int (u^4 - u^6) \, du.

Question 4

If In=∫0π/2sin⁡nx dxI_n = \int_0^{\pi/2} \sin^n x \, dx, which relationship correctly connects I4I_4 and I2I_2?

  1. I4=12I2I_4 = \frac{1}{2} I_2
  2. I4=23I2I_4 = \frac{2}{3} I_2
  3. I4=34I2I_4 = \frac{3}{4} I_2 (correct answer)
  4. I4=43I2I_4 = \frac{4}{3} I_2
Explanation: When you encounter integrals of the form ∫0π/2sin⁡nx dx\int_0^{\pi/2} \sin^n x \, dx, you're dealing with a classic reduction formula problem. These integrals follow a recursive pattern that connects higher powers to lower powers. To find the relationship between I4I_4 and I2I_2, use integration by parts on I4=∫0π/2sin⁡4x dxI_4 = \int_0^{\pi/2} \sin^4 x \, dx. Rewrite this as ∫0π/2sin⁡3x⋅sin⁡x dx\int_0^{\pi/2} \sin^3 x \cdot \sin x \, dx. Let u=sin⁡3xu = \sin^3 x and dv=sin⁡x dxdv = \sin x \, dx, so du=3sin⁡2xcos⁡x dxdu = 3\sin^2 x \cos x \, dx and v=−cos⁡xv = -\cos x. After applying integration by parts and using the identity sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x, you'll derive the reduction formula: In=n−1nIn−2I_n = \frac{n-1}{n} I_{n-2} for n≥2n \geq 2. Applying this formula: I4=4−14I2=34I2I_4 = \frac{4-1}{4} I_2 = \frac{3}{4} I_2. This confirms answer choice C. Looking at the wrong answers: A gives I4=12I2I_4 = \frac{1}{2} I_2, which would correspond to the coefficient 2−12\frac{2-1}{2} from I2I_2, not I4I_4. B gives 23\frac{2}{3}, which is the reciprocal of the correct fraction—a common error when misremembering the reduction formula. D gives 43\frac{4}{3}, which incorrectly uses nn−1\frac{n}{n-1} instead of n−1n\frac{n-1}{n}. Study tip: Memorize the reduction formula In=n−1nIn−2I_n = \frac{n-1}{n} I_{n-2} for sine integrals. Notice that the coefficient is always less than 1, which makes physical sense since higher powers of sine create "sharper" functions with smaller areas.

Question 5

For the integral ∫sin⁡7x dx\int \sin^7 x \, dx, after the first substitution step, which integral must be evaluated?

  1. ∫(1−u2)3 du\int (1-u^2)^3 \, du where u=sin⁡xu = \sin x
  2. ∫u6(1−u2) du\int u^6(1-u^2) \, du where u=sin⁡xu = \sin x
  3. −∫u6(1−u2) du-\int u^6(1-u^2) \, du where u=cos⁡xu = \cos x
  4. −∫(1−u2)3 du-\int (1-u^2)^3 \, du where u=cos⁡xu = \cos x (correct answer)
Explanation: 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 ∫sin⁡7x dx\int \sin^7 x \, dx, rewrite this as ∫sin⁡6x⋅sin⁡x dx\int \sin^6 x \cdot \sin x \, dx. Since the power of sine is odd, you can express sin⁡6x=(sin⁡2x)3=(1−cos⁡2x)3\sin^6 x = (\sin^2 x)^3 = (1 - \cos^2 x)^3 using the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. The integral becomes ∫(1−cos⁡2x)3sin⁡x dx\int (1 - \cos^2 x)^3 \sin x \, dx. Now substitute u=cos⁡xu = \cos x, so du=−sin⁡x dxdu = -\sin x \, dx, which means sin⁡x dx=−du\sin x \, dx = -du. Substituting: ∫(1−u2)3sin⁡x dx=∫(1−u2)3(−du)=−∫(1−u2)3 du\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⁡xu = \sin x but keeps the original power structure u6(1−u2)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⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. Watch your signs carefully when substituting—du=−sin⁡x dxdu = -\sin x \, dx when u=cos⁡xu = \cos x.

Question 6

Evaluate ∫−π/2π/2sin⁡3(x)cos⁡2(x) dx\int_{-\pi/2}^{\pi/2} \sin^3(x)\cos^2(x) \, dx.

  1. 00 (correct answer)
  2. 215\frac{2}{15}
  3. 415\frac{4}{15}
  4. −215-\frac{2}{15}
Explanation: The integrand is an odd function, and the integral is over a symmetric interval. Let f(x)=sin⁡3(x)cos⁡2(x)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)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][-a, a] is always zero. Therefore, the value of the integral is 0.

Question 7

Which integral requires the use of integration by parts rather than standard sine-cosine techniques?

  1. ∫sin⁡4xcos⁡6x dx\int \sin^4 x \cos^6 x \, dx
  2. ∫xsin⁡2x dx\int x \sin^2 x \, dx (correct answer)
  3. ∫sin⁡3xcos⁡2x dx\int \sin^3 x \cos^2 x \, dx
  4. ∫sin⁡2xcos⁡4x dx\int \sin^2 x \cos^4 x \, dx
Explanation: Choice B contains the factor xx, which cannot be handled by standard sine-cosine power techniques (odd power substitution or even power reduction). It requires integration by parts with u=xu = x and dv=sin⁡2x dxdv = \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⁡xu = \cos x substitution since sine has odd power, and Choice D uses power reduction since both powers are even.

Question 8

Find an antiderivative for f(x)=sin⁡3(x)cos⁡2(x)f(x) = \frac{\sin^3(x)}{\cos^2(x)}.

  1. sec⁡(x)−cos⁡(x)+C\sec(x) - \cos(x) + C
  2. tan⁡(x)−sin⁡(x)+C\tan(x) - \sin(x) + C
  3. sec⁡(x)+cos⁡(x)+C\sec(x) + \cos(x) + C (correct answer)
  4. 13tan⁡3(x)+C\frac{1}{3}\tan^3(x) + C
Explanation: The power of sine is odd. Rewrite the integral as ∫sin⁡2(x)sin⁡(x)cos⁡2(x)dx=∫(1−cos⁡2(x))sin⁡(x)cos⁡2(x)dx\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)u = \cos(x), so du=−sin⁡(x)dxdu = -\sin(x)dx. The integral becomes ∫1−u2u2(−du)=∫u2−1u2du=∫(1−u−2)du\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 uu gives u−u−1−1+C=u+u−1+Cu - \frac{u^{-1}}{-1} + C = u + u^{-1} + C. Substituting back u=cos⁡(x)u = \cos(x) yields cos⁡(x)+1cos⁡(x)+C=cos⁡(x)+sec⁡(x)+C\cos(x) + \frac{1}{\cos(x)} + C = \cos(x) + \sec(x) + C.

Question 9

Which of the following describes the first step in the most efficient method to evaluate ∫sin⁡4(x)cos⁡4(x)dx\int \sin^4(x)\cos^4(x) dx?

  1. Use the substitution u=sin⁡(x)u = \sin(x) after converting cos⁡4(x)\cos^4(x) using cos⁡2(x)=1−sin⁡2(x)\cos^2(x)=1-\sin^2(x).
  2. Apply the half-angle identity sin⁡2(x)=1−cos⁡(2x)2\sin^2(x) = \frac{1-\cos(2x)}{2} to the sin⁡4(x)\sin^4(x) term.
  3. Use the double-angle identity sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x) to simplify the integrand to 116sin⁡4(2x)\frac{1}{16}\sin^4(2x). (correct answer)
  4. Use integration by parts with u=sin⁡4(x)u = \sin^4(x) and dv=cos⁡4(x)dxdv = \cos^4(x)dx.
Explanation: The integrand can be written as (sin⁡(x)cos⁡(x))4(\sin(x)\cos(x))^4. Using the identity sin⁡(x)cos⁡(x)=12sin⁡(2x)\sin(x)\cos(x) = \frac{1}{2}\sin(2x), the expression becomes (12sin⁡(2x))4=116sin⁡4(2x)(\frac{1}{2}\sin(2x))^4 = \frac{1}{16}\sin^4(2x). This simplifies the problem to integrating 116∫sin⁡4(2x)dx\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)\sin^4(x) and cos⁡4(x)\cos^4(x) and multiplying the resulting complex expressions.

Question 10

To evaluate ∫sin⁡m(x)cos⁡n(x)dx\int \sin^m(x)\cos^n(x)dx, a standard strategy is to use substitution. If mm is odd, we use u=cos⁡(x)u=\cos(x). If nn is odd, we use u=sin⁡(x)u=\sin(x). If both mm and nn are even, we use half-angle identities.

Given the strategies in the passage, which integral would require the most applications of half-angle identities to evaluate?

  1. ∫sin⁡2(x)cos⁡4(x)dx\int \sin^2(x)\cos^4(x) dx
  2. ∫sin⁡3(x)cos⁡5(x)dx\int \sin^3(x)\cos^5(x) dx
  3. ∫cos⁡6(x)dx\int \cos^6(x) dx (correct answer)
  4. ∫sin⁡4(x)dx\int \sin^4(x) dx
Explanation: We only need half-angle identities when both powers are even. This eliminates option B. For option A, ∫sin⁡2(x)cos⁡4(x)dx=∫(1−cos⁡(2x)2)(1+cos⁡(2x)2)2dx\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)\cos(2x), \cos^2(2x), \cos^3(2x). The cos⁡2(2x)\cos^2(2x) term needs one half-angle identity. The cos⁡3(2x)\cos^3(2x) term can be solved with substitution. For option D, ∫sin⁡4(x)dx=∫(1−cos⁡(2x)2)2dx=14∫(1−2cos⁡(2x)+cos⁡2(2x))dx\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)\cos^2(2x) term. For option C, ∫cos⁡6(x)dx=∫(1+cos⁡(2x)2)3dx=18∫(1+3cos⁡(2x)+3cos⁡2(2x)+cos⁡3(2x))dx\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)\cos^2(2x) term requires a half-angle identity, leading to a cos⁡(4x)\cos(4x) term. The cos⁡3(2x)\cos^3(2x) term can be solved with substitution. However, if one consistently applies half-angle identities, ∫cos⁡6(x)dx\int\cos^6(x)dx will require more steps. For instance, after the first step, you get cos⁡2(2x)\cos^2(2x) which becomes cos⁡(4x)\cos(4x). And another term. Then ∫cos⁡8(x)dx\int\cos^8(x)dx would need more. Comparing the options, cos⁡6(x)\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.

Question 11

To evaluate ∫sin⁡3xcos⁡4x dx\int \sin^3 x \cos^4 x \, dx, which substitution and resulting integral is correct?

  1. u=cos⁡xu = \cos x, du=−sin⁡x dxdu = -\sin x \, dx, giving −∫u4(1−u2) du-\int u^4(1-u^2) \, du (correct answer)
  2. u=sin⁡xu = \sin x, du=cos⁡x dxdu = \cos x \, dx, giving ∫u3(1−u2)2 du\int u^3(1-u^2)^2 \, du
  3. u=cos⁡xu = \cos x, du=−sin⁡x dxdu = -\sin x \, dx, giving −∫u4(1−u2)2 du-\int u^4(1-u^2)^2 \, du
  4. u=sin⁡xu = \sin x, du=cos⁡x dxdu = \cos x \, dx, giving ∫u3(1−u2) du\int u^3(1-u^2) \, du
Explanation: Since the power of sine is odd (3), we save one factor of sine and convert the rest using sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x. We have sin⁡3x=sin⁡2x⋅sin⁡x=(1−cos⁡2x)sin⁡x\sin^3 x = \sin^2 x \cdot \sin x = (1-\cos^2 x)\sin x. With u=cos⁡xu = \cos x, du=−sin⁡x dxdu = -\sin x \, dx, the integral becomes −∫(1−u2)u4 du=−∫u4(1−u2) du-\int (1-u^2)u^4 \, du = -\int u^4(1-u^2) \, du. Choice B uses the wrong substitution for odd sine power. Choice C has the wrong converted form. Choice D uses wrong substitution and wrong form.

Question 12

What is the value of ∫0π/2sin⁡2xcos⁡2x dx\int_0^{\pi/2} \sin^2 x \cos^2 x \, dx?

  1. π16\frac{\pi}{16} (correct answer)
  2. π8\frac{\pi}{8}
  3. π4\frac{\pi}{4}
  4. π32\frac{\pi}{32}
Explanation: Since both powers are even, use the identity sin⁡2xcos⁡2x=14sin⁡2(2x)\sin^2 x \cos^2 x = \frac{1}{4}\sin^2(2x). Then sin⁡2(2x)=1−cos⁡(4x)2\sin^2(2x) = \frac{1-\cos(4x)}{2}, so the integral becomes 14∫0π/21−cos⁡(4x)2 dx=18∫0π/2(1−cos⁡(4x)) dx=18[x−sin⁡(4x)4]0π/2=18⋅π2=π16\frac{1}{4} \int_0^{\pi/2} \frac{1-\cos(4x)}{2} \, dx = \frac{1}{8} \int_0^{\pi/2} (1-\cos(4x)) \, dx = \frac{1}{8}\left[x - \frac{\sin(4x)}{4}\right]_0^{\pi/2} = \frac{1}{8} \cdot \frac{\pi}{2} = \frac{\pi}{16}. Choice B forgets the factor of 2 in the denominator. Choice C uses wrong power reduction. Choice D has an extra factor of 2 in denominator.

Question 13

Evaluate the integral ∫cos⁡(5x)cos⁡(3x) dx\int \cos(5x)\cos(3x) \, dx.

  1. sin⁡(2x)2+sin⁡(8x)8+C\frac{\sin(2x)}{2} + \frac{\sin(8x)}{8} + C
  2. −cos⁡(2x)4−cos⁡(8x)16+C-\frac{\cos(2x)}{4} - \frac{\cos(8x)}{16} + C
  3. sin⁡(2x)4+sin⁡(8x)16+C\frac{\sin(2x)}{4} + \frac{\sin(8x)}{16} + C (correct answer)
  4. sin⁡(2x)4−sin⁡(8x)16+C\frac{\sin(2x)}{4} - \frac{\sin(8x)}{16} + C
Explanation: This integral is solved using the product-to-sum identity cos⁡(A)cos⁡(B)=12[cos⁡(A−B)+cos⁡(A+B)]\cos(A)\cos(B) = \frac{1}{2}[\cos(A-B) + \cos(A+B)]. With A=5xA=5x and B=3xB=3x, the integrand becomes 12[cos⁡(2x)+cos⁡(8x)]\frac{1}{2}[\cos(2x) + \cos(8x)]. Integrating this expression gives ∫12(cos⁡(2x)+cos⁡(8x)) dx=12[sin⁡(2x)2+sin⁡(8x)8]+C=sin⁡(2x)4+sin⁡(8x)16+C\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.

Question 14

What is ∫0πsin⁡2xcos⁡2x dx\int_0^{\pi} \sin^2 x \cos^2 x \, dx?

  1. 00
  2. π4\frac{\pi}{4}
  3. π16\frac{\pi}{16}
  4. π8\frac{\pi}{8} (correct answer)
Explanation: When you encounter products of powers of sine and cosine, trigonometric identities are your key tools for simplification. This integral requires the power-reduction formulas to convert the fourth-degree trigonometric expression into something manageable. Start by using the identity sin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2} and cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1 + \cos 2x}{2}. Therefore: sin⁡2xcos⁡2x=1−cos⁡2x2⋅1+cos⁡2x2=(1−cos⁡2x)(1+cos⁡2x)4=1−cos⁡22x4\sin^2 x \cos^2 x = \frac{1 - \cos 2x}{2} \cdot \frac{1 + \cos 2x}{2} = \frac{(1 - \cos 2x)(1 + \cos 2x)}{4} = \frac{1 - \cos^2 2x}{4} Since 1−cos⁡22x=sin⁡22x1 - \cos^2 2x = \sin^2 2x, we have sin⁡2xcos⁡2x=sin⁡22x4\sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}. Now apply the power-reduction formula again: sin⁡22x=1−cos⁡4x2\sin^2 2x = \frac{1 - \cos 4x}{2}, giving us: sin⁡2xcos⁡2x=14⋅1−cos⁡4x2=1−cos⁡4x8\sin^2 x \cos^2 x = \frac{1}{4} \cdot \frac{1 - \cos 4x}{2} = \frac{1 - \cos 4x}{8} The integral becomes: ∫0π1−cos⁡4x8dx=18[x−sin⁡4x4]0π=18[π−0]=π8\int_0^{\pi} \frac{1 - \cos 4x}{8} dx = \frac{1}{8} \left[ x - \frac{\sin 4x}{4} \right]_0^{\pi} = \frac{1}{8}[\pi - 0] = \frac{\pi}{8} Choice A (00) ignores that we have a squared term that's always non-negative. Choice B (π4\frac{\pi}{4}) likely comes from forgetting the second power reduction. Choice C (π16\frac{\pi}{16}) probably results from an arithmetic error in the denominators. Remember: products of even powers of sine and cosine almost always require repeated applications of power-reduction formulas. Work systematically through each reduction step.

Question 15

Which statement about ∫sin⁡mxcos⁡nx dx\int \sin^m x \cos^n x \, dx is always true when both mm and nn are positive even integers?

  1. The integral can be evaluated using substitution u=sin⁡xu = \sin x
  2. The integral requires repeated application of power reduction formulas (correct answer)
  3. The integral can be evaluated using substitution u=cos⁡xu = \cos x
  4. The integral equals zero over any symmetric interval
Explanation: When both mm and nn are even, neither substitution method works (since we need at least one odd power for substitution). We must use power reduction formulas sin⁡2x=1−cos⁡(2x)2\sin^2 x = \frac{1-\cos(2x)}{2} and cos⁡2x=1+cos⁡(2x)2\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.

Question 16

What is the average value of the function f(x)=cos⁡3(x)f(x) = \cos^3(x) on the interval [0,π/2][0, \pi/2]?

  1. 23π\frac{2}{3\pi}
  2. 43π\frac{4}{3\pi} (correct answer)
  3. 23\frac{2}{3}
  4. 12π\frac{1}{2\pi}
Explanation: The average value is 1b−a∫abf(x)dx\frac{1}{b-a}\int_a^b f(x) dx. Here, it is 1π/2−0∫0π/2cos⁡3(x)dx=2π∫0π/2cos⁡3(x)dx\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 ∫cos⁡2(x)cos⁡(x)dx=∫(1−sin⁡2(x))cos⁡(x)dx\int \cos^2(x)\cos(x) dx = \int(1-\sin^2(x))\cos(x)dx. Let u=sin⁡(x)u=\sin(x), so du=cos⁡(x)dxdu=\cos(x)dx. The bounds change from 0,π/20, \pi/2 to 0,10, 1. The integral is ∫01(1−u2)du=[u−u33]01=1−13=23\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 2π⋅23=43π\frac{2}{\pi} \cdot \frac{2}{3} = \frac{4}{3\pi}.

Question 17

Evaluate the integral ∫sin⁡3(x)cos⁡4(x) dx\int \frac{\sin^3(x)}{\cos^4(x)} \, dx.

  1. sec⁡3(x)3−sec⁡(x)+C\frac{\sec^3(x)}{3} - \sec(x) + C (correct answer)
  2. sec⁡3(x)3+sec⁡(x)+C\frac{\sec^3(x)}{3} + \sec(x) + C
  3. sec⁡(x)−sec⁡3(x)3+C\sec(x) - \frac{\sec^3(x)}{3} + C
  4. tan⁡4(x)4+C\frac{\tan^4(x)}{4} + C
Explanation: This can be written as ∫tan⁡3(x)sec⁡(x) dx\int \tan^3(x) \sec(x) \, dx. The standard method for this form is to factor out sec⁡(x)tan⁡(x)\sec(x)\tan(x). ∫tan⁡2(x)(sec⁡(x)tan⁡(x)) dx\int \tan^2(x) (\sec(x)\tan(x)) \, dx. Then use the identity tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1. The integral becomes ∫(sec⁡2(x)−1)sec⁡(x)tan⁡(x) dx\int (\sec^2(x) - 1) \sec(x)\tan(x) \, dx. Let u=sec⁡(x)u = \sec(x), so du=sec⁡(x)tan⁡(x) dxdu = \sec(x)\tan(x) \, dx. The integral in uu is ∫(u2−1) du=u33−u+C\int (u^2 - 1) \, du = \frac{u^3}{3} - u + C. Substituting back gives sec⁡3(x)3−sec⁡(x)+C\frac{\sec^3(x)}{3} - \sec(x) + C.

Question 18

Evaluate the definite integral ∫π/4π/2sin⁡4(x) dx\int_{\pi/4}^{\pi/2} \sin^4(x) \, dx.

  1. 3π16\frac{3\pi}{16}
  2. 3π32+14\frac{3\pi}{32} + \frac{1}{4} (correct answer)
  3. 3π32−14\frac{3\pi}{32} - \frac{1}{4}
  4. 3π16−12\frac{3\pi}{16} - \frac{1}{2}
Explanation: First, find the antiderivative of sin⁡4(x)\sin^4(x). ∫sin⁡4(x)dx=∫(sin⁡2(x))2dx=∫(1−cos⁡(2x)2)2dx=14∫(1−2cos⁡(2x)+cos⁡2(2x))dx\int \sin^4(x) dx = \int (\sin^2(x))^2 dx = \int (\frac{1-\cos(2x)}{2})^2 dx = \frac{1}{4} \int (1 - 2\cos(2x) + \cos^2(2x)) dx. Use the half-angle identity again for cos⁡2(2x)=1+cos⁡(4x)2\cos^2(2x) = \frac{1+\cos(4x)}{2}. The integral becomes 14∫(1−2cos⁡(2x)+1+cos⁡(4x)2)dx=14∫(32−2cos⁡(2x)+12cos⁡(4x))dx=14[32x−sin⁡(2x)+18sin⁡(4x)]\frac{1}{4} \int (1 - 2\cos(2x) + \frac{1+\cos(4x)}{2}) dx = \frac{1}{4} \int (\frac{3}{2} - 2\cos(2x) + \frac{1}{2}\cos(4x)) dx = \frac{1}{4} [\frac{3}{2}x - \sin(2x) + \frac{1}{8}\sin(4x)]. Now evaluate from π/4\pi/4 to π/2\pi/2. At x=π/2x=\pi/2: 14[3π4−sin⁡(π)+18sin⁡(2π)]=3π16\frac{1}{4}[\frac{3\pi}{4} - \sin(\pi) + \frac{1}{8}\sin(2\pi)] = \frac{3\pi}{16}. At x=π/4x=\pi/4: 14[3π8−sin⁡(π/2)+18sin⁡(π)]=14[3π8−1]=3π32−14\frac{1}{4}[\frac{3\pi}{8} - \sin(\pi/2) + \frac{1}{8}\sin(\pi)] = \frac{1}{4}[\frac{3\pi}{8} - 1] = \frac{3\pi}{32} - \frac{1}{4}. The result is 3π16−(3π32−14)=6π−3π32+14=3π32+14\frac{3\pi}{16} - (\frac{3\pi}{32} - \frac{1}{4}) = \frac{6\pi - 3\pi}{32} + \frac{1}{4} = \frac{3\pi}{32} + \frac{1}{4}.

Question 19

Which of the following is equivalent to ∫1sin⁡2(x)cos⁡2(x)dx\int \frac{1}{\sin^2(x)\cos^2(x)} dx?

  1. tan⁡(x)+cot⁡(x)+C\tan(x) + \cot(x) + C
  2. −2cot⁡(2x)+C-2\cot(2x) + C (correct answer)
  3. sec⁡(x)tan⁡(x)−csc⁡(x)cot⁡(x)+C\sec(x)\tan(x) - \csc(x)\cot(x) + C
  4. 13sin⁡3(x)cos⁡3(x)+C\frac{1}{3\sin^3(x)\cos^3(x)} + C
Explanation: One method is to use the identity sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x), so sin⁡(x)cos⁡(x)=12sin⁡(2x)\sin(x)\cos(x) = \frac{1}{2}\sin(2x). The integrand is 1(12sin⁡(2x))2=4sin⁡2(2x)=4csc⁡2(2x)\frac{1}{(\frac{1}{2}\sin(2x))^2} = \frac{4}{\sin^2(2x)} = 4\csc^2(2x). The integral is ∫4csc⁡2(2x)dx\int 4\csc^2(2x) dx. Using the substitution u=2xu=2x, du=2dxdu=2dx, we get ∫4csc⁡2(u)du2=2∫csc⁡2(u)du=−2cot⁡(u)+C=−2cot⁡(2x)+C\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)1 = \sin^2(x) + \cos^2(x) in the numerator, giving ∫sin⁡2(x)+cos⁡2(x)sin⁡2(x)cos⁡2(x)dx=∫(1cos⁡2(x)+1sin⁡2(x))dx=∫(sec⁡2(x)+csc⁡2(x))dx=tan⁡(x)−cot⁡(x)+C\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, −2cot⁡(2x)-2\cot(2x) and tan⁡(x)−cot⁡(x)\tan(x) - \cot(x), are equivalent via double angle identities.

Question 20

The integral ∫sin⁡2(x)cos⁡(2x) dx\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?

  1. Replace cos⁡(2x)\cos(2x) with cos⁡2(x)−sin⁡2(x)\cos^2(x) - \sin^2(x) to get an integral entirely in terms of powers of sin⁡(x)\sin(x) and cos⁡(x)\cos(x).
  2. Replace sin⁡2(x)\sin^2(x) with 1−cos⁡(2x)2\frac{1-\cos(2x)}{2} to get an integral in terms of powers of cos⁡(2x)\cos(2x). (correct answer)
  3. Replace sin⁡2(x)\sin^2(x) with 1−cos⁡2(x)1-\cos^2(x) and then use integration by parts.
  4. Use the product-to-sum formula sin⁡(A)cos⁡(B)=12[sin⁡(A−B)+sin⁡(A+B)]\sin(A)\cos(B) = \frac{1}{2}[\sin(A-B)+\sin(A+B)].
Explanation: By replacing sin⁡2(x)\sin^2(x) with the half-angle identity 1−cos⁡(2x)2\frac{1-\cos(2x)}{2}, the integral becomes ∫1−cos⁡(2x)2cos⁡(2x) dx=12∫(cos⁡(2x)−cos⁡2(2x)) dx\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)\cos^2(2x), requires one more application of the half-angle identity. This approach systematically reduces the complexity. Approach A results in ∫sin⁡2(x)(cos⁡2(x)−sin⁡2(x))dx=∫(sin⁡2(x)cos⁡2(x)−sin⁡4(x))dx\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.