Calculus 2 Quiz: Algebraic And Trig Simplification
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Algebraic And Trig SimplificationQuestion 1 of 20

To integrate ∫sec⁡xtan⁡x−1dx\int \frac{\sec x}{\tan x - 1} dx, it is useful to rewrite the integrand entirely in terms of sin⁡x\sin x and cos⁡x\cos x. What is the resulting expression?

cos⁡xsin⁡x−cos⁡x\frac{\cos x}{\sin x - \cos x}
1sin⁡x−cos⁡x\frac{1}{\sin x - \cos x}
1sin⁡x−1\frac{1}{\sin x - 1}
cos⁡xsin⁡x−1\frac{\cos x}{\sin x - 1}
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Calculus 2 Quiz

Calculus 2 Quiz: Algebraic And Trig Simplification

Practice Algebraic And Trig Simplification 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 Algebraic And Trig Simplification, 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

To integrate ∫sec⁡xtan⁡x−1dx\int \frac{\sec x}{\tan x - 1} dx, it is useful to rewrite the integrand entirely in terms of sin⁡x\sin x and cos⁡x\cos x. What is the resulting expression?

  1. cos⁡xsin⁡x−cos⁡x\frac{\cos x}{\sin x - \cos x}
  2. 1sin⁡x−cos⁡x\frac{1}{\sin x - \cos x} (correct answer)
  3. 1sin⁡x−1\frac{1}{\sin x - 1}
  4. cos⁡xsin⁡x−1\frac{\cos x}{\sin x - 1}
Explanation: When you encounter trigonometric integrands with multiple trig functions, converting everything to sine and cosine often simplifies the expression and reveals substitution opportunities. To rewrite sec⁡xtan⁡x−1\frac{\sec x}{\tan x - 1} in terms of sine and cosine, recall that sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x} and tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}. Substituting these definitions: sec⁡xtan⁡x−1=1cos⁡xsin⁡xcos⁡x−1\frac{\sec x}{\tan x - 1} = \frac{\frac{1}{\cos x}}{\frac{\sin x}{\cos x} - 1} To simplify the complex fraction, multiply both numerator and denominator by cos⁡x\cos x: 1cos⁡x⋅cos⁡x(sin⁡xcos⁡x−1)⋅cos⁡x=1sin⁡x−cos⁡x\frac{\frac{1}{\cos x} \cdot \cos x}{(\frac{\sin x}{\cos x} - 1) \cdot \cos x} = \frac{1}{\sin x - \cos x} This gives us answer choice B. Let's examine why the other options are incorrect. Choice A, cos⁡xsin⁡x−cos⁡x\frac{\cos x}{\sin x - \cos x}, has an extra cos⁡x\cos x in the numerator that doesn't belong—this would result from incorrectly leaving cos⁡x\cos x in the numerator after clearing the complex fraction. Choice C, 1sin⁡x−1\frac{1}{\sin x - 1}, appears if you mistakenly write tan⁡x−1=sin⁡xcos⁡x−1=sin⁡x−1cos⁡x\tan x - 1 = \frac{\sin x}{\cos x} - 1 = \frac{\sin x - 1}{\cos x}, forgetting that subtracting 1 requires a common denominator. Choice D, cos⁡xsin⁡x−1\frac{\cos x}{\sin x - 1}, combines both errors from choices A and C. Study tip: When simplifying complex fractions involving trig functions, always multiply numerator and denominator by the same expression to clear nested fractions. Converting to sine and cosine first makes algebraic manipulation much clearer.

Question 2

To simplify ∫1−tan⁡2x1+tan⁡2xdx\int \frac{1 - \tan^2 x}{1 + \tan^2 x} dx, what is the most direct simplification of the integrand using trigonometric identities?

  1. cos⁡2x−sin⁡4x\cos^2 x - \sin^4 x
  2. sec⁡(2x)\sec(2x)
  3. 1−sin⁡2x1 - \sin^2 x
  4. cos⁡(2x)\cos(2x) (correct answer)
Explanation: When you encounter integrals with expressions involving tan⁡2x\tan^2 x, look for opportunities to use the Pythagorean identity 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x. This identity often simplifies complex-looking trigonometric expressions. Let's work with the integrand 1−tan⁡2x1+tan⁡2x\frac{1 - \tan^2 x}{1 + \tan^2 x}. Since 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x, we can rewrite this as: 1−tan⁡2xsec⁡2x=(1−tan⁡2x)cos⁡2x\frac{1 - \tan^2 x}{\sec^2 x} = (1 - \tan^2 x) \cos^2 x Expanding this gives us: cos⁡2x−tan⁡2xcos⁡2x=cos⁡2x−sin⁡2xcos⁡2x⋅cos⁡2x=cos⁡2x−sin⁡2x\cos^2 x - \tan^2 x \cos^2 x = \cos^2 x - \frac{\sin^2 x}{\cos^2 x} \cdot \cos^2 x = \cos^2 x - \sin^2 x This expression cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x is the double angle formula for cosine: cos⁡(2x)\cos(2x). So our integrand simplifies directly to cos⁡(2x)\cos(2x), making choice D correct. Choice A (cos⁡2x−sin⁡4x\cos^2 x - \sin^4 x) incorrectly squares the sin⁡2x\sin^2 x term during simplification. Choice B (sec⁡(2x)\sec(2x)) confuses the double angle identity—there's no standard identity that would lead to this form. Choice C (1−sin⁡2x1 - \sin^2 x) stops at the Pythagorean identity cos⁡2x\cos^2 x without recognizing the complete double angle pattern. Study tip: Memorize the double angle formulas, especially cos⁡(2x)=cos⁡2x−sin⁡2x\cos(2x) = \cos^2 x - \sin^2 x. When you see expressions with 1+tan⁡2x1 + \tan^2 x in denominators, immediately think of substituting sec⁡2x\sec^2 x to unlock simplification opportunities.

Question 3

What is the correct form of the partial fraction decomposition for the integrand 5x−3(x−2)2(x+1)\frac{5x-3}{(x-2)^2(x+1)}?

  1. Ax−2+B(x−2)2+Cx+1\frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+1} (correct answer)
  2. A(x−2)2+Bx+1\frac{A}{(x-2)^2} + \frac{B}{x+1}
  3. Ax+B(x−2)2+Cx+1\frac{Ax+B}{(x-2)^2} + \frac{C}{x+1}
  4. Ax−2+Bx−2+Cx+1\frac{A}{x-2} + \frac{B}{x-2} + \frac{C}{x+1}
Explanation: The denominator has a repeated linear factor (x−2)2(x-2)^2 and a distinct linear factor (x+1)(x+1). For a repeated linear factor (ax+b)n(ax+b)^n, the decomposition must include a term for each power from 1 to n. Thus, for (x−2)2(x-2)^2, we need terms Ax−2\frac{A}{x-2} and B(x−2)2\frac{B}{(x-2)^2}. Choice B omits the term for the single power of the repeated factor. Choice C uses the incorrect form for a repeated linear factor. Choice D is redundant and incorrectly notated.

Question 4

The integral ∫e2x−1e2x+1dx\int \frac{e^{2x} - 1}{e^{2x} + 1} dx can be simplified by dividing the numerator and denominator by exe^x. Which standard integrable function is the result?

  1. tanh⁡(2x)\tanh(2x)
  2. coth⁡(x)\coth(x)
  3. tanh⁡(x)\tanh(x) (correct answer)
  4. 1−e−2x1 - e^{-2x}
Explanation: When you encounter a rational function involving exponentials, algebraic manipulation often reveals hidden hyperbolic functions. The key insight here is recognizing that dividing both numerator and denominator by the same expression can transform the integrand into a standard form. Let's divide both the numerator and denominator by exe^x: e2x−1e2x+1=e2x−1exe2x+1ex=ex−e−xex+e−x\frac{e^{2x} - 1}{e^{2x} + 1} = \frac{\frac{e^{2x} - 1}{e^x}}{\frac{e^{2x} + 1}{e^x}} = \frac{e^x - e^{-x}}{e^x + e^{-x}} This is precisely the definition of tanh⁡(x)=ex−e−xex+e−x\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}, which means our integral becomes ∫tanh⁡(x)dx\int \tanh(x) dx. This confirms that answer C is correct. Looking at the wrong answers: Choice A, tanh⁡(2x)\tanh(2x), would have the form e2x−e−2xe2x+e−2x\frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}, which doesn't match our simplified expression. Choice B, coth⁡(x)\coth(x), equals ex+e−xex−e−x\frac{e^x + e^{-x}}{e^x - e^{-x}}—notice the numerator and denominator are flipped compared to our result. Choice D, 1−e−2x1 - e^{-2x}, isn't even a hyperbolic function and would require polynomial long division to obtain. Study tip: Memorize the definitions of hyperbolic functions in terms of exponentials: sinh⁡(x)=ex−e−x2\sinh(x) = \frac{e^x - e^{-x}}{2}, cosh⁡(x)=ex+e−x2\cosh(x) = \frac{e^x + e^{-x}}{2}, and tanh⁡(x)=sinh⁡(x)cosh⁡(x)=ex−e−xex+e−x\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}}. Recognizing these patterns will help you spot hyperbolic functions disguised in exponential form.

Question 5

When evaluating ∫sec⁡3xtan⁡2x+1dx\int \frac{\sec^3 x}{\tan^2 x + 1} dx, which identity should be applied to achieve the most direct simplification?

  1. Use tan⁡2x+1=sec⁡2x\tan^2 x + 1 = \sec^2 x to get ∫sec⁡3xsec⁡2xdx=∫sec⁡xdx\int \frac{\sec^3 x}{\sec^2 x} dx = \int \sec x dx (correct answer)
  2. Use sec⁡3x=sec⁡x⋅sec⁡2x\sec^3 x = \sec x \cdot \sec^2 x and tan⁡2x+1=sec⁡2x\tan^2 x + 1 = \sec^2 x separately
  3. Convert everything to sines and cosines: sec⁡3xtan⁡2x+1=1cos⁡3x⋅1sin⁡2xcos⁡2x+1\frac{\sec^3 x}{\tan^2 x + 1} = \frac{1}{\cos^3 x} \cdot \frac{1}{\frac{\sin^2 x}{\cos^2 x} + 1}
  4. Use sec⁡3x=sec⁡x(tan⁡2x+1)\sec^3 x = \sec x(\tan^2 x + 1) to create a common factor with the denominator
Explanation: The Pythagorean identity tan⁡2x+1=sec⁡2x\tan^2 x + 1 = \sec^2 x immediately simplifies the integral to ∫sec⁡3xsec⁡2xdx=∫sec⁡xdx\int \frac{\sec^3 x}{\sec^2 x} dx = \int \sec x dx, which has the known antiderivative ln⁡∣sec⁡x+tan⁡x∣+C\ln|\sec x + \tan x| + C. Choice B is redundant since both transformations lead to the same result as A. Choice C unnecessarily complicates the expression. Choice D uses an incorrect identity (sec⁡3x≠sec⁡x(tan⁡2x+1)\sec^3 x \neq \sec x(\tan^2 x + 1)).

Question 6

To simplify ∫x2+6x+13x+3dx\int \frac{\sqrt{x^2 + 6x + 13}}{x + 3} dx, what algebraic manipulation should be performed first?

  1. Complete the square: x2+6x+13=(x+3)2+4x^2 + 6x + 13 = (x + 3)^2 + 4, then substitute u=x+3u = x + 3 (correct answer)
  2. Factor the numerator and cancel with denominator terms directly
  3. Use substitution u=x+3u = x + 3 first, then complete the square in the resulting expression
  4. Rationalize by multiplying by x2+6x+13x2+6x+13\frac{\sqrt{x^2 + 6x + 13}}{\sqrt{x^2 + 6x + 13}} and simplify the denominator
Explanation: Completing the square first: x2+6x+13=x2+6x+9+4=(x+3)2+4x^2 + 6x + 13 = x^2 + 6x + 9 + 4 = (x + 3)^2 + 4. The integral becomes ∫(x+3)2+4x+3dx\int \frac{\sqrt{(x+3)^2 + 4}}{x + 3} dx. Then substitute u=x+3u = x + 3, du=dxdu = dx to get ∫u2+4udu\int \frac{\sqrt{u^2 + 4}}{u} du, which can be solved using trigonometric substitution u=2tan⁡θu = 2\tan \theta. Choice B is impossible since the numerator doesn't factor nicely. Choice C works but is less direct. Choice D doesn't help simplify the integral.

Question 7

Which expression represents the correct partial fraction decomposition for the rational function f(x)=x2+1(x−1)(x2+4)f(x) = \frac{x^2 + 1}{(x-1)(x^2+4)}?

  1. Ax−1+Bx2+4\frac{A}{x-1} + \frac{B}{x^2+4}
  2. Ax−1+Bx+2+Cx−2\frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-2}
  3. Ax−1+Bx+Cx2+4\frac{A}{x-1} + \frac{Bx+C}{x^2+4} (correct answer)
  4. Ax+Bx−1+Cx2+4\frac{Ax+B}{x-1} + \frac{C}{x^2+4}
Explanation: The denominator contains a distinct linear factor (x−1)(x-1) and an irreducible quadratic factor (x2+4)(x^2+4). The correct form for the decomposition is a constant numerator for the linear factor and a linear numerator for the irreducible quadratic factor. Therefore, the form is Ax−1+Bx+Cx2+4\frac{A}{x-1} + \frac{Bx+C}{x^2+4}. Choice A incorrectly uses a constant for the quadratic factor. Choice B incorrectly factors x2+4x^2+4 over the real numbers. Choice D incorrectly uses a linear numerator for the linear factor.

Question 8

To evaluate ∫11−cos⁡xdx\int \frac{1}{1 - \cos x} dx, a useful simplification is to multiply the numerator and denominator by the conjugate of the denominator. What is the resulting integrand after this step and a subsequent simplification using a Pythagorean identity?

  1. 1+cos⁡xsin⁡2x\frac{1+\cos x}{\sin^2 x} (correct answer)
  2. 1−cos⁡xsin⁡2x\frac{1-\cos x}{\sin^2 x}
  3. 1+cos⁡xcos⁡2x\frac{1+\cos x}{\cos^2 x}
  4. 1+cos⁡x1−cos⁡2x\frac{1+\cos x}{1-\cos^2 x}
Explanation: Multiplying the numerator and denominator by the conjugate, 1+cos⁡x1+\cos x, gives 1+cos⁡x(1−cos⁡x)(1+cos⁡x)=1+cos⁡x1−cos⁡2x\frac{1+\cos x}{(1-\cos x)(1+\cos x)} = \frac{1+\cos x}{1-\cos^2 x}. Applying the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, we know 1−cos⁡2x=sin⁡2x1-\cos^2 x = \sin^2 x. The resulting integrand is 1+cos⁡xsin⁡2x\frac{1+\cos x}{\sin^2 x}. Choice D shows the correct multiplication but omits the identity application. Choices B and C result from incorrect conjugate multiplication or identity usage.

Question 9

The Weierstrass substitution t=tan⁡(x/2)t = \tan(x/2) is used to transform rational functions of trigonometric functions into rational functions of tt. Using this substitution, how is cos⁡(x)\cos(x) expressed in terms of tt?

  1. 1+t21−t2\frac{1+t^2}{1-t^2}
  2. 2t1+t2\frac{2t}{1+t^2}
  3. 2t1−t2\frac{2t}{1-t^2}
  4. 1−t21+t2\frac{1-t^2}{1+t^2} (correct answer)
Explanation: The Weierstrass substitution is a powerful technique for integrating rational functions of trigonometric functions. When you encounter t=tan⁡(x/2)t = \tan(x/2), you need to express sin⁡(x)\sin(x), cos⁡(x)\cos(x), and dxdx in terms of tt and dtdt. To find cos⁡(x)\cos(x) in terms of tt, start with the double angle formula: cos⁡(x)=cos⁡(2⋅x2)=cos⁡2(x2)−sin⁡2(x2)\cos(x) = \cos(2 \cdot \frac{x}{2}) = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}). Since t=tan⁡(x/2)t = \tan(x/2), we know that cos⁡2(x/2)=11+t2\cos^2(x/2) = \frac{1}{1+t^2} and sin⁡2(x/2)=t21+t2\sin^2(x/2) = \frac{t^2}{1+t^2} (from the fundamental trigonometric identity 1+tan⁡2(θ)=sec⁡2(θ)1 + \tan^2(\theta) = \sec^2(\theta)). Therefore: cos⁡(x)=11+t2−t21+t2=1−t21+t2\cos(x) = \frac{1}{1+t^2} - \frac{t^2}{1+t^2} = \frac{1-t^2}{1+t^2} This confirms answer D is correct. Looking at the wrong answers: A) 1+t21−t2\frac{1+t^2}{1-t^2} has the numerator and denominator swapped and would actually give sec⁡(x)\sec(x) under different conditions. B) 2t1+t2\frac{2t}{1+t^2} is actually the Weierstrass formula for sin⁡(x)\sin(x), not cos⁡(x)\cos(x). C) 2t1−t2\frac{2t}{1-t^2} represents tan⁡(x)\tan(x) in the Weierstrass substitution. Study tip: Memorize all three Weierstrass formulas as a set: sin⁡(x)=2t1+t2\sin(x) = \frac{2t}{1+t^2}, cos⁡(x)=1−t21+t2\cos(x) = \frac{1-t^2}{1+t^2}, and tan⁡(x)=2t1−t2\tan(x) = \frac{2t}{1-t^2}. Notice the pattern in numerators and denominators to avoid mixing them up.

Question 10

The integral ∫sec⁡2xtan⁡2x−9dx\int \frac{\sec^2 x}{\tan^2 x - 9} dx can be simplified by a substitution. Which substitution directly transforms the integral into a standard form involving logarithms or inverse hyperbolic functions?

  1. u=sec⁡xu = \sec x
  2. u=tan⁡xu = \tan x (correct answer)
  3. u=tan⁡2xu = \tan^2 x
  4. x=3sec⁡θx = 3\sec \theta
Explanation: When you encounter integrals involving trigonometric functions, look for substitutions that simplify the expression by leveraging trigonometric relationships and derivatives. Let's examine what happens with u=tan⁡xu = \tan x. When we substitute u=tan⁡xu = \tan x, we get du=sec⁡2x dxdu = \sec^2 x \, dx. This is perfect because sec⁡2x\sec^2 x appears in our numerator! The integral becomes: ∫sec⁡2xtan⁡2x−9dx=∫1u2−9du\int \frac{\sec^2 x}{\tan^2 x - 9} dx = \int \frac{1}{u^2 - 9} du This transforms into the standard form ∫1u2−a2du\int \frac{1}{u^2 - a^2} du, which integrates to 12aln⁡∣u−au+a∣+C\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right| + C or can be expressed using inverse hyperbolic functions. Option A (u=sec⁡xu = \sec x) fails because du=sec⁡xtan⁡x dxdu = \sec x \tan x \, dx, which doesn't match our numerator and leaves us with a complex expression involving both sec⁡x\sec x and tan⁡x\tan x. Option C (u=tan⁡2xu = \tan^2 x) gives du=2tan⁡xsec⁡2x dxdu = 2\tan x \sec^2 x \, dx, which introduces an unwanted tan⁡x\tan x factor that doesn't cancel with anything in our original integral. Option D (x=3sec⁡θx = 3\sec \theta) is a trigonometric substitution typically used for expressions like x2−9\sqrt{x^2 - 9}, not for integrals already in trigonometric form. Strategy tip: When you see sec⁡2x\sec^2 x in the numerator of a trigonometric integral, immediately consider u=tan⁡xu = \tan x since ddx[tan⁡x]=sec⁡2x\frac{d}{dx}[\tan x] = \sec^2 x. This substitution often transforms complex trigonometric integrals into standard algebraic forms.

Question 11

Before applying partial fraction decomposition, the integrand of ∫x3+2xx2−1dx\int \frac{x^3 + 2x}{x^2 - 1} dx must be rewritten because the degree of the numerator is not less than the degree of the denominator. Which of the following is the correct result of performing polynomial long division?

  1. x+3xx2−1x + \frac{3x}{x^2 - 1} (correct answer)
  2. x+xx2−1x + \frac{x}{x^2 - 1}
  3. 1+2x+1x2−11 + \frac{2x+1}{x^2 - 1}
  4. x−1+2x+1x2−1x - 1 + \frac{2x+1}{x^2-1}
Explanation: Performing polynomial long division of (x3+2x)(x^3 + 2x) by (x2−1)(x^2 - 1) gives a quotient of xx and a remainder of 3x3x. Specifically, x(x2−1)=x3−xx(x^2 - 1) = x^3 - x, and (x3+2x)−(x3−x)=3x(x^3 + 2x) - (x^3 - x) = 3x. Thus, the expression is rewritten as x+3xx2−1x + \frac{3x}{x^2 - 1}. The other options result from common errors in the long division process, such as incorrect subtraction of terms or miscalculation of the remainder.

Question 12

To prepare the integral ∫dx8+2x−x2\int \frac{dx}{\sqrt{8+2x-x^2}} for a trigonometric substitution, the expression inside the square root must be rewritten by completing the square. Which expression is the correct transformation?

  1. 9−(x−1)29 - (x-1)^2 (correct answer)
  2. (x−1)2+9(x-1)^2 + 9
  3. 8−(x−1)28-(x-1)^2
  4. 9−(x+1)29-(x+1)^2
Explanation: To complete the square for 8+2x−x28+2x-x^2, first factor out the negative from the x-terms: 8−(x2−2x)8 - (x^2 - 2x). To complete the square for x2−2xx^2 - 2x, we add and subtract 1 inside the parentheses: 8−(x2−2x+1−1)8 - (x^2 - 2x + 1 - 1). This becomes 8−((x−1)2−1)8 - ((x-1)^2 - 1), which simplifies to 8−(x−1)2+1=9−(x−1)28 - (x-1)^2 + 1 = 9 - (x-1)^2. The other choices result from sign errors or incorrect application of the completing the square algorithm.

Question 13

To evaluate ∫cos⁡2(3x)sin⁡5(3x)dx\int \cos^2(3x) \sin^5(3x) dx, which substitution method is most appropriate after applying trigonometric identities?

  1. Apply power reduction to both terms: cos⁡2(3x)=1+cos⁡(6x)2\cos^2(3x) = \frac{1+\cos(6x)}{2} and sin⁡5(3x)\sin^5(3x)
  2. Use power reduction on cos⁡2(3x)\cos^2(3x) first, then substitute u=sin⁡(3x)u = \sin(3x)
  3. Factor as cos⁡2(3x)sin⁡4(3x)⋅sin⁡(3x)\cos^2(3x) \sin^4(3x) \cdot \sin(3x) and substitute u=cos⁡(3x)u = \cos(3x)
  4. Use sin⁡5(3x)=sin⁡(3x)(1−cos⁡2(3x))2\sin^5(3x) = \sin(3x)(1-\cos^2(3x))^2 and substitute u=cos⁡(3x)u = \cos(3x) (correct answer)
Explanation: When you encounter integrals involving powers of sine and cosine, your strategy depends on which functions have odd versus even powers. The key insight is that odd powers allow you to "peel off" one factor to use as part of your differential when substituting. Since sin⁡5(3x)\sin^5(3x) has an odd power, you can rewrite it as sin⁡5(3x)=sin⁡(3x)⋅sin⁡4(3x)=sin⁡(3x)⋅(sin⁡2(3x))2\sin^5(3x) = \sin(3x) \cdot \sin^4(3x) = \sin(3x) \cdot (\sin^2(3x))^2. Using the Pythagorean identity sin⁡2(3x)=1−cos⁡2(3x)\sin^2(3x) = 1 - \cos^2(3x), this becomes sin⁡(3x)(1−cos⁡2(3x))2\sin(3x)(1-\cos^2(3x))^2. Now if you substitute u=cos⁡(3x)u = \cos(3x), then du=−3sin⁡(3x)dxdu = -3\sin(3x)dx, so sin⁡(3x)dx=−13du\sin(3x)dx = -\frac{1}{3}du. Your integral transforms into ∫cos⁡2(3x)sin⁡5(3x)dx=∫u2(1−u2)2⋅(−13)du\int \cos^2(3x) \sin^5(3x) dx = \int u^2(1-u^2)^2 \cdot (-\frac{1}{3})du, which expands to a polynomial that's easy to integrate. Option A would make the problem unnecessarily complex by applying power reduction to both terms. Option B starts with power reduction on the even power, which is backwards—you only need power reduction when all powers are even. Option C factors incorrectly and chooses the wrong substitution; since cos⁡2(3x)\cos^2(3x) has an even power, substituting u=cos⁡(3x)u = \cos(3x) won't work cleanly because you'd need sin⁡(3x)dx\sin(3x)dx for the differential, but you'd have leftover even powers of sine. Strategy tip: For ∫sin⁡m(x)cos⁡n(x)dx\int \sin^m(x)\cos^n(x)dx, if either exponent is odd, use substitution with the function having the even exponent. Save power reduction for when both exponents are even.

Question 14

For the rational function 2x3−x2+3x−1x4−x2\frac{2x^3 - x^2 + 3x - 1}{x^4 - x^2}, what is the correct partial fraction decomposition setup?

  1. Ax+Bx2+Cx2−1\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^2 - 1}
  2. Ax+Bx+Cx2+Dx−1+Ex+1\frac{A}{x} + \frac{Bx + C}{x^2} + \frac{D}{x-1} + \frac{E}{x+1}
  3. Ax+Bx2+Cx+Dx2−1\frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 - 1}
  4. Ax+Bx2+Cx−1+Dx+1\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{x+1} (correct answer)
Explanation: When you encounter partial fraction decomposition, you need to factor the denominator completely and match each factor with the appropriate partial fraction form. This technique breaks complex rational functions into simpler pieces that are easier to integrate or manipulate. First, factor the denominator: x4−x2=x2(x2−1)=x2(x−1)(x+1)x^4 - x^2 = x^2(x^2 - 1) = x^2(x-1)(x+1). Now you can see the complete factorization involves three distinct factors: x2x^2, (x−1)(x-1), and (x+1)(x+1). For partial fractions, each factor gets its own term. The repeated factor x2x^2 requires two terms: Ax\frac{A}{x} for the linear factor xx and Bx2\frac{B}{x^2} for the quadratic factor x2x^2. The distinct linear factors (x−1)(x-1) and (x+1)(x+1) each contribute one term with constant numerators: Cx−1\frac{C}{x-1} and Dx+1\frac{D}{x+1}. This gives us answer choice D. Choice A incorrectly treats (x2−1)(x^2-1) as irreducible, missing that it factors further. Choice B uses Bx+Cx2\frac{Bx+C}{x^2} which is wrong for a repeated linear factor - you only use polynomial numerators for irreducible quadratic factors. Choice C makes the same error as A by not factoring x2−1x^2-1 completely, then compounds it with the incorrect polynomial numerator. Remember: always factor the denominator completely first, then assign constant numerators to linear factors and polynomial numerators only to irreducible quadratic factors. For repeated factors, include a term for each power up to the highest.

Question 15

The integral ∫ln⁡(x3)dx\int \ln(x^3) dx can be simplified before applying integration by parts. Which of the following is an equivalent integral after this simplification?

  1. ∫(ln⁡x)3dx\int (\ln x)^3 dx
  2. 3∫ln⁡(x)dx3\int \ln(x) dx (correct answer)
  3. 13∫ln⁡(x)dx\frac{1}{3}\int \ln(x) dx
  4. The integral cannot be simplified and requires immediate application of integration by parts.
Explanation: When you encounter integrals involving logarithmic expressions, always check if you can apply logarithm properties to simplify before attempting integration techniques like integration by parts. The key insight here is recognizing that ln⁡(x3)\ln(x^3) can be simplified using the logarithm power rule: ln⁡(an)=nln⁡(a)\ln(a^n) = n \ln(a). Applying this property, ln⁡(x3)=3ln⁡(x)\ln(x^3) = 3\ln(x). Therefore, ∫ln⁡(x3)dx=∫3ln⁡(x)dx=3∫ln⁡(x)dx\int \ln(x^3) dx = \int 3\ln(x) dx = 3\int \ln(x) dx, which matches answer choice B. Let's examine why the other options are incorrect. Choice A suggests ∫(ln⁡x)3dx\int (\ln x)^3 dx, which represents the cube of the natural logarithm function, not the logarithm of x3x^3. This is a completely different expression that would require more complex integration techniques. Choice C gives 13∫ln⁡(x)dx\frac{1}{3}\int \ln(x) dx, which incorrectly applies the coefficient—this would result from misremembering the logarithm property or confusing it with a substitution rule. Choice D claims no simplification is possible, but this ignores the fundamental logarithm properties that should be applied before attempting integration by parts. Remember this strategy: whenever you see logarithms with exponents, powers, products, or quotients inside the argument, always apply logarithm properties first to simplify. This approach often transforms a complex integral into a much more manageable one, saving you time and reducing the chance of errors in lengthy integration by parts calculations.

Question 16

To evaluate ∫x3+2x2−x+1x2+x−2dx\int \frac{x^3 + 2x^2 - x + 1}{x^2 + x - 2} dx, what should be the result after polynomial long division?

  1. ∫(x+1+2x−1x2+x−2)dx\int \left(x + 1 + \frac{2x - 1}{x^2 + x - 2}\right) dx
  2. ∫(x+1+2x+3x2+x−2)dx\int \left(x + 1 + \frac{2x + 3}{x^2 + x - 2}\right) dx
  3. ∫(x+1+x+3x2+x−2)dx\int \left(x + 1 + \frac{x + 3}{x^2 + x - 2}\right) dx (correct answer)
  4. ∫(x+2+x−3x2+x−2)dx\int \left(x + 2 + \frac{x - 3}{x^2 + x - 2}\right) dx
Explanation: Performing polynomial long division: x3+2x2−x+1=(x2+x−2)(x+1)+(x+3)x^3 + 2x^2 - x + 1 = (x^2 + x - 2)(x + 1) + (x + 3). This can be verified by expanding: (x2+x−2)(x+1)=x3+x2+x2+x−2x−2=x3+2x2−x−2(x^2 + x - 2)(x + 1) = x^3 + x^2 + x^2 + x - 2x - 2 = x^3 + 2x^2 - x - 2. Adding the remainder: x3+2x2−x−2+x+3=x3+2x2−x+1x^3 + 2x^2 - x - 2 + x + 3 = x^3 + 2x^2 - x + 1 ✓. Choice A has wrong remainder. Choice B has wrong remainder. Choice D has wrong quotient and remainder.

Question 17

To evaluate the integral ∫sin⁡3xcos⁡2xdx\int \frac{\sin^3 x}{\cos^2 x} dx, which of the following is the most effective first step in rewriting the integrand to prepare for a u-substitution?

  1. Rewrite sin⁡3x\sin^3 x as sin⁡x(sin⁡2x)\sin x (\sin^2 x) and then as sin⁡x(1−cos⁡2x)\sin x (1-\cos^2 x). (correct answer)
  2. Rewrite the integrand as tan⁡2xsin⁡x\tan^2 x \sin x and use the identity tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1.
  3. Apply the half-angle identity to sin⁡2x\sin^2 x after rewriting sin⁡3x\sin^3 x as sin⁡xsin⁡2x\sin x \sin^2 x.
  4. Convert the entire expression to sines and cosines, which is sin⁡3x1−sin⁡2x\frac{\sin^3 x}{1-\sin^2 x}.
Explanation: The correct strategy is to prepare for a u-substitution with u=cos⁡xu = \cos x. To do this, we need a single sin⁡x\sin x factor left over for du=−sin⁡xdxdu = -\sin x dx. By rewriting sin⁡3x\sin^3 x as sin⁡x(1−cos⁡2x)\sin x (1-\cos^2 x), the integrand becomes sin⁡x(1−cos⁡2x)cos⁡2x\frac{\sin x (1-\cos^2 x)}{\cos^2 x}, which is perfectly set up for this substitution. Choice B is a valid path but less direct. Choice C introduces a double angle, unnecessarily complicating the integral. Choice D correctly uses an identity but does not simplify the integrand in a useful way for substitution.

Question 18

To simplify the integral ∫dxsin⁡(x)cos⁡(x)\int \frac{dx}{\sin(x)\cos(x)}, which manipulation based on a trigonometric identity is most effective?

  1. Multiply the numerator and denominator by 2 to get ∫22sin⁡(x)cos⁡(x)dx=∫2sin⁡(2x)dx\int \frac{2}{2\sin(x)\cos(x)} dx = \int \frac{2}{\sin(2x)} dx. (correct answer)
  2. Use the Pythagorean identity to replace cos⁡(x)\cos(x) with 1−sin⁡2(x)\sqrt{1-\sin^2(x)}, creating a more complex radical.
  3. Rewrite the integrand as ∫cot⁡(x)sec⁡2(x)dx\int \cot(x)\sec^2(x) dx and attempt integration by parts.
  4. Split the fraction into ∫(1sin⁡x−1cos⁡x)dx\int (\frac{1}{\sin x} - \frac{1}{\cos x}) dx, which is an invalid algebraic step.
Explanation: The most effective simplification is to recognize that the denominator sin⁡(x)cos⁡(x)\sin(x)\cos(x) is half of sin⁡(2x)\sin(2x). By multiplying the numerator and denominator by 2, the integrand becomes 2sin⁡(2x)=2csc⁡(2x)\frac{2}{\sin(2x)} = 2\csc(2x), which is a standard integral form. The other options either complicate the integral unnecessarily or are based on invalid mathematical operations.

Question 19

Which of the following is an equivalent integrand to x+1x2+1\frac{x+1}{x^2+1} that is rewritten as a sum of two terms, allowing for direct integration using basic rules?

  1. The fraction cannot be split and must be integrated using a single, more advanced technique.
  2. 1x+1x2+1\frac{1}{x} + \frac{1}{x^2+1}
  3. x+1x2+x+11\frac{x+1}{x^2} + \frac{x+1}{1}
  4. xx2+1+1x2+1\frac{x}{x^2+1} + \frac{1}{x^2+1} (correct answer)
Explanation: When you encounter a rational function where the numerator has multiple terms, consider whether you can split it into simpler fractions that integrate easily using basic antiderivative formulas. The key insight here is recognizing that x+1x2+1\frac{x+1}{x^2+1} can be rewritten as the sum xx2+1+1x2+1\frac{x}{x^2+1} + \frac{1}{x^2+1}. This works because when you add fractions with the same denominator, you simply add the numerators: xx2+1+1x2+1=x+1x2+1\frac{x}{x^2+1} + \frac{1}{x^2+1} = \frac{x+1}{x^2+1}. This splitting is powerful because each term now has a standard antiderivative: ∫xx2+1dx=12ln⁡(x2+1)+C\int \frac{x}{x^2+1} dx = \frac{1}{2}\ln(x^2+1) + C (using substitution with u=x2+1u = x^2+1), and ∫1x2+1dx=arctan⁡(x)+C\int \frac{1}{x^2+1} dx = \arctan(x) + C (a fundamental inverse trigonometric integral). Choice A is wrong because the fraction can indeed be split—there's no need for advanced techniques. Choice B incorrectly changes the denominators; 1x+1x2+1\frac{1}{x} + \frac{1}{x^2+1} would equal x2+1+xx(x2+1)\frac{x^2+1+x}{x(x^2+1)}, which doesn't match our original expression. Choice C completely destroys the structure by breaking up the denominator incorrectly, creating x+1x2+(x+1)\frac{x+1}{x^2} + (x+1), which is algebraically unrelated to the original. Study tip: When you see a rational function with addition in the numerator, try splitting it into separate fractions with the same denominator. This often transforms a complex integration into two simpler, recognizable forms.

Question 20

To integrate ∫(sin⁡x+cos⁡x)2dx\int (\sin x + \cos x)^2 dx, the most effective first step is to expand the binomial. What is the simplified integrand after expansion and application of trigonometric identities?

  1. 1+2sin⁡xcos⁡x1 + 2\sin x \cos x
  2. sin⁡2x+cos⁡2x\sin^2 x + \cos^2 x
  3. 11
  4. 1+sin⁡(2x)1 + \sin(2x) (correct answer)
Explanation: When you encounter integrands with squared trigonometric expressions like (sin⁡x+cos⁡x)2(\sin x + \cos x)^2, expanding the binomial and applying fundamental identities is often the key to simplification. Let's expand this binomial systematically: (sin⁡x+cos⁡x)2=sin⁡2x+2sin⁡xcos⁡x+cos⁡2x(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x Now apply two crucial trigonometric identities. First, the Pythagorean identity: sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. Second, the double angle identity: 2sin⁡xcos⁡x=sin⁡(2x)2\sin x \cos x = \sin(2x). Substituting these identities gives us: sin⁡2x+2sin⁡xcos⁡x+cos⁡2x=(sin⁡2x+cos⁡2x)+2sin⁡xcos⁡x=1+sin⁡(2x)\sin^2 x + 2\sin x \cos x + \cos^2 x = (\sin^2 x + \cos^2 x) + 2\sin x \cos x = 1 + \sin(2x) This matches answer choice D. Now let's examine why the other options fall short. Choice A gives 1+2sin⁡xcos⁡x1 + 2\sin x \cos x but fails to apply the double angle identity to simplify 2sin⁡xcos⁡x2\sin x \cos x into sin⁡(2x)\sin(2x). Choice B shows only the first two terms from the expansion, completely omitting the crucial cross-term 2sin⁡xcos⁡x2\sin x \cos x. Choice C incorrectly suggests that 2sin⁡xcos⁡x=02\sin x \cos x = 0, which would only be true at specific values, not generally. Study tip: When you see squared trigonometric expressions, immediately think "Pythagorean identity" for sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 and "double angle identity" for 2sin⁡xcos⁡x=sin⁡(2x)2\sin x \cos x = \sin(2x). These two identities together transform many complex-looking integrands into much simpler forms.