Calculus 2 Quiz: Trig Substitution
20 questions · exam conditions
0:00
Trig SubstitutionQuestion 1 of 20

If the substitution x−2=5tan⁡θx-2 = \sqrt{5}\tan\theta is used to evaluate an integral, which of the following expressions is equivalent to x2−4x+9\sqrt{x^2-4x+9}?

5tan⁡θ\sqrt{5}\tan\theta
5sec⁡θ5\sec\theta
5sec⁡θ\sqrt{5}\sec\theta
5tan⁡θ5\tan\theta
← Back to quizzes

Calculus 2 Quiz

Calculus 2 Quiz: Trig Substitution

Practice Trig Substitution 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 Trig Substitution, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

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.

All questions

Question 1

If the substitution x−2=5tan⁡θx-2 = \sqrt{5}\tan\theta is used to evaluate an integral, which of the following expressions is equivalent to x2−4x+9\sqrt{x^2-4x+9}?

  1. 5tan⁡θ\sqrt{5}\tan\theta
  2. 5sec⁡θ5\sec\theta
  3. 5sec⁡θ\sqrt{5}\sec\theta (correct answer)
  4. 5tan⁡θ5\tan\theta
Explanation: When you encounter trigonometric substitution problems, the key is recognizing which substitution works for different radical forms. The expression x2−4x+9\sqrt{x^2-4x+9} needs to be rewritten using the given substitution x−2=5tan⁡θx-2 = \sqrt{5}\tan\theta. First, solve for xx: x=2+5tan⁡θx = 2 + \sqrt{5}\tan\theta. Now substitute this into the original expression: x2−4x+9=(2+5tan⁡θ)2−4(2+5tan⁡θ)+9\sqrt{x^2-4x+9} = \sqrt{(2 + \sqrt{5}\tan\theta)^2 - 4(2 + \sqrt{5}\tan\theta) + 9} Expanding the squared term: (2+5tan⁡θ)2=4+45tan⁡θ+5tan⁡2θ(2 + \sqrt{5}\tan\theta)^2 = 4 + 4\sqrt{5}\tan\theta + 5\tan^2\theta Substituting everything: 4+45tan⁡θ+5tan⁡2θ−8−45tan⁡θ+9\sqrt{4 + 4\sqrt{5}\tan\theta + 5\tan^2\theta - 8 - 4\sqrt{5}\tan\theta + 9} The middle terms cancel: 5+5tan⁡2θ=5(1+tan⁡2θ)\sqrt{5 + 5\tan^2\theta} = \sqrt{5(1 + \tan^2\theta)} Using the fundamental trigonometric identity 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta: 5sec⁡2θ=5∣sec⁡θ∣=5sec⁡θ\sqrt{5\sec^2\theta} = \sqrt{5}|\sec\theta| = \sqrt{5}\sec\theta (We drop the absolute value since sec⁡θ>0\sec\theta > 0 in the typical domain for this substitution.) Choice A (5tan⁡θ\sqrt{5}\tan\theta) ignores the trigonometric identity entirely. Choice B (5sec⁡θ5\sec\theta) incorrectly brings the 5 outside the square root without taking its square root. Choice D (5tan⁡θ5\tan\theta) makes both errors—wrong trigonometric function and incorrect handling of the constant. Study tip: For trigonometric substitutions, always complete the square first, then apply the fundamental trigonometric identities. Remember that a2=∣a∣\sqrt{a^2} = |a|, and consider the appropriate sign based on your domain.

Question 2

Consider the integral ∫x3x2+4dx\int \frac{x^3}{\sqrt{x^2+4}} dx. After making the substitution x=2tan⁡θx=2\tan\theta, the integral is converted to an integral in θ\theta. Which of the following is the correct antiderivative in terms of θ\theta?

  1. 8sec⁡3θ−8sec⁡θ+C8\sec^3\theta - 8\sec\theta + C
  2. 83sec⁡3θ−8sec⁡θ+C\frac{8}{3}\sec^3\theta - 8\sec\theta + C (correct answer)
  3. 83tan⁡3θ−8tan⁡θ+C\frac{8}{3}\tan^3\theta - 8\tan\theta + C
  4. 8ln⁡∣sec⁡θ+tan⁡θ∣−8sin⁡θ+C8\ln|\sec\theta+\tan\theta| - 8\sin\theta + C
Explanation: With x=2tan⁡θx=2\tan\theta, we have dx=2sec⁡2θ dθdx=2\sec^2\theta \, d\theta and x2+4=4tan⁡2θ+4=2sec⁡θ\sqrt{x^2+4}=\sqrt{4\tan^2\theta+4}=2\sec\theta. The integral becomes ∫(2tan⁡θ)32sec⁡θ(2sec⁡2θ)dθ=∫8tan⁡3θ2sec⁡θ(2sec⁡2θ)dθ=8∫tan⁡3θsec⁡θ dθ\int \frac{(2\tan\theta)^3}{2\sec\theta}(2\sec^2\theta) d\theta = \int \frac{8\tan^3\theta}{2\sec\theta}(2\sec^2\theta) d\theta = 8\int \tan^3\theta \sec\theta \, d\theta. To solve this, rewrite as 8∫tan⁡2θ(sec⁡θtan⁡θ) dθ=8∫(sec⁡2θ−1)(sec⁡θtan⁡θ) dθ8\int \tan^2\theta (\sec\theta\tan\theta) \, d\theta = 8\int (\sec^2\theta-1)(\sec\theta\tan\theta) \, d\theta. Let u=sec⁡θu=\sec\theta, so du=sec⁡θtan⁡θ dθdu=\sec\theta\tan\theta \, d\theta. The integral becomes 8∫(u2−1)du=8(u33−u)+C=83sec⁡3θ−8sec⁡θ+C8\int (u^2-1) du = 8(\frac{u^3}{3}-u) + C = \frac{8}{3}\sec^3\theta - 8\sec\theta + C.

Question 3

The evaluation of ∫9x2−1xdx\int \frac{\sqrt{9x^2-1}}{x} dx using a trigonometric substitution leads to the expression tan⁡θ−θ+C\tan\theta - \theta + C in terms of θ\theta. Which of the following is the correct expression for the integral in terms of xx?

  1. 9x2−1−arcsec⁡(3x)+C\sqrt{9x^2-1} - \operatorname{arcsec}(3x) + C (correct answer)
  2. 9x2−1−arctan⁡(3x)+C\sqrt{9x^2-1} - \arctan(3x) + C
  3. 9x2−13−arcsec⁡(3x)+C\frac{\sqrt{9x^2-1}}{3} - \operatorname{arcsec}(3x) + C
  4. 9x2−13x−arcsin⁡(13x)+C\frac{\sqrt{9x^2-1}}{3x} - \arcsin(\frac{1}{3x}) + C
Explanation: The form (3x)2−12\sqrt{(3x)^2-1^2} suggests the substitution 3x=sec⁡θ3x = \sec\theta. From this, we can find expressions for tan⁡θ\tan\theta and θ\theta. We have θ=arcsec⁡(3x)\theta = \operatorname{arcsec}(3x). To find tan⁡θ\tan\theta, we use the identity 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta = \sec^2\theta, so tan⁡θ=sec⁡2θ−1=(3x)2−1=9x2−1\tan\theta = \sqrt{\sec^2\theta-1} = \sqrt{(3x)^2-1} = \sqrt{9x^2-1}. Substituting these back into the given expression tan⁡θ−θ+C\tan\theta - \theta + C yields 9x2−1−arcsec⁡(3x)+C\sqrt{9x^2-1} - \operatorname{arcsec}(3x) + C.

Question 4

Evaluate the definite integral ∫1edxx(1+(ln⁡x)2)\int_{1}^{e} \frac{dx}{x(1+(\ln x)^2)}. A student correctly identifies that this can be solved with a u-substitution, but it can also be solved with a trigonometric substitution after the initial u-substitution. What is the value of the integral?

  1. ln⁡(2)\ln(2)
  2. π2\frac{\pi}{2}
  3. π4\frac{\pi}{4} (correct answer)
  4. π3\frac{\pi}{3}
Explanation: When you encounter an integral with logarithms in the denominator, u-substitution is often your first move. Here, the presence of ln⁡x\ln x suggests letting u=ln⁡xu = \ln x, which gives du=1xdxdu = \frac{1}{x}dx. Let's work through this systematically. With u=ln⁡xu = \ln x, when x=1x = 1, u=ln⁡(1)=0u = \ln(1) = 0, and when x=ex = e, u=ln⁡(e)=1u = \ln(e) = 1. The integral becomes: ∫01du1+u2\int_{0}^{1} \frac{du}{1+u^2} This is now a standard arctangent integral! You should recognize that ∫du1+u2=arctan⁡(u)+C\int \frac{du}{1+u^2} = \arctan(u) + C. Therefore: ∫01du1+u2=arctan⁡(1)−arctan⁡(0)=π4−0=π4\int_{0}^{1} \frac{du}{1+u^2} = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} Looking at the wrong answers: Choice (A) ln⁡(2)\ln(2) might tempt you if you incorrectly tried to integrate using logarithmic functions instead of recognizing the arctangent form. Choice (B) π2\frac{\pi}{2} could result from confusing this with arctan⁡(∞)\arctan(\infty) or mixing up common arctangent values. Choice (D) π3\frac{\pi}{3} corresponds to arctan⁡(3)\arctan(\sqrt{3}), which has no connection to this problem. The correct answer is (C) π4\frac{\pi}{4}. Key strategy: When you see 11+u2\frac{1}{1+u^2} after substitution, immediately think arctangent. Memorize that arctan⁡(1)=π4\arctan(1) = \frac{\pi}{4} and arctan⁡(0)=0\arctan(0) = 0 — these values appear frequently in calculus problems.

Question 5

When evaluating ∫dxx2+a2\int \frac{dx}{\sqrt{x^2+a^2}} using a trigonometric substitution, the final result is ln⁡∣x2+a2+x∣+C\ln|\sqrt{x^2+a^2}+x| + C. Which intermediate integral in terms of θ\theta produces this result?

  1. ∫sec⁡θtan⁡θ dθ\int \sec\theta\tan\theta \, d\theta
  2. ∫csc⁡θ dθ\int \csc\theta \, d\theta
  3. ∫sec⁡2θ dθ\int \sec^2\theta \, d\theta
  4. ∫sec⁡θ dθ\int \sec\theta \, d\theta (correct answer)
Explanation: When you encounter an integral with the form ∫dxx2+a2\int \frac{dx}{\sqrt{x^2+a^2}}, this signals a trigonometric substitution problem. The presence of x2+a2x^2+a^2 under the square root tells you to use the substitution x=atan⁡θx = a\tan\theta, which transforms the radical into a simpler trigonometric expression. With x=atan⁡θx = a\tan\theta, we have dx=asec⁡2θ dθdx = a\sec^2\theta \, d\theta and x2+a2=a2tan⁡2θ+a2=atan⁡2θ+1=asec⁡θ\sqrt{x^2+a^2} = \sqrt{a^2\tan^2\theta + a^2} = a\sqrt{\tan^2\theta + 1} = a\sec\theta. Substituting these into the original integral: ∫dxx2+a2=∫asec⁡2θ dθasec⁡θ=∫sec⁡θ dθ\int \frac{dx}{\sqrt{x^2+a^2}} = \int \frac{a\sec^2\theta \, d\theta}{a\sec\theta} = \int \sec\theta \, d\theta This is answer choice D. The integral ∫sec⁡θ dθ=ln⁡∣sec⁡θ+tan⁡θ∣+C\int \sec\theta \, d\theta = \ln|\sec\theta + \tan\theta| + C, which when converted back to xx gives ln⁡∣x2+a2+x∣+C\ln|\sqrt{x^2+a^2}+x| + C as stated. Choice A, ∫sec⁡θtan⁡θ dθ=sec⁡θ+C\int \sec\theta\tan\theta \, d\theta = \sec\theta + C, would come from a different substitution setup. Choice B, ∫csc⁡θ dθ\int \csc\theta \, d\theta, appears in integrals involving a2−x2\sqrt{a^2-x^2} with x=asin⁡θx = a\sin\theta. Choice C, ∫sec⁡2θ dθ=tan⁡θ+C\int \sec^2\theta \, d\theta = \tan\theta + C, is the derivative of tangent but doesn't match our substitution result. Study tip: Memorize the three standard trig substitution patterns: x2+a2x^2+a^2 uses x=atan⁡θx = a\tan\theta, a2−x2a^2-x^2 uses x=asin⁡θx = a\sin\theta, and x2−a2x^2-a^2 uses x=asec⁡θx = a\sec\theta.

Question 6

The integral ∫1x2−a2dx\int \frac{1}{x^2-a^2} dx is typically solved using partial fractions. However, it can also be evaluated using the trigonometric substitution x=asec⁡θx=a\sec\theta. If this method is used, what is the resulting expression in terms of xx?

  1. 1aln⁡∣sec⁡θ∣+C\frac{1}{a} \ln|\sec\theta| + C
  2. 12aln⁡∣x−ax+a∣+C\frac{1}{2a} \ln|\frac{x-a}{x+a}| + C (correct answer)
  3. 1aarctan⁡(xa)+C\frac{1}{a}\arctan(\frac{x}{a}) + C
  4. 12aln⁡∣a+xa−x∣+C\frac{1}{2a} \ln|\frac{a+x}{a-x}| + C
Explanation: Let x=asec⁡θx=a\sec\theta, so dx=asec⁡θtan⁡θ dθdx = a\sec\theta\tan\theta \, d\theta. The denominator is x2−a2=a2sec⁡2θ−a2=a2tan⁡2θx^2-a^2 = a^2\sec^2\theta-a^2 = a^2\tan^2\theta. The integral becomes ∫asec⁡θtan⁡θa2tan⁡2θdθ=1a∫sec⁡θtan⁡θdθ=1a∫csc⁡θ dθ\int \frac{a\sec\theta\tan\theta}{a^2\tan^2\theta} d\theta = \frac{1}{a}\int \frac{\sec\theta}{\tan\theta} d\theta = \frac{1}{a}\int \csc\theta \, d\theta. The antiderivative is 1aln⁡∣csc⁡θ−cot⁡θ∣+C\frac{1}{a} \ln|\csc\theta - \cot\theta| + C. Now, we must convert back to xx. From x=asec⁡θx=a\sec\theta, we have sec⁡θ=x/a\sec\theta = x/a. The triangle has hypotenuse xx and adjacent side aa, so the opposite side is x2−a2\sqrt{x^2-a^2}. Then csc⁡θ=xx2−a2\csc\theta = \frac{x}{\sqrt{x^2-a^2}} and cot⁡θ=ax2−a2\cot\theta = \frac{a}{\sqrt{x^2-a^2}}. The expression is 1aln⁡∣x−ax2−a2∣+C=1aln⁡∣x−a(x−a)(x+a)∣+C=1aln⁡∣x−ax+a∣+C=12aln⁡∣x−ax+a∣+C\frac{1}{a}\ln|\frac{x-a}{\sqrt{x^2-a^2}}| + C = \frac{1}{a}\ln|\frac{x-a}{\sqrt{(x-a)(x+a)}}| + C = \frac{1}{a}\ln|\sqrt{\frac{x-a}{x+a}}| + C = \frac{1}{2a}\ln|\frac{x-a}{x+a}| + C. This matches the result from partial fractions.

Question 7

The integral ∫dxx24−x2\int \frac{dx}{x^2 \sqrt{4-x^2}} is evaluated using the substitution x=2sin⁡θx=2\sin\theta. The resulting antiderivative in θ\theta is −14cot⁡θ+C-\frac{1}{4}\cot\theta + C. What is the antiderivative in terms of xx?

  1. −4−x24x+C-\frac{\sqrt{4-x^2}}{4x} + C (correct answer)
  2. −x44−x2+C-\frac{x}{4\sqrt{4-x^2}} + C
  3. −4x4−x2+C-\frac{4x}{\sqrt{4-x^2}} + C
  4. −4−x2x+C-\frac{\sqrt{4-x^2}}{x} + C
Explanation: From the substitution x=2sin⁡θx=2\sin\theta, we have sin⁡θ=x/2\sin\theta = x/2. We can form a right triangle where the opposite side is xx and the hypotenuse is 2. The adjacent side is 22−x2=4−x2\sqrt{2^2-x^2} = \sqrt{4-x^2}. The cotangent function is the ratio of the adjacent side to the opposite side, so cot⁡θ=4−x2x\cot\theta = \frac{\sqrt{4-x^2}}{x}. Substituting this into the given result −14cot⁡θ+C-\frac{1}{4}\cot\theta + C yields −144−x2x+C=−4−x24x+C-\frac{1}{4} \frac{\sqrt{4-x^2}}{x} + C = -\frac{\sqrt{4-x^2}}{4x} + C.

Question 8

To evaluate the integral ∫dx(x2+4x+13)2\int \frac{dx}{(x^2+4x+13)^2}, one must first complete the square in the denominator. After doing so and making the appropriate trigonometric substitution, which of the following integrals in θ\theta is obtained?

  1. 127∫cos⁡2θ dθ\frac{1}{27} \int \cos^2\theta \, d\theta (correct answer)
  2. 19∫sec⁡θ dθ\frac{1}{9} \int \sec\theta \, d\theta
  3. 181∫cos⁡2θ dθ\frac{1}{81} \int \cos^2\theta \, d\theta
  4. 127∫cos⁡3θ dθ\frac{1}{27} \int \cos^3\theta \, d\theta
Explanation: First, complete the square for the quadratic in the denominator: x2+4x+13=(x2+4x+4)+9=(x+2)2+9x^2+4x+13 = (x^2+4x+4)+9 = (x+2)^2+9. The integral becomes ∫dx((x+2)2+9)2\int \frac{dx}{((x+2)^2+9)^2}. Let u=x+2u=x+2, so du=dxdu=dx. The integral is ∫du(u2+9)2\int \frac{du}{(u^2+9)^2}. This suggests the substitution u=3tan⁡θu=3\tan\theta, so du=3sec⁡2θ dθdu=3\sec^2\theta \, d\theta. The denominator becomes ((3tan⁡θ)2+9)2=(9tan⁡2θ+9)2=(9sec⁡2θ)2=81sec⁡4θ( (3\tan\theta)^2+9)^2 = (9\tan^2\theta+9)^2 = (9\sec^2\theta)^2 = 81\sec^4\theta. Substituting into the integral gives ∫3sec⁡2θ81sec⁡4θdθ=381∫1sec⁡2θdθ=127∫cos⁡2θ dθ\int \frac{3\sec^2\theta}{81\sec^4\theta} d\theta = \frac{3}{81} \int \frac{1}{\sec^2\theta} d\theta = \frac{1}{27} \int \cos^2\theta \, d\theta.

Question 9

Consider the integral ∫dxx2x2+3\int \frac{dx}{x^2\sqrt{x^2+3}}. After making the substitution x=3tan⁡θx=\sqrt{3}\tan\theta, the integral is transformed into an integral with respect to θ\theta. After integrating, the result is −csc⁡θ3+C-\frac{\csc\theta}{3} + C. Which of the following is the final answer in terms of xx?

  1. −x2+33x+C-\frac{\sqrt{x^2+3}}{3x} + C (correct answer)
  2. −x3x2+3+C-\frac{x}{3\sqrt{x^2+3}} + C
  3. −3xx2+3+C-\frac{3x}{\sqrt{x^2+3}} + C
  4. −x2+3x+C-\frac{\sqrt{x^2+3}}{x} + C
Explanation: Given the substitution x=3tan⁡θx=\sqrt{3}\tan\theta, we have tan⁡θ=x3\tan\theta = \frac{x}{\sqrt{3}}. We can construct a right triangle where the opposite side is xx and the adjacent side is 3\sqrt{3}. The hypotenuse is then x2+(3)2=x2+3\sqrt{x^2+(\sqrt{3})^2} = \sqrt{x^2+3}. The cosecant function is the ratio of the hypotenuse to the opposite side, so csc⁡θ=x2+3x\csc\theta = \frac{\sqrt{x^2+3}}{x}. Substituting this into the given result −csc⁡θ3+C-\frac{\csc\theta}{3} + C gives −x2+33x+C-\frac{\sqrt{x^2+3}}{3x} + C.

Question 10

Consider the integral ∫dx(x2+6x+13)3/2\int \frac{dx}{(x^2 + 6x + 13)^{3/2}}. After completing the square and applying the appropriate trigonometric substitution, the integral can be expressed in which of the following forms?

  1. x+34x2+6x+13+C\frac{x+3}{4\sqrt{x^2+6x+13}} + C (correct answer)
  2. x+32x2+6x+13+C\frac{x+3}{2\sqrt{x^2+6x+13}} + C
  3. x−34x2+6x+13+C\frac{x-3}{4\sqrt{x^2+6x+13}} + C
  4. 2(x+3)x2+6x+13+C\frac{2(x+3)}{\sqrt{x^2+6x+13}} + C
Explanation: First, complete the square: x2+6x+13=(x+3)2+4x^2 + 6x + 13 = (x+3)^2 + 4. Let u=x+3u = x+3, so the integral becomes ∫du(u2+4)3/2\int \frac{du}{(u^2 + 4)^{3/2}}. Using the substitution u=2tan⁡θu = 2\tan\theta, we have du=2sec⁡2θ dθdu = 2\sec^2\theta \, d\theta and u2+4=4tan⁡2θ+4=4sec⁡2θu^2 + 4 = 4\tan^2\theta + 4 = 4\sec^2\theta. The integral becomes ∫2sec⁡2θ dθ(4sec⁡2θ)3/2=∫2sec⁡2θ dθ8sec⁡3θ=14∫cos⁡θ dθ=14sin⁡θ+C\int \frac{2\sec^2\theta \, d\theta}{(4\sec^2\theta)^{3/2}} = \int \frac{2\sec^2\theta \, d\theta}{8\sec^3\theta} = \frac{1}{4}\int \cos\theta \, d\theta = \frac{1}{4}\sin\theta + C. Since tan⁡θ=u2=x+32\tan\theta = \frac{u}{2} = \frac{x+3}{2}, we have sin⁡θ=tan⁡θsec⁡θ=u/21+(u/2)2=u/2(u2+4)/4=uu2+4=x+3x2+6x+13\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{u/2}{\sqrt{1+(u/2)^2}} = \frac{u/2}{\sqrt{(u^2+4)/4}} = \frac{u}{\sqrt{u^2+4}} = \frac{x+3}{\sqrt{x^2+6x+13}}. Therefore, the answer is 14⋅x+3x2+6x+13+C=x+34x2+6x+13+C\frac{1}{4} \cdot \frac{x+3}{\sqrt{x^2+6x+13}} + C = \frac{x+3}{4\sqrt{x^2+6x+13}} + C.

Question 11

The integral ∫dxxx2−16\int \frac{dx}{x\sqrt{x^2-16}} where x>4x > 4 requires a trigonometric substitution. After applying the appropriate substitution and simplifying, which expression represents the antiderivative?

  1. 14sec⁡−1(x4)+C\frac{1}{4}\sec^{-1}\left(\frac{x}{4}\right) + C (correct answer)
  2. −14sec⁡−1(x4)+C-\frac{1}{4}\sec^{-1}\left(\frac{x}{4}\right) + C
  3. 116sec⁡−1(x4)+C\frac{1}{16}\sec^{-1}\left(\frac{x}{4}\right) + C
  4. 14cos⁡−1(4x)+C\frac{1}{4}\cos^{-1}\left(\frac{4}{x}\right) + C
Explanation: For x2−16\sqrt{x^2-16} with x>4x > 4, use x=4sec⁡θx = 4\sec\theta where 0<θ<π20 < \theta < \frac{\pi}{2}. Then dx=4sec⁡θtan⁡θ dθdx = 4\sec\theta\tan\theta \, d\theta and x2−16=4tan⁡θ\sqrt{x^2-16} = 4\tan\theta. The integral becomes ∫4sec⁡θtan⁡θ dθ4sec⁡θ⋅4tan⁡θ=∫dθ4=θ4+C\int \frac{4\sec\theta\tan\theta \, d\theta}{4\sec\theta \cdot 4\tan\theta} = \int \frac{d\theta}{4} = \frac{\theta}{4} + C. Since sec⁡θ=x4\sec\theta = \frac{x}{4}, we have θ=sec⁡−1(x4)\theta = \sec^{-1}\left(\frac{x}{4}\right). Therefore, the answer is 14sec⁡−1(x4)+C\frac{1}{4}\sec^{-1}\left(\frac{x}{4}\right) + C. Note that sec⁡−1(x4)=cos⁡−1(4x)\sec^{-1}\left(\frac{x}{4}\right) = \cos^{-1}\left(\frac{4}{x}\right), but the coefficient differs between choices A and D.

Question 12

A student attempts to evaluate ∫x16+x2 dx\int \frac{x}{\sqrt{16+x^2}} \, dx using the trigonometric substitution x=4tan⁡θx = 4\tan\theta. However, this integral can be solved more efficiently using a different method. What is the correct antiderivative using the most efficient approach?

  1. 16+x2+C\sqrt{16+x^2} + C (correct answer)
  2. 4ln⁡∣sec⁡θ+tan⁡θ∣+C4\ln|\sec\theta + \tan\theta| + C where tan⁡θ=x4\tan\theta = \frac{x}{4}
  3. x16+x22+8ln⁡∣x+16+x2∣+C\frac{x\sqrt{16+x^2}}{2} + 8\ln|x + \sqrt{16+x^2}| + C
  4. 12ln⁡(16+x2)+C\frac{1}{2}\ln(16+x^2) + C
Explanation: While trigonometric substitution would work, this integral is much more efficiently solved using ordinary substitution. Let u=16+x2u = 16 + x^2, then du=2x dxdu = 2x \, dx, so x dx=12dux \, dx = \frac{1}{2} du. The integral becomes ∫1u⋅12 du=12∫u−1/2 du=12⋅2u1/2+C=u+C=16+x2+C\int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du = \frac{1}{2} \int u^{-1/2} \, du = \frac{1}{2} \cdot 2u^{1/2} + C = \sqrt{u} + C = \sqrt{16+x^2} + C. Choice B represents the result if the student incorrectly tried to use trigonometric substitution on this problem (which would be unnecessarily complicated). Choice C looks like the result for ∫16+x2 dx\int \sqrt{16+x^2} \, dx, and Choice D would be incorrect due to the square root in the denominator.

Question 13

What is the area of the region enclosed by the hyperbola 4x2−y2=44x^2 - y^2 = 4 and the line x=2x=\sqrt{2}?

  1. 2−ln⁡(1+2)\sqrt{2} - \ln(1+\sqrt{2})
  2. 22−2ln⁡(1+2)2\sqrt{2} - 2\ln(1+\sqrt{2}) (correct answer)
  3. 22+2ln⁡(1+2)2\sqrt{2} + 2\ln(1+\sqrt{2})
  4. 2+ln⁡(1+2)\sqrt{2} + \ln(1+\sqrt{2})
Explanation: The hyperbola is y2=4x2−4y^2 = 4x^2-4, so y=±2x2−1y=\pm2\sqrt{x^2-1}. The hyperbola intersects the x-axis at x=±1x=\pm1. The area is given by the integral of the upper branch minus the lower branch, from x=1x=1 to x=2x=\sqrt{2}. Area A=∫122x2−1−(−2x2−1)dx=4∫12x2−1dxA = \int_{1}^{\sqrt{2}} 2\sqrt{x^2-1} - (-2\sqrt{x^2-1}) dx = 4\int_{1}^{\sqrt{2}} \sqrt{x^2-1} dx. Let x=sec⁡θx=\sec\theta, so dx=sec⁡θtan⁡θ dθdx = \sec\theta\tan\theta \, d\theta. Limits: x=1⇒sec⁡θ=1⇒θ=0x=1 \Rightarrow \sec\theta=1 \Rightarrow \theta=0. x=2⇒sec⁡θ=2⇒θ=π/4x=\sqrt{2} \Rightarrow \sec\theta=\sqrt{2} \Rightarrow \theta=\pi/4. The integral becomes 4∫0π/4sec⁡2θ−1(sec⁡θtan⁡θ)dθ=4∫0π/4tan⁡2θsec⁡θdθ=4∫0π/4(sec⁡2θ−1)sec⁡θdθ=4∫0π/4(sec⁡3θ−sec⁡θ)dθ4\int_{0}^{\pi/4} \sqrt{\sec^2\theta-1} (\sec\theta\tan\theta) d\theta = 4\int_{0}^{\pi/4} \tan^2\theta\sec\theta d\theta = 4\int_{0}^{\pi/4} (\sec^2\theta-1)\sec\theta d\theta = 4\int_{0}^{\pi/4} (\sec^3\theta - \sec\theta) d\theta. The antiderivative is 4[12(sec⁡θtan⁡θ+ln⁡∣sec⁡θ+tan⁡θ∣)−ln⁡∣sec⁡θ+tan⁡θ∣]=2sec⁡θtan⁡θ−2ln⁡∣sec⁡θ+tan⁡θ∣4[\frac{1}{2}(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|) - \ln|\sec\theta+\tan\theta|] = 2\sec\theta\tan\theta - 2\ln|\sec\theta+\tan\theta|. Evaluating from 0 to π/4\pi/4 gives [2(2)(1)−2ln⁡∣2+1∣]−[0−0]=22−2ln⁡(1+2)[2(\sqrt{2})(1) - 2\ln|\sqrt{2}+1|] - [0-0] = 2\sqrt{2} - 2\ln(1+\sqrt{2}).

Question 14

An ellipse is described by the equation 9x2+16y2=1449x^2+16y^2=144. What is the area of the region in the first quadrant bounded by this ellipse?

  1. 12π12\pi
  2. 6π6\pi
  3. 3π3\pi (correct answer)
  4. 24π24\pi
Explanation: The equation of the ellipse can be written in standard form as x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1. This shows the semi-axes are a=4a=4 and b=3b=3. The total area of the ellipse is πab=π(4)(3)=12π\pi ab = \pi(4)(3) = 12\pi. The question asks for the area in the first quadrant, which is one-fourth of the total area. Therefore, the area is 12π4=3π\frac{12\pi}{4} = 3\pi. This can be verified by setting up the integral A=∫04y dx=∫0431−x216dx=34∫0416−x2dxA = \int_0^4 y \, dx = \int_0^4 3\sqrt{1-\frac{x^2}{16}} dx = \frac{3}{4}\int_0^4 \sqrt{16-x^2} dx. The integral ∫0416−x2dx\int_0^4 \sqrt{16-x^2} dx represents the area of a quarter circle of radius 4, which is 14π(42)=4π\frac{1}{4}\pi(4^2) = 4\pi. So the area is 34(4π)=3π\frac{3}{4}(4\pi) = 3\pi. The trigonometric substitution x=4sin⁡θx=4\sin\theta would be used to evaluate this integral formally.

Question 15

Which trigonometric substitution is most appropriate to begin evaluating the integral ∫7−2x2dx\int \sqrt{7-2x^2} dx?

  1. x=7/2sin⁡θx = \sqrt{7/2} \sin\theta (correct answer)
  2. x=7sin⁡θx = \sqrt{7} \sin\theta
  3. x=7/2tan⁡θx = \sqrt{7/2} \tan\theta
  4. 2x=7sec⁡θ2x = \sqrt{7} \sec\theta
Explanation: The integral contains a term of the form a2−u2\sqrt{a^2-u^2}. We first rewrite the expression as 7−2x2=2(72−x2)\sqrt{7-2x^2} = \sqrt{2(\frac{7}{2}-x^2)} - this is incorrect. We must factor out the coefficient of x2x^2 from the whole expression inside the radical. Or, better, identify u2u^2 and a2a^2. The expression is (7)2−(2x)2\sqrt{(\sqrt{7})^2 - (\sqrt{2}x)^2}. This is of the form a2−u2\sqrt{a^2-u^2} with a=7a=\sqrt{7} and u=2xu=\sqrt{2}x. The appropriate substitution is u=asin⁡θu=a\sin\theta, which translates to 2x=7sin⁡θ\sqrt{2}x = \sqrt{7}\sin\theta. Solving for xx gives x=72sin⁡θ=7/2sin⁡θx = \frac{\sqrt{7}}{\sqrt{2}}\sin\theta = \sqrt{7/2} \sin\theta.

Question 16

To compute the integral ∫e2te2t+1dt\int \frac{e^{2t}}{\sqrt{e^{2t}+1}} dt, a student first uses the substitution x=etx=e^t. Which trigonometric substitution should be applied next?

  1. x2=tan⁡θx^2=\tan\theta
  2. x=sin⁡θx=\sin\theta
  3. x=sec⁡θx=\sec\theta
  4. x=tan⁡θx=\tan\theta (correct answer)
Explanation: When you encounter integrals involving expressions like a2+x2\sqrt{a^2 + x^2}, a2−x2\sqrt{a^2 - x^2}, or x2−a2\sqrt{x^2 - a^2}, trigonometric substitution is your go-to technique. The key is recognizing which trigonometric identity matches your expression's form. After the substitution x=etx = e^t, your integral becomes ∫x2x2+1dx\int \frac{x^2}{\sqrt{x^2 + 1}} dx. This expression contains x2+1\sqrt{x^2 + 1}, which has the form x2+a2\sqrt{x^2 + a^2} where a=1a = 1. For expressions of the form x2+a2\sqrt{x^2 + a^2}, you should use the substitution x=atan⁡θx = a\tan\theta. Since a=1a = 1 here, the correct substitution is x=tan⁡θx = \tan\theta. This works because of the identity 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta, which transforms x2+1=tan⁡2θ+1=sec⁡2θ=sec⁡θ\sqrt{x^2 + 1} = \sqrt{\tan^2\theta + 1} = \sqrt{\sec^2\theta} = \sec\theta. Looking at the wrong answers: Choice A) x2=tan⁡θx^2 = \tan\theta gives you x=tan⁡θx = \sqrt{\tan\theta}, which doesn't simplify your square root expression. Choice B) x=sin⁡θx = \sin\theta is used for expressions like a2−x2\sqrt{a^2 - x^2}, not x2+a2\sqrt{x^2 + a^2}. Choice C) x=sec⁡θx = \sec\theta is used for expressions like x2−a2\sqrt{x^2 - a^2}. Remember this pattern: x2+a2→x=atan⁡θ\sqrt{x^2 + a^2} \rightarrow x = a\tan\theta, a2−x2→x=asin⁡θ\sqrt{a^2 - x^2} \rightarrow x = a\sin\theta, and x2−a2→x=asec⁡θ\sqrt{x^2 - a^2} \rightarrow x = a\sec\theta. Matching the form of your expression to the correct trigonometric identity is essential for successful trigonometric substitution.

Question 17

Find the volume of the solid generated by revolving the region bounded by y=1x2+1y=\frac{1}{\sqrt{x^2+1}}, y=0y=0, x=0x=0, and x=1x=1 about the x-axis.

  1. π22\frac{\pi^2}{2}
  2. π2\frac{\pi}{2}
  3. π\pi
  4. π24\frac{\pi^2}{4} (correct answer)
Explanation: When you encounter a problem asking for the volume of a solid of revolution about the x-axis, you'll use the disk method. The volume formula is V=π∫ab[f(x)]2dxV = \pi \int_a^b [f(x)]^2 dx, where the region is bounded by y=f(x)y = f(x) and the x-axis. Here, you're revolving the region bounded by y=1x2+1y = \frac{1}{\sqrt{x^2+1}}, y=0y = 0, x=0x = 0, and x=1x = 1 about the x-axis. Setting up the integral: V=π∫01(1x2+1)2dx=π∫011x2+1dxV = \pi \int_0^1 \left(\frac{1}{\sqrt{x^2+1}}\right)^2 dx = \pi \int_0^1 \frac{1}{x^2+1} dx The antiderivative of 1x2+1\frac{1}{x^2+1} is arctan⁡(x)\arctan(x), so: V=π[arctan⁡(x)]01=π(arctan⁡(1)−arctan⁡(0))=π(π4−0)=π24V = \pi[\arctan(x)]_0^1 = \pi(\arctan(1) - \arctan(0)) = \pi\left(\frac{\pi}{4} - 0\right) = \frac{\pi^2}{4} This confirms answer choice D is correct. Choice A (π22\frac{\pi^2}{2}) likely results from incorrectly evaluating arctan⁡(1)\arctan(1) as π2\frac{\pi}{2} instead of π4\frac{\pi}{4}. Choice B (π2\frac{\pi}{2}) comes from forgetting to square the function in the disk method formula. Choice C (π\pi) represents the error of both forgetting to square the function AND incorrectly evaluating the arctangent. Remember: always square the function in the disk method, and memorize that arctan⁡(1)=π4\arctan(1) = \frac{\pi}{4}. These are two of the most common mistakes on volume of revolution problems.

Question 18

Evaluate ∫dx(1+x2)3/2\int \frac{dx}{(1+x^2)^{3/2}}.

  1. 11+x2+C\frac{1}{\sqrt{1+x^2}} + C
  2. x1+x2+C\frac{x}{\sqrt{1+x^2}} + C (correct answer)
  3. −11+x2+C\frac{-1}{\sqrt{1+x^2}} + C
  4. −x1+x2+C\frac{-x}{\sqrt{1+x^2}} + C
Explanation: Use the trigonometric substitution x=tan⁡θx=\tan\theta, so dx=sec⁡2θ dθdx=\sec^2\theta \, d\theta. The denominator becomes (1+tan⁡2θ)3/2=(sec⁡2θ)3/2=sec⁡3θ(1+\tan^2\theta)^{3/2} = (\sec^2\theta)^{3/2} = \sec^3\theta. The integral transforms to ∫sec⁡2θsec⁡3θdθ=∫1sec⁡θdθ=∫cos⁡θ dθ=sin⁡θ+C\int \frac{\sec^2\theta}{\sec^3\theta} d\theta = \int \frac{1}{\sec\theta} d\theta = \int \cos\theta \, d\theta = \sin\theta + C. To convert back to xx, we use the relation x=tan⁡θx=\tan\theta. In a right triangle, tan⁡θ=x/1\tan\theta=x/1, so the opposite side is xx, the adjacent side is 1, and the hypotenuse is x2+1\sqrt{x^2+1}. Therefore, sin⁡θ=oppositehypotenuse=xx2+1\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2+1}}. The final answer is x1+x2+C\frac{x}{\sqrt{1+x^2}} + C.

Question 19

An attempt to evaluate ∫dxx2−16\int \frac{dx}{\sqrt{x^2-16}} using the substitution x=4sin⁡θx=4\sin\theta would fail. What is the primary reason for this failure?

  1. The derivative dx=4cos⁡θ dθdx = 4\cos\theta\,d\theta would introduce a cosine.
  2. The expression sin⁡2θ−1\sin^2\theta-1 does not simplify to a single positive trigonometric function.
  3. The domain of x=4sin⁡θx=4\sin\theta is [−4,4][-4, 4], which does not match the domain of the integrand. (correct answer)
  4. The resulting integral in θ\theta would be more complex than the original integral.
Explanation: The integrand 1x2−16\frac{1}{\sqrt{x^2-16}} is defined for x2>16x^2 > 16, i.e., x>4x>4 or x<−4x<-4. The substitution x=4sin⁡θx=4\sin\theta restricts xx to the interval [−4,4][-4, 4], since sin⁡θ\sin\theta is between -1 and 1. The domain of the substitution and the domain of the function to be integrated do not align. Furthermore, using this substitution would lead to 16sin⁡2θ−16=−16cos⁡2θ\sqrt{16\sin^2\theta-16} = \sqrt{-16\cos^2\theta}, which is not a real number for real θ\theta. Option B is related but C is the more fundamental reason related to the domains of the functions.

Question 20

To evaluate the integral ∫46dxx2−8x+25\int_{4}^{6} \frac{dx}{\sqrt{x^2-8x+25}}, which of the following is the most appropriate first step?

  1. Substitute x=5tan⁡θx=5\tan\theta because the form resembles x2+a2x^2+a^2.
  2. Complete the square on the quadratic expression in the denominator. (correct answer)
  3. Substitute u=x2−8x+25u=x^2-8x+25 and use the power rule for integration.
  4. Substitute x=5sec⁡θx=5\sec\theta because of the presence of x2x^2 and 2525 under the radical.
Explanation: The integrand contains a quadratic expression under a radical. Before a trigonometric substitution can be applied, the quadratic x2−8x+25x^2-8x+25 must be rewritten to match one of the standard forms. This is done by completing the square: x2−8x+25=(x2−8x+16)−16+25=(x−4)2+9x^2-8x+25 = (x^2-8x+16) - 16 + 25 = (x-4)^2+9. After this step, a substitution like x−4=3tan⁡θx-4=3\tan\theta can be made. The other options are incorrect because they either misidentify the form of the integral or suggest an inappropriate technique.