Calculus 2 Quiz: Integrating Tan And Sec Products
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Integrating Tan And Sec ProductsQuestion 1 of 20

Evaluate the integral ∫tan⁡3(x)sec⁡(x) dx\int \tan^3(x) \sec(x) \, dx.

13tan⁡3(x)−tan⁡(x)+x+C\frac{1}{3}\tan^3(x) - \tan(x) + x + C
13sec⁡3(x)+sec⁡(x)+C\frac{1}{3}\sec^3(x) + \sec(x) + C
14tan⁡4(x)+C\frac{1}{4}\tan^4(x) + C
13sec⁡3(x)−sec⁡(x)+C\frac{1}{3}\sec^3(x) - \sec(x) + C
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Calculus 2 Quiz

Calculus 2 Quiz: Integrating Tan And Sec Products

Practice Integrating Tan And Sec 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 Tan And Sec Products, 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

Evaluate the integral ∫tan⁡3(x)sec⁡(x) dx\int \tan^3(x) \sec(x) \, dx.

  1. 13tan⁡3(x)−tan⁡(x)+x+C\frac{1}{3}\tan^3(x) - \tan(x) + x + C
  2. 13sec⁡3(x)+sec⁡(x)+C\frac{1}{3}\sec^3(x) + \sec(x) + C
  3. 14tan⁡4(x)+C\frac{1}{4}\tan^4(x) + C
  4. 13sec⁡3(x)−sec⁡(x)+C\frac{1}{3}\sec^3(x) - \sec(x) + C (correct answer)
Explanation: When you encounter integrals involving powers of tangent and secant, look for substitution opportunities using their derivatives. Since ddx[tan⁡(x)]=sec⁡2(x)\frac{d}{dx}[\tan(x)] = \sec^2(x) and ddx[sec⁡(x)]=sec⁡(x)tan⁡(x)\frac{d}{dx}[\sec(x)] = \sec(x)\tan(x), the presence of sec⁡(x)\sec(x) in your integral suggests using u=sec⁡(x)u = \sec(x). Let u=sec⁡(x)u = \sec(x), so du=sec⁡(x)tan⁡(x) dxdu = \sec(x)\tan(x) \, dx. Now you need to express tan⁡3(x)sec⁡(x)\tan^3(x) \sec(x) in terms of uu and dudu. Since tan⁡2(x)=sec⁡2(x)−1=u2−1\tan^2(x) = \sec^2(x) - 1 = u^2 - 1, you have: tan⁡3(x)sec⁡(x)=tan⁡2(x)⋅tan⁡(x)sec⁡(x)=(u2−1)⋅du\tan^3(x) \sec(x) = \tan^2(x) \cdot \tan(x) \sec(x) = (u^2 - 1) \cdot du The integral becomes: ∫(u2−1) du=u33−u+C=13sec⁡3(x)−sec⁡(x)+C\int (u^2 - 1) \, du = \frac{u^3}{3} - u + C = \frac{1}{3}\sec^3(x) - \sec(x) + C Option A uses tangent functions instead of secant, which would result from a different substitution approach that doesn't work cleanly here. Option B has the wrong sign—it adds sec⁡(x)\sec(x) instead of subtracting it, likely from incorrectly handling the −1-1 term in (u2−1)(u^2 - 1). Option C suggests using u=tan⁡(x)u = \tan(x), but this substitution fails because you'd need sec⁡2(x)\sec^2(x) in the integral, not just sec⁡(x)\sec(x). Study tip: For tangent-secant integrals, try u=sec⁡(x)u = \sec(x) when you have an odd power of tangent with sec⁡(x)\sec(x), and u=tan⁡(x)u = \tan(x) when you have an even power of tangent with sec⁡2(x)\sec^2(x).

Question 2

The integral ∫tan⁡2(x)sec⁡3(x) dx\int \tan^2(x) \sec^3(x) \, dx can be rewritten using the identity tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1. What integral results from this substitution?

  1. 13tan⁡3(x)⋅14sec⁡4(x)+C\frac{1}{3}\tan^3(x) \cdot \frac{1}{4}\sec^4(x) + C
  2. ∫(sec⁡5(x)+sec⁡3(x)) dx\int (\sec^5(x) + \sec^3(x)) \, dx
  3. ∫(sec⁡4(x)−sec⁡2(x)) dx\int (\sec^4(x) - \sec^2(x)) \, dx
  4. ∫(sec⁡5(x)−sec⁡3(x)) dx\int (\sec^5(x) - \sec^3(x)) \, dx (correct answer)
Explanation: When you encounter integrals involving powers of trigonometric functions, using trigonometric identities to rewrite the integrand is often the key to simplification. This problem tests your ability to apply the Pythagorean identity tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1 correctly. Starting with ∫tan⁡2(x)sec⁡3(x) dx\int \tan^2(x) \sec^3(x) \, dx, substitute the identity: ∫(sec⁡2(x)−1)sec⁡3(x) dx\int (\sec^2(x) - 1) \sec^3(x) \, dx Using the distributive property: ∫sec⁡2(x)⋅sec⁡3(x) dx−∫1⋅sec⁡3(x) dx\int \sec^2(x) \cdot \sec^3(x) \, dx - \int 1 \cdot \sec^3(x) \, dx This simplifies to: ∫sec⁡5(x) dx−∫sec⁡3(x) dx=∫(sec⁡5(x)−sec⁡3(x)) dx\int \sec^5(x) \, dx - \int \sec^3(x) \, dx = \int (\sec^5(x) - \sec^3(x)) \, dx This confirms answer D is correct. Looking at the incorrect options: Answer A represents a common misconception where students think they can integrate each factor separately and multiply the results—this violates basic integration rules. Answer B has the wrong sign; it shows addition instead of subtraction, likely from misapplying the identity or making an algebraic error. Answer C incorrectly reduces the powers of secant, possibly from confusion about how exponents combine during multiplication. Remember this pattern: when you see products of trigonometric functions with different powers, look for identities that can split one function into simpler parts. The key is careful algebraic manipulation after substitution—distribute thoroughly and watch your signs.

Question 3

Let f′(x)=tan⁡(x)sec⁡3(x)f'(x) = \tan(x) \sec^3(x). If f(0)=2f(0) = 2, what is the value of f(π/4)f(\pi/4)?

  1. 223\frac{2\sqrt{2}}{3}
  2. 22+53\frac{2\sqrt{2}+5}{3} (correct answer)
  3. 22+73\frac{2\sqrt{2}+7}{3}
  4. 53\frac{5}{3}
Explanation: First, find the antiderivative f(x)=∫tan⁡(x)sec⁡3(x) dxf(x) = \int \tan(x) \sec^3(x)\,dx. Save a factor of sec⁡(x)tan⁡(x)\sec(x)\tan(x): ∫sec⁡2(x)(sec⁡(x)tan⁡(x)) dx\int \sec^2(x) (\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 becomes ∫u2 du=u33+C=13sec⁡3(x)+C\int u^2\,du = \frac{u^3}{3} + C = \frac{1}{3}\sec^3(x) + C. Now use the initial condition f(0)=2f(0) = 2: 2=13sec⁡3(0)+C  ⟹  2=13(1)3+C  ⟹  C=2−13=532 = \frac{1}{3}\sec^3(0) + C \implies 2 = \frac{1}{3}(1)^3 + C \implies C = 2 - \frac{1}{3} = \frac{5}{3}. So, f(x)=13sec⁡3(x)+53f(x) = \frac{1}{3}\sec^3(x) + \frac{5}{3}. Finally, evaluate at x=π/4x=\pi/4: f(π/4)=13sec⁡3(π/4)+53=13(2)3+53=223+53=22+53f(\pi/4) = \frac{1}{3}\sec^3(\pi/4) + \frac{5}{3} = \frac{1}{3}(\sqrt{2})^3 + \frac{5}{3} = \frac{2\sqrt{2}}{3} + \frac{5}{3} = \frac{2\sqrt{2}+5}{3}.

Question 4

Consider the integral ∫sec⁡5(x) dx\int \sec^5(x) \, dx. Using the reduction formula In=1n−1sec⁡n−2(x)tan⁡(x)+n−2n−1In−2I_n = \frac{1}{n-1}\sec^{n-2}(x)\tan(x) + \frac{n-2}{n-1}I_{n-2}, where In=∫sec⁡n(x) dxI_n = \int \sec^n(x)\,dx, what is the integral I5I_5?

  1. 15sec⁡4(x)tan⁡(x)+45sec⁡2(x)tan⁡(x)+C\frac{1}{5}\sec^4(x)\tan(x) + \frac{4}{5}\sec^2(x)\tan(x) + C
  2. 14sec⁡3(x)tan⁡(x)+34ln⁡∣sec⁡(x)+tan⁡(x)∣+C\frac{1}{4}\sec^3(x)\tan(x) + \frac{3}{4}\ln|\sec(x)+\tan(x)| + C
  3. 14sec⁡3(x)tan⁡(x)+38sec⁡(x)tan⁡(x)+38ln⁡∣sec⁡(x)+tan⁡(x)∣+C\frac{1}{4}\sec^3(x)\tan(x) + \frac{3}{8}\sec(x)\tan(x) + \frac{3}{8}\ln|\sec(x)+\tan(x)| + C (correct answer)
  4. 14sec⁡3(x)tan⁡(x)−38sec⁡(x)tan⁡(x)−38ln⁡∣sec⁡(x)+tan⁡(x)∣+C\frac{1}{4}\sec^3(x)\tan(x) - \frac{3}{8}\sec(x)\tan(x) - \frac{3}{8}\ln|\sec(x)+\tan(x)| + C
Explanation: When you encounter integrals of powers of secant, reduction formulas are your best tool because direct integration becomes extremely complex for higher powers. The key is applying the formula systematically and knowing the base cases. Starting with the given reduction formula for I5=∫sec⁡5(x) dxI_5 = \int \sec^5(x) \, dx, substitute n=5n = 5: I5=15−1sec⁡5−2(x)tan⁡(x)+5−25−1I5−2=14sec⁡3(x)tan⁡(x)+34I3I_5 = \frac{1}{5-1}\sec^{5-2}(x)\tan(x) + \frac{5-2}{5-1}I_{5-2} = \frac{1}{4}\sec^3(x)\tan(x) + \frac{3}{4}I_3 Now you need I3=∫sec⁡3(x) dxI_3 = \int \sec^3(x) \, dx. Apply the formula again with n=3n = 3: I3=13−1sec⁡3−2(x)tan⁡(x)+3−23−1I3−2=12sec⁡(x)tan⁡(x)+12I1I_3 = \frac{1}{3-1}\sec^{3-2}(x)\tan(x) + \frac{3-2}{3-1}I_{3-2} = \frac{1}{2}\sec(x)\tan(x) + \frac{1}{2}I_1 The base case is I1=∫sec⁡(x) dx=ln⁡∣sec⁡(x)+tan⁡(x)∣+CI_1 = \int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C. Substituting back: I3=12sec⁡(x)tan⁡(x)+12ln⁡∣sec⁡(x)+tan⁡(x)∣+CI_3 = \frac{1}{2}\sec(x)\tan(x) + \frac{1}{2}\ln|\sec(x) + \tan(x)| + C Therefore: I5=14sec⁡3(x)tan⁡(x)+34⋅[12sec⁡(x)tan⁡(x)+12ln⁡∣sec⁡(x)+tan⁡(x)∣]+CI_5 = \frac{1}{4}\sec^3(x)\tan(x) + \frac{3}{4} \cdot \left[\frac{1}{2}\sec(x)\tan(x) + \frac{1}{2}\ln|\sec(x) + \tan(x)|\right] + C =14sec⁡3(x)tan⁡(x)+38sec⁡(x)tan⁡(x)+38ln⁡∣sec⁡(x)+tan⁡(x)∣+C= \frac{1}{4}\sec^3(x)\tan(x) + \frac{3}{8}\sec(x)\tan(x) + \frac{3}{8}\ln|\sec(x) + \tan(x)| + C This matches answer C. Answer A stops at the first reduction step without continuing to I3I_3. Answer B incorrectly computes the coefficients in the reduction. Answer D has incorrect signs, likely from algebraic errors during substitution. Always work reduction formulas step-by-step down to known base cases, and double-check your arithmetic at each substitution.

Question 5

To evaluate ∫sec⁡2(x)tan⁡(x) dx\int \frac{\sec^2(x)}{\sqrt{\tan(x)}} \, dx, a u-substitution is performed. What is the resulting integral in terms of uu?

  1. ∫u−1/2 du\int u^{-1/2} \, du (correct answer)
  2. ∫u1/2 du\int u^{1/2} \, du
  3. ∫1usec⁡(x) du\int \frac{1}{\sqrt{u}} \sec(x) \, du
  4. ∫(u2+1)−1/4 du\int (u^2+1)^{-1/4} \, du
Explanation: The integrand can be written as ∫(tan⁡(x))−1/2sec⁡2(x) dx\int (\tan(x))^{-1/2} \sec^2(x) \, dx. The presence of tan⁡(x)\tan(x) and its derivative sec⁡2(x)\sec^2(x) suggests the substitution u=tan⁡(x)u = \tan(x). The differential is du=sec⁡2(x) dxdu = \sec^2(x)\,dx. Substituting uu and dudu into the integral directly yields ∫u−1/2 du\int u^{-1/2} \, du.

Question 6

Evaluate the integral ∫0π/4sec⁡3(x) dx\int_0^{\pi/4} \sec^3(x) \, dx.

  1. 12(2+ln⁡(2+1))\frac{1}{2}(\sqrt{2} + \ln(\sqrt{2}+1)) (correct answer)
  2. 12(2−ln⁡(2+1))\frac{1}{2}(\sqrt{2} - \ln(\sqrt{2}+1))
  3. 2+ln⁡(2+1)\sqrt{2} + \ln(\sqrt{2}+1)
  4. 22\frac{\sqrt{2}}{2}
Explanation: The antiderivative of sec⁡3(x)\sec^3(x) is a standard result found using integration by parts: ∫sec⁡3(x) dx=12(sec⁡(x)tan⁡(x)+ln⁡∣sec⁡(x)+tan⁡(x)∣)+C\int \sec^3(x)\,dx = \frac{1}{2}(\sec(x)\tan(x) + \ln|\sec(x) + \tan(x)|) + C. We evaluate this from 0 to π/4\pi/4. At x=π/4x=\pi/4, we have 12(sec⁡(π/4)tan⁡(π/4)+ln⁡∣sec⁡(π/4)+tan⁡(π/4)∣)=12(2⋅1+ln⁡∣2+1∣)\frac{1}{2}(\sec(\pi/4)\tan(\pi/4) + \ln|\sec(\pi/4) + \tan(\pi/4)|) = \frac{1}{2}(\sqrt{2} \cdot 1 + \ln|\sqrt{2} + 1|). At x=0x=0, we have 12(sec⁡(0)tan⁡(0)+ln⁡∣sec⁡(0)+tan⁡(0)∣)=12(1⋅0+ln⁡∣1+0∣)=12(0+0)=0\frac{1}{2}(\sec(0)\tan(0) + \ln|\sec(0) + \tan(0)|) = \frac{1}{2}(1 \cdot 0 + \ln|1 + 0|) = \frac{1}{2}(0 + 0) = 0. The final result is 12(2+ln⁡(2+1))\frac{1}{2}(\sqrt{2} + \ln(\sqrt{2}+1)).

Question 7

Which of the following represents the correct approach to evaluate ∫tan⁡5xsec⁡3x dx\int \tan^5 x \sec^3 x \, dx?

  1. Use substitution u=tan⁡xu = \tan x and rewrite sec⁡3x=sec⁡x(1+tan⁡2x)\sec^3 x = \sec x(1 + \tan^2 x)
  2. Use substitution u=sec⁡xu = \sec x and rewrite tan⁡5x=(sec⁡2x−1)2⋅tan⁡x\tan^5 x = (\sec^2 x - 1)^2 \cdot \tan x (correct answer)
  3. Use integration by parts with u=tan⁡4xu = \tan^4 x and dv=tan⁡xsec⁡3x dxdv = \tan x \sec^3 x \, dx
  4. Use substitution u=sin⁡xu = \sin x and convert everything to sine and cosine functions
Explanation: For ∫tan⁡5xsec⁡3x dx\int \tan^5 x \sec^3 x \, dx, we have an odd power of tangent and an odd power of secant. The standard approach is to use u=sec⁡xu = \sec x, du=sec⁡xtan⁡x dxdu = \sec x \tan x \, dx. We need to factor out sec⁡xtan⁡x\sec x \tan x and express the remaining tan⁡4xsec⁡2x\tan^4 x \sec^2 x in terms of secant: tan⁡4x=(sec⁡2x−1)2\tan^4 x = (\sec^2 x - 1)^2. Choice A won't work because we can't easily express sec⁡3x\sec^3 x in terms of tangent. Choice C is unnecessarily complicated. Choice D converts to sine/cosine but isn't the most efficient method for this integral type.

Question 8

Evaluate the definite integral ∫0π/8sec⁡4(2x)tan⁡(2x) dx\int_0^{\pi/8} \sec^4(2x) \tan(2x) \, dx.

  1. 38\frac{3}{8} (correct answer)
  2. 34\frac{3}{4}
  3. 12\frac{1}{2}
  4. 716\frac{7}{16}
Explanation: Let w=2xw = 2x, so dw=2dxdw = 2dx or dx=12dwdx = \frac{1}{2}dw. The integral becomes 12∫0π/4sec⁡4(w)tan⁡(w) dw\frac{1}{2}\int_{0}^{\pi/4} \sec^4(w) \tan(w) \, dw. This fits both standard cases. Let's use u=sec⁡(w)u=\sec(w), so du=sec⁡(w)tan⁡(w) dwdu = \sec(w)\tan(w)\,dw. The integral is 12∫sec⁡3(w)(sec⁡(w)tan⁡(w)) dw\frac{1}{2}\int \sec^3(w) (\sec(w)\tan(w))\,dw. In terms of uu, this is 12∫u3 du=12u44=18u4=18sec⁡4(w)\frac{1}{2} \int u^3\,du = \frac{1}{2} \frac{u^4}{4} = \frac{1}{8}u^4 = \frac{1}{8}\sec^4(w). Evaluating from w=0w=0 to w=π/4w=\pi/4: [18sec⁡4(w)]0π/4=18(sec⁡4(π/4)−sec⁡4(0))=18((2)4−14)=18(4−1)=38[\frac{1}{8}\sec^4(w)]_0^{\pi/4} = \frac{1}{8}(\sec^4(\pi/4) - \sec^4(0)) = \frac{1}{8}((\sqrt{2})^4 - 1^4) = \frac{1}{8}(4 - 1) = \frac{3}{8}.

Question 9

Evaluate the definite integral ∫0π/4tan⁡3(x)sec⁡4(x) dx\int_0^{\pi/4} \tan^3(x) \sec^4(x) \, dx

  1. 512\frac{5}{12} (correct answer)
  2. 112\frac{1}{12}
  3. −112-\frac{1}{12}
  4. 712\frac{7}{12}
Explanation: The power of secant is even, so we save a sec⁡2(x)\sec^2(x) factor and use the identity sec⁡2(x)=1+tan⁡2(x)\sec^2(x) = 1 + \tan^2(x). The integral becomes ∫0π/4tan⁡3(x)(1+tan⁡2(x))sec⁡2(x) dx\int_0^{\pi/4} \tan^3(x)(1+\tan^2(x))\sec^2(x)\,dx. Let u=tan⁡(x)u = \tan(x), so du=sec⁡2(x) dxdu = \sec^2(x)\,dx. The bounds change: when x=0,u=tan⁡(0)=0x=0, u=\tan(0)=0; when x=π/4,u=tan⁡(π/4)=1x=\pi/4, u=\tan(\pi/4)=1. The integral in uu is ∫01u3(1+u2) du=∫01(u3+u5) du\int_0^1 u^3(1+u^2)\,du = \int_0^1 (u^3+u^5)\,du. Evaluating this gives [u44+u66]01=(14+16)−0=3+212=512[\frac{u^4}{4} + \frac{u^6}{6}]_0^1 = (\frac{1}{4} + \frac{1}{6}) - 0 = \frac{3+2}{12} = \frac{5}{12}.

Question 10

To evaluate an integral of the form ∫tan⁡m(x)sec⁡n(x) dx\int \tan^m(x) \sec^n(x) \, dx, the substitution u=sec⁡(x)u = \sec(x) is most effective under which general conditions for the integers mm and nn?

  1. mm is any integer and nn is a positive even integer.
  2. mm is a positive odd integer and n≥1n \ge 1. (correct answer)
  3. mm is a positive even integer and nn is a positive odd integer.
  4. mm and nn are both positive even integers.
Explanation: The substitution u=sec⁡(x)u = \sec(x) requires the differential du=sec⁡(x)tan⁡(x) dxdu = \sec(x)\tan(x)\,dx. To use this, we must be able to factor out one sec⁡(x)\sec(x) term and one tan⁡(x)\tan(x) term from the integrand. This is possible if the power of tangent, mm, is at least 1 and odd, and the power of secant, nn, is at least 1. If mm is odd, we can factor out tan⁡(x)\tan(x), and the remaining even power tan⁡m−1(x)\tan^{m-1}(x) can be converted to secants using tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1.

Question 11

Evaluate the integral ∫−π/4π/4tan⁡5(x)sec⁡2(x) dx\int_{-\pi/4}^{\pi/4} \tan^5(x) \sec^2(x) \, dx.

  1. 00 (correct answer)
  2. 13\frac{1}{3}
  3. 16\frac{1}{6}
  4. −13-\frac{1}{3}
Explanation: The integrand f(x)=tan⁡5(x)sec⁡2(x)f(x) = \tan^5(x) \sec^2(x) is an odd function because f(−x)=(tan⁡(−x))5(sec⁡(−x))2=(−tan⁡(x))5(sec⁡(x))2=−tan⁡5(x)sec⁡2(x)=−f(x)f(-x) = (\tan(-x))^5 (\sec(-x))^2 = (-\tan(x))^5 (\sec(x))^2 = -\tan^5(x) \sec^2(x) = -f(x). The integral of an odd function over a symmetric interval [−a,a][-a, a] is always zero. Alternatively, one can compute the integral using u=tan⁡(x)u=\tan(x), du=sec⁡2(x) dxdu=\sec^2(x)\,dx. The bounds become u(−π/4)=−1u(-\pi/4) = -1 and u(π/4)=1u(\pi/4) = 1. The integral is ∫−11u5 du=[u66]−11=166−(−1)66=16−16=0\int_{-1}^1 u^5\,du = [\frac{u^6}{6}]_{-1}^1 = \frac{1^6}{6} - \frac{(-1)^6}{6} = \frac{1}{6} - \frac{1}{6} = 0.

Question 12

The integral ∫xsec⁡(x)tan⁡(x) dx\int x \sec(x) \tan(x) \, dx is best evaluated using which technique?

  1. Integration by parts with u=xu = x and dv=sec⁡(x)tan⁡(x) dxdv = \sec(x)\tan(x)\,dx. (correct answer)
  2. Integration by parts with u=sec⁡(x)tan⁡(x)u = \sec(x)\tan(x) and dv=x dxdv = x\,dx.
  3. U-substitution with u=sec⁡(x)u = \sec(x).
  4. U-substitution with u=xu = x.
Explanation: The integral contains a product of an algebraic function (xx) and a trigonometric function (sec⁡(x)tan⁡(x)\sec(x)\tan(x)). This structure suggests integration by parts. The choice of uu and dvdv is guided by the LIATE rule (Log, Inverse, Algebraic, Trig, Exponential). We choose the algebraic term u=xu=x, as its derivative simplifies to du=dxdu=dx. The remaining part, dv=sec⁡(x)tan⁡(x) dxdv = \sec(x)\tan(x)\,dx, is a standard derivative, and its integral is v=sec⁡(x)v=\sec(x). This choice leads to a simpler integral to solve. The alternative choice in B would lead to a more complex integral.

Question 13

The evaluation of ∫tan⁡m(x)sec⁡n(x) dx\int \tan^m(x) \sec^n(x) \, dx where mm is a non-negative even integer and nn is a positive odd integer typically requires which sequence of steps?

  1. A single u-substitution where u=tan⁡(x)u = \tan(x), after factoring out sec⁡2(x)\sec^2(x).
  2. A single u-substitution where u=sec⁡(x)u = \sec(x), after factoring out sec⁡(x)tan⁡(x)\sec(x)\tan(x).
  3. Using identities to express the integrand in terms of sec⁡(x)\sec(x) alone, followed by integration by parts. (correct answer)
  4. Using identities to express the integrand in terms of tan⁡(x)\tan(x) alone, followed by polynomial integration.
Explanation: This case (even tangent, odd secant) is the most complex scenario. Neither of the standard u-substitutions works directly. The typical strategy is to use the identity tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1 to convert all even powers of tangent into powers of secant. This results in a sum of integrals of the form ∫sec⁡k(x) dx\int \sec^k(x)\,dx where kk is odd. These integrals are then solved using integration by parts, often via a reduction formula.

Question 14

Which of the following integrals is equivalent to ∫tan⁡3(x) dx\int \tan^3(x) \, dx?

  1. 14tan⁡4(x)+C\frac{1}{4}\tan^4(x) + C
  2. ∫(sec⁡2(x)+1)tan⁡(x) dx\int (\sec^2(x) + 1)\tan(x) \, dx
  3. ∫(sec⁡2(x)−1)tan⁡(x) dx\int (\sec^2(x) - 1)\tan(x) \, dx (correct answer)
  4. ∫(sec⁡2(x))tan⁡(x) dx−x+C\int (\sec^2(x))\tan(x) \, dx - x + C
Explanation: When you encounter integrals of powers of trigonometric functions, the key strategy is to use trigonometric identities to rewrite the integrand in a more manageable form. For tan⁡3(x)\tan^3(x), you need to break it down using the fundamental identity tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1. Starting with ∫tan⁡3(x) dx\int \tan^3(x) \, dx, rewrite this as ∫tan⁡(x)⋅tan⁡2(x) dx\int \tan(x) \cdot \tan^2(x) \, dx. Now substitute the identity: ∫tan⁡(x)⋅(sec⁡2(x)−1) dx=∫(sec⁡2(x)−1)tan⁡(x) dx\int \tan(x) \cdot (\sec^2(x) - 1) \, dx = \int (\sec^2(x) - 1)\tan(x) \, dx. This matches option C exactly. Looking at the wrong answers: Option A suggests 14tan⁡4(x)+C\frac{1}{4}\tan^4(x) + C, which would be correct if we were integrating tan⁡3(x)sec⁡2(x)\tan^3(x) \sec^2(x) using substitution, but that's not our original integral. Option B has (sec⁡2(x)+1)tan⁡(x)(\sec^2(x) + 1)\tan(x), which incorrectly adds 1 instead of subtracting it from sec⁡2(x)\sec^2(x) in the identity. Option D includes −x+C-x + C at the end, suggesting someone partially integrated (sec⁡2(x)−1)tan⁡(x)(\sec^2(x) - 1)\tan(x) and remembered that ∫(−1) dx=−x\int (-1) \, dx = -x, but this represents a completed integral, not an equivalent form of the original. The correct answer is C because it shows the proper application of the Pythagorean identity to transform tan⁡3(x)\tan^3(x) into an integrable form. Remember: when dealing with odd powers of tangent, always factor out one tan⁡(x)\tan(x) and convert the remaining even power using tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1.

Question 15

The area of the region bounded by the curve y=sec⁡2(x)tan⁡(x)y = \sec^2(x)\tan(x), the x-axis, and the lines x=0x=0 and x=π/3x=\pi/3 is given by which value?

  1. 1
  2. 32\frac{3}{2} (correct answer)
  3. 2
  4. 12\frac{1}{2}
Explanation: The area is given by the definite integral ∫0π/3sec⁡2(x)tan⁡(x) dx\int_0^{\pi/3} \sec^2(x)\tan(x) \, dx. This integral can be solved with the substitution u=tan⁡(x)u = \tan(x), which gives du=sec⁡2(x) dxdu = \sec^2(x)\,dx. The bounds of integration change from x=0x=0 to u=tan⁡(0)=0u = \tan(0) = 0 and from x=π/3x=\pi/3 to u=tan⁡(π/3)=3u = \tan(\pi/3) = \sqrt{3}. The integral becomes ∫03u du\int_0^{\sqrt{3}} u \, du. Evaluating this gives [u22]03=(3)22−022=32[\frac{u^2}{2}]_0^{\sqrt{3}} = \frac{(\sqrt{3})^2}{2} - \frac{0^2}{2} = \frac{3}{2}.

Question 16

Evaluate the indefinite integral ∫sec⁡4(x)tan⁡2(x) dx\int \frac{\sec^4(x)}{\tan^2(x)} \, dx.

  1. −csc⁡(x)+sec⁡(x)+C-\csc(x) + \sec(x) + C
  2. cot⁡(x)−tan⁡(x)+C\cot(x) - \tan(x) + C
  3. −cot⁡(x)+tan⁡(x)+C-\cot(x) + \tan(x) + C (correct answer)
  4. ln⁡∣tan⁡(x)∣+12tan⁡2(x)+C\ln|\tan(x)| + \frac{1}{2}\tan^2(x) + C
Explanation: When you encounter integrals involving powers of trigonometric functions, look for opportunities to rewrite them using fundamental identities and algebraic manipulation before attempting substitution or integration by parts. Start by rewriting this integral using the identity sec⁡2(x)=1+tan⁡2(x)\sec^2(x) = 1 + \tan^2(x): ∫sec⁡4(x)tan⁡2(x) dx=∫sec⁡2(x)⋅sec⁡2(x)tan⁡2(x) dx=∫(1+tan⁡2(x))sec⁡2(x)tan⁡2(x) dx\int \frac{\sec^4(x)}{\tan^2(x)} \, dx = \int \frac{\sec^2(x) \cdot \sec^2(x)}{\tan^2(x)} \, dx = \int \frac{(1 + \tan^2(x)) \sec^2(x)}{\tan^2(x)} \, dx Split this into two separate integrals: ∫sec⁡2(x)tan⁡2(x) dx+∫tan⁡2(x)sec⁡2(x)tan⁡2(x) dx=∫sec⁡2(x)tan⁡2(x) dx+∫sec⁡2(x) dx\int \frac{\sec^2(x)}{\tan^2(x)} \, dx + \int \frac{\tan^2(x) \sec^2(x)}{\tan^2(x)} \, dx = \int \frac{\sec^2(x)}{\tan^2(x)} \, dx + \int \sec^2(x) \, dx The second integral is straightforward: ∫sec⁡2(x) dx=tan⁡(x)\int \sec^2(x) \, dx = \tan(x) For the first integral, use the identity sec⁡2(x)tan⁡2(x)=1sin⁡2(x)=csc⁡2(x)\frac{\sec^2(x)}{\tan^2(x)} = \frac{1}{\sin^2(x)} = \csc^2(x) Therefore: ∫csc⁡2(x) dx=−cot⁡(x)\int \csc^2(x) \, dx = -\cot(x) Combining both parts: −cot⁡(x)+tan⁡(x)+C-\cot(x) + \tan(x) + C, which is answer C. Answer A incorrectly involves secant and cosecant functions that don't appear in the correct antiderivative. Answer B has the wrong sign on the cotangent term. Answer D suggests a logarithmic form that would only arise from a rational function in tangent, not from this particular combination. Strategy tip: When dealing with quotients of trigonometric powers, always try rewriting using Pythagorean identities first. This often transforms complex expressions into recognizable standard forms.

Question 17

Which u-substitution would be the most effective first step for evaluating ∫tan⁡(x)sec⁡4(x) dx\int \sqrt{\tan(x)} \sec^4(x) \, dx?

  1. u=tan⁡(x)u = \sqrt{\tan(x)}
  2. u=sec⁡(x)u = \sec(x)
  3. u=sec⁡2(x)u = \sec^2(x)
  4. u=tan⁡(x)u = \tan(x) (correct answer)
Explanation: When evaluating integrals involving trigonometric functions with radicals, your goal is to find a substitution that simultaneously simplifies the radical and creates a manageable derivative relationship with the remaining terms. Let's examine what happens with u=tan⁡(x)u = \tan(x) (choice D). This gives us du=sec⁡2(x) dxdu = \sec^2(x) \, dx, so dx=dusec⁡2(x)dx = \frac{du}{\sec^2(x)}. The integral becomes: ∫usec⁡4(x)⋅dusec⁡2(x)=∫usec⁡2(x) du\int \sqrt{u} \sec^4(x) \cdot \frac{du}{\sec^2(x)} = \int \sqrt{u} \sec^2(x) \, du Since u=tan⁡(x)u = \tan(x), we know that sec⁡2(x)=1+tan⁡2(x)=1+u2\sec^2(x) = 1 + \tan^2(x) = 1 + u^2. This transforms our integral into: ∫u(1+u2) du=∫(u1/2+u5/2) du\int \sqrt{u}(1 + u^2) \, du = \int (u^{1/2} + u^{5/2}) \, du This is now a straightforward polynomial integration! Choice A (u=tan⁡(x)u = \sqrt{\tan(x)}) creates du=sec⁡2(x)2tan⁡(x)dxdu = \frac{\sec^2(x)}{2\sqrt{\tan(x)}} dx, leading to a more complex relationship. Choice B (u=sec⁡(x)u = \sec(x)) gives du=sec⁡(x)tan⁡(x) dxdu = \sec(x)\tan(x) \, dx, which doesn't align well with our integrand structure. Choice C (u=sec⁡2(x)u = \sec^2(x)) produces du=2sec⁡2(x)tan⁡(x) dxdu = 2\sec^2(x)\tan(x) \, dx, requiring us to express everything in terms of secant, which complicates the square root. Key strategy: When you see radicals involving trigonometric functions, try substituting the expression inside the radical first. This often eliminates the radical while creating favorable derivative relationships with the remaining terms.

Question 18

Evaluate the integral ∫dxcos⁡2(x)tan⁡(x)\int \frac{dx}{\cos^2(x)\tan(x)}.

  1. −cot⁡(x)+C-\cot(x) + C
  2. ln⁡∣sin⁡(x)∣+C\ln|\sin(x)| + C
  3. sec⁡(x)+C\sec(x) + C
  4. ln⁡∣tan⁡(x)∣+C\ln|\tan(x)| + C (correct answer)
Explanation: When you encounter an integral with trigonometric functions like this, your first step should be to simplify the integrand using trigonometric identities. This will often reveal a more manageable form. Start by rewriting the integrand: 1cos⁡2(x)tan⁡(x)=1cos⁡2(x)⋅1tan⁡(x)=sec⁡2(x)⋅cot⁡(x)\frac{1}{\cos^2(x)\tan(x)} = \frac{1}{\cos^2(x)} \cdot \frac{1}{\tan(x)} = \sec^2(x) \cdot \cot(x) Since cot⁡(x)=cos⁡(x)sin⁡(x)\cot(x) = \frac{\cos(x)}{\sin(x)}, you can write this as: sec⁡2(x)⋅cot⁡(x)=1cos⁡2(x)⋅cos⁡(x)sin⁡(x)=1cos⁡(x)sin⁡(x)\sec^2(x) \cdot \cot(x) = \frac{1}{\cos^2(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{1}{\cos(x)\sin(x)} Now use the substitution u=tan⁡(x)u = \tan(x), so du=sec⁡2(x)dxdu = \sec^2(x)dx. Notice that 1cos⁡(x)sin⁡(x)=sec⁡2(x)tan⁡(x)\frac{1}{\cos(x)\sin(x)} = \frac{\sec^2(x)}{\tan(x)}, which becomes 1udu\frac{1}{u}du. This integrates to ln⁡∣u∣+C=ln⁡∣tan⁡(x)∣+C\ln|u| + C = \ln|\tan(x)| + C, confirming answer D. Let's examine why the other options are incorrect: A would result from incorrectly thinking the derivative of cot⁡(x)\cot(x) is positive rather than −csc⁡2(x)-\csc^2(x). B comes from misapplying trigonometric identities and confusing this with ∫cot⁡(x)dx\int \cot(x)dx. C represents the antiderivative of sec⁡(x)tan⁡(x)\sec(x)\tan(x), not our integrand. Strategy tip: When facing complex trigonometric integrals, always simplify using identities first, then look for substitutions involving u=tan⁡(x)u = \tan(x), u=sin⁡(x)u = \sin(x), or u=cos⁡(x)u = \cos(x). The form 1u\frac{1}{u} should immediately signal ln⁡∣u∣\ln|u| as the antiderivative.

Question 19

Find the particular solution to the differential equation dydx=tan⁡3(x)\frac{dy}{dx} = \tan^3(x) given the initial condition y(0)=1y(0) = 1.

  1. y=sec⁡2(x)−ln⁡∣sec⁡(x)∣y = \sec^2(x) - \ln|\sec(x)|
  2. y=12tan⁡2(x)−ln⁡∣sec⁡(x)∣+1y = \frac{1}{2}\tan^2(x) - \ln|\sec(x)| + 1
  3. y=14tan⁡4(x)+1y = \frac{1}{4}\tan^4(x) + 1
  4. y=12tan⁡2(x)+ln⁡∣cos⁡(x)∣+1y = \frac{1}{2}\tan^2(x) + \ln|\cos(x)| + 1 (correct answer)
Explanation: When you encounter a differential equation like dydx=tan⁡3(x)\frac{dy}{dx} = \tan^3(x), you need to integrate the right side and then apply the initial condition to find the particular solution. To integrate tan⁡3(x)\tan^3(x), use the identity tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1 to rewrite it as: tan⁡3(x)=tan⁡(x)⋅tan⁡2(x)=tan⁡(x)(sec⁡2(x)−1)=tan⁡(x)sec⁡2(x)−tan⁡(x)\tan^3(x) = \tan(x) \cdot \tan^2(x) = \tan(x)(\sec^2(x) - 1) = \tan(x)\sec^2(x) - \tan(x) Now integrate term by term:
  • ∫tan⁡(x)sec⁡2(x)dx=12tan⁡2(x)\int \tan(x)\sec^2(x) dx = \frac{1}{2}\tan^2(x) (using substitution u=tan⁡(x)u = \tan(x), du=sec⁡2(x)dxdu = \sec^2(x)dx)
  • ∫tan⁡(x)dx=ln⁡∣sec⁡(x)∣=−ln⁡∣cos⁡(x)∣\int \tan(x) dx = \ln|\sec(x)| = -\ln|\cos(x)|
So y=12tan⁡2(x)−(−ln⁡∣cos⁡(x)∣)+C=12tan⁡2(x)+ln⁡∣cos⁡(x)∣+Cy = \frac{1}{2}\tan^2(x) - (-\ln|\cos(x)|) + C = \frac{1}{2}\tan^2(x) + \ln|\cos(x)| + C Apply the initial condition y(0)=1y(0) = 1: Since tan⁡(0)=0\tan(0) = 0 and cos⁡(0)=1\cos(0) = 1, we get 1=0+ln⁡(1)+C=C1 = 0 + \ln(1) + C = C. Therefore, the particular solution is y=12tan⁡2(x)+ln⁡∣cos⁡(x)∣+1y = \frac{1}{2}\tan^2(x) + \ln|\cos(x)| + 1, which is choice D. Choice A uses sec⁡2(x)\sec^2(x) instead of 12tan⁡2(x)\frac{1}{2}\tan^2(x) and has the wrong sign on the logarithm. Choice B has the correct 12tan⁡2(x)\frac{1}{2}\tan^2(x) term but uses −ln⁡∣sec⁡(x)∣-\ln|\sec(x)| instead of +ln⁡∣cos⁡(x)∣+\ln|\cos(x)|. Choice C completely misses the tan⁡(x)\tan(x) integration, showing only the tan⁡(x)sec⁡2(x)\tan(x)\sec^2(x) result. Remember: when integrating powers of tangent, break them down using tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1 to create manageable integrals.

Question 20

Evaluate ∫sin⁡3(x)cos⁡6(x) dx\int \frac{\sin^3(x)}{\cos^6(x)} \, dx.

  1. 15sec⁡5(x)−13sec⁡3(x)+C\frac{1}{5}\sec^5(x) - \frac{1}{3}\sec^3(x) + C (correct answer)
  2. 13sec⁡3(x)−15sec⁡5(x)+C\frac{1}{3}\sec^3(x) - \frac{1}{5}\sec^5(x) + C
  3. 14tan⁡4(x)+16tan⁡6(x)+C\frac{1}{4}\tan^4(x) + \frac{1}{6}\tan^6(x) + C
  4. 14sec⁡4(x)−12sec⁡2(x)+C\frac{1}{4}\sec^4(x) - \frac{1}{2}\sec^2(x) + C
Explanation: First, rewrite the integrand in terms of tangent and secant: ∫sin⁡3(x)cos⁡3(x)⋅1cos⁡3(x) dx=∫tan⁡3(x)sec⁡3(x) dx\int \frac{\sin^3(x)}{\cos^3(x)} \cdot \frac{1}{\cos^3(x)} \, dx = \int \tan^3(x) \sec^3(x) \, dx. Since the power of tangent is odd, save a sec⁡(x)tan⁡(x)\sec(x)\tan(x) factor and let u=sec⁡(x)u = \sec(x). The integral becomes ∫tan⁡2(x)sec⁡2(x)(sec⁡(x)tan⁡(x)) dx\int \tan^2(x)\sec^2(x)(\sec(x)\tan(x))\,dx. Substitute tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1 to get ∫(sec⁡2(x)−1)sec⁡2(x)(sec⁡(x)tan⁡(x)) dx\int (\sec^2(x)-1)\sec^2(x)(\sec(x)\tan(x))\,dx. With u=sec⁡(x)u = \sec(x) and du=sec⁡(x)tan⁡(x) dxdu = \sec(x)\tan(x)\,dx, this becomes ∫(u2−1)u2 du=∫(u4−u2) du=u55−u33+C\int (u^2-1)u^2\,du = \int (u^4 - u^2)\,du = \frac{u^5}{5} - \frac{u^3}{3} + C. Substituting back gives 15sec⁡5(x)−13sec⁡3(x)+C\frac{1}{5}\sec^5(x) - \frac{1}{3}\sec^3(x) + C.