Integrating tan & sec Products - Calculus 2

1. Evaluate the integral $\int \frac{dx}{\cos^2(x)\tan(x)}$.
   Answer: $\ln|\tan(x)| + C$ 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: $\frac{1}{\cos^2(x)\tan(x)} = \frac{1}{\cos^2(x)} \cdot \frac{1}{\tan(x)} = \sec^2(x) \cdot \cot(x)$ Since $\cot(x) = \frac{\cos(x)}{\sin(x)}$, you can write this as: $\sec^2(x) \cdot \cot(x) = \frac{1}{\cos^2(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{1}{\cos(x)\sin(x)}$ Now use the substitution $u = \tan(x)$, so $du = \sec^2(x)dx$. Notice that $\frac{1}{\cos(x)\sin(x)} = \frac{\sec^2(x)}{\tan(x)}$, which becomes $\frac{1}{u}du$. This integrates to $\ln|u| + C = \ln|\tan(x)| + C$, confirming answer D. Let's examine why the other options are incorrect: A would result from incorrectly thinking the derivative of $\cot(x)$ is positive rather than $-\csc^2(x)$. B comes from misapplying trigonometric identities and confusing this with $\int \cot(x)dx$. C represents the antiderivative of $\sec(x)\tan(x)$, not our integrand. Strategy tip: When facing complex trigonometric integrals, always simplify using identities first, then look for substitutions involving $u = \tan(x)$, $u = \sin(x)$, or $u = \cos(x)$. The form $\frac{1}{u}$ should immediately signal $\ln|u|$ as the antiderivative.
2. The integral $\int \tan^2(x) \sec^3(x) \, dx$ can be rewritten using the identity $\tan^2(x) = \sec^2(x) - 1$. What integral results from this substitution?
   Answer: $\int (\sec^5(x) - \sec^3(x)) \, dx$ 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 $\tan^2(x) = \sec^2(x) - 1$ correctly. Starting with $\int \tan^2(x) \sec^3(x) \, dx$, substitute the identity: $\int (\sec^2(x) - 1) \sec^3(x) \, dx$ Using the distributive property: $\int \sec^2(x) \cdot \sec^3(x) \, dx - \int 1 \cdot \sec^3(x) \, dx$ This simplifies to: $\int \sec^5(x) \, dx - \int \sec^3(x) \, dx = \int (\sec^5(x) - \sec^3(x)) \, dx$ 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.
3. To evaluate an integral of the form $\int \tan^m(x) \sec^n(x) \, dx$, the substitution $u = \sec(x)$ is most effective under which general conditions for the integers $m$ and $n$?
   Answer: $m$ is a positive odd integer and $n \ge 1$. The substitution $u = \sec(x)$ requires the differential $du = \sec(x)\tan(x)\,dx$. To use this, we must be able to factor out one $\sec(x)$ term and one $\tan(x)$ term from the integrand. This is possible if the power of tangent, $m$, is at least 1 and odd, and the power of secant, $n$, is at least 1. If $m$ is odd, we can factor out $\tan(x)$, and the remaining even power $\tan^{m-1}(x)$ can be converted to secants using $\tan^2(x) = \sec^2(x) - 1$.
4. Which of the following is an antiderivative of $f(x) = \tan^2(x) \sec(x)$?
   Answer: $\frac{1}{2}\sec(x)\tan(x) - \frac{1}{2}\ln|\sec(x) + \tan(x)| + C$ First, use the identity $\tan^2(x) = \sec^2(x) - 1$. The integral becomes $\int (\sec^2(x) - 1)\sec(x)\,dx = \int (\sec^3(x) - \sec(x))\,dx$. This separates into $\int \sec^3(x)\,dx - \int \sec(x)\,dx$. Using the known formulas for these integrals, we get $(\frac{1}{2}\sec(x)\tan(x) + \frac{1}{2}\ln|\sec(x) + \tan(x)|) - (\ln|\sec(x) + \tan(x)|) + C$. Combining the logarithmic terms gives the final answer: $\frac{1}{2}\sec(x)\tan(x) - \frac{1}{2}\ln|\sec(x) + \tan(x)| + C$.
5. The evaluation of $\int \tan^m(x) \sec^n(x) \, dx$ where $m$ is a non-negative even integer and $n$ is a positive odd integer typically requires which sequence of steps?
   Answer: Using identities to express the integrand in terms of $\sec(x)$ alone, followed by integration by parts. 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 $\tan^2(x) = \sec^2(x) - 1$ to convert all even powers of tangent into powers of secant. This results in a sum of integrals of the form $\int \sec^k(x)\,dx$ where $k$ is odd. These integrals are then solved using integration by parts, often via a reduction formula.
6. The area of the region bounded by the curve $y = \sec^2(x)\tan(x)$, the x-axis, and the lines $x=0$ and $x=\pi/3$ is given by which value?
   Answer: $\frac{3}{2}$ The area is given by the definite integral $\int_0^{\pi/3} \sec^2(x)\tan(x) \, dx$. This integral can be solved with the substitution $u = \tan(x)$, which gives $du = \sec^2(x)\,dx$. The bounds of integration change from $x=0$ to $u = \tan(0) = 0$ and from $x=\pi/3$ to $u = \tan(\pi/3) = \sqrt{3}$. The integral becomes $\int_0^{\sqrt{3}} u \, du$. Evaluating this gives $[\frac{u^2}{2}]_0^{\sqrt{3}} = \frac{(\sqrt{3})^2}{2} - \frac{0^2}{2} = \frac{3}{2}$.
7. Which of the following integrals is equivalent to $\int \tan^3(x) \, dx$?
   Answer: $\int (\sec^2(x) - 1)\tan(x) \, dx$ 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 $\tan^3(x)$, you need to break it down using the fundamental identity $\tan^2(x) = \sec^2(x) - 1$. Starting with $\int \tan^3(x) \, dx$, rewrite this as $\int \tan(x) \cdot \tan^2(x) \, dx$. Now substitute the identity: $\int \tan(x) \cdot (\sec^2(x) - 1) \, dx = \int (\sec^2(x) - 1)\tan(x) \, dx$. This matches option C exactly. Looking at the wrong answers: Option A suggests $\frac{1}{4}\tan^4(x) + C$, which would be correct if we were integrating $\tan^3(x) \sec^2(x)$ using substitution, but that's not our original integral. Option B has $(\sec^2(x) + 1)\tan(x)$, which incorrectly adds 1 instead of subtracting it from $\sec^2(x)$ in the identity. Option D includes $-x + C$ at the end, suggesting someone partially integrated $(\sec^2(x) - 1)\tan(x)$ and remembered that $\int (-1) \, dx = -x$, 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 $\tan^3(x)$ into an integrable form. Remember: when dealing with odd powers of tangent, always factor out one $\tan(x)$ and convert the remaining even power using $\tan^2(x) = \sec^2(x) - 1$.
8. Which of the following represents the correct approach to evaluate $\int \tan^5 x \sec^3 x \, dx$?
   Answer: Use substitution $u = \sec x$ and rewrite $\tan^5 x = (\sec^2 x - 1)^2 \cdot \tan x$ For $\int \tan^5 x \sec^3 x \, dx$, we have an odd power of tangent and an odd power of secant. The standard approach is to use $u = \sec x$, $du = \sec x \tan x \, dx$. We need to factor out $\sec x \tan x$ and express the remaining $\tan^4 x \sec^2 x$ in terms of secant: $\tan^4 x = (\sec^2 x - 1)^2$. Choice A won't work because we can't easily express $\sec^3 x$ 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.
9. Evaluate the integral $\int \tan^4(x) \, dx$.
   Answer: $\frac{1}{3}\tan^3(x) - \tan(x) + x + C$ Rewrite $\tan^4(x)$ as $\tan^2(x) \tan^2(x)$ and substitute $\tan^2(x) = \sec^2(x) - 1$ for one of the factors. This gives $\int \tan^2(x)(\sec^2(x) - 1) \, dx = \int (\tan^2(x)\sec^2(x) - \tan^2(x)) \, dx$. The integral separates into $\int \tan^2(x)\sec^2(x)\,dx - \int \tan^2(x)\,dx$. For the first part, let $u = \tan(x)$ to get $\int u^2\,du = \frac{u^3}{3} = \frac{1}{3}\tan^3(x)$. For the second part, $\int \tan^2(x)\,dx = \int (\sec^2(x) - 1)\,dx = \tan(x) - x$. Combining these results gives $\frac{1}{3}\tan^3(x) - (\tan(x) - x) + C = \frac{1}{3}\tan^3(x) - \tan(x) + x + C$.
10. Find the particular solution to the differential equation $\frac{dy}{dx} = \tan^3(x)$ given the initial condition $y(0) = 1$.
    Answer: $y = \frac{1}{2}\tan^2(x) + \ln|\cos(x)| + 1$ When you encounter a differential equation like $\frac{dy}{dx} = \tan^3(x)$, you need to integrate the right side and then apply the initial condition to find the particular solution. To integrate $\tan^3(x)$, use the identity $\tan^2(x) = \sec^2(x) - 1$ to rewrite it as: $\tan^3(x) = \tan(x) \cdot \tan^2(x) = \tan(x)(\sec^2(x) - 1) = \tan(x)\sec^2(x) - \tan(x)$ Now integrate term by term: - $\int \tan(x)\sec^2(x) dx = \frac{1}{2}\tan^2(x)$ (using substitution $u = \tan(x)$, $du = \sec^2(x)dx$) - $\int \tan(x) dx = \ln|\sec(x)| = -\ln|\cos(x)|$ So $y = \frac{1}{2}\tan^2(x) - (-\ln|\cos(x)|) + C = \frac{1}{2}\tan^2(x) + \ln|\cos(x)| + C$ Apply the initial condition $y(0) = 1$: Since $\tan(0) = 0$ and $\cos(0) = 1$, we get $1 = 0 + \ln(1) + C = C$. Therefore, the particular solution is $y = \frac{1}{2}\tan^2(x) + \ln|\cos(x)| + 1$, which is choice D. Choice A uses $\sec^2(x)$ instead of $\frac{1}{2}\tan^2(x)$ and has the wrong sign on the logarithm. Choice B has the correct $\frac{1}{2}\tan^2(x)$ term but uses $-\ln|\sec(x)|$ instead of $+\ln|\cos(x)|$. Choice C completely misses the $\tan(x)$ integration, showing only the $\tan(x)\sec^2(x)$ result. Remember: when integrating powers of tangent, break them down using $\tan^2(x) = \sec^2(x) - 1$ to create manageable integrals.
11. Evaluate $\int \tan^3 x \sec^4 x \, dx$ using an appropriate substitution technique.
    Answer: $\frac{1}{4}\tan^4 x + \frac{1}{6}\tan^6 x + C$ Since we have an even power of secant, we can use the substitution $u = \tan x$, so $du = \sec^2 x \, dx$. We rewrite $\sec^4 x = \sec^2 x \cdot \sec^2 x = \sec^2 x(1 + \tan^2 x) = \sec^2 x(1 + u^2)$. The integral becomes $\int u^3(1 + u^2) \, du = \int (u^3 + u^5) \, du = \frac{u^4}{4} + \frac{u^6}{6} + C = \frac{1}{4}\tan^4 x + \frac{1}{6}\tan^6 x + C$. Choice B uses incorrect powers (missing the squared term expansion). Choice C incorrectly substitutes for secant instead of tangent. Choice D uses wrong exponents from integration errors.
12. Consider the integral $\int \sec^5(x) \, dx$. Using the reduction formula $I_n = \frac{1}{n-1}\sec^{n-2}(x)\tan(x) + \frac{n-2}{n-1}I_{n-2}$, where $I_n = \int \sec^n(x)\,dx$, what is the integral $I_5$?
    Answer: $\frac{1}{4}\sec^3(x)\tan(x) + \frac{3}{8}\sec(x)\tan(x) + \frac{3}{8}\ln|\sec(x)+\tan(x)| + C$ 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 $I_5 = \int \sec^5(x) \, dx$, substitute $n = 5$: $I_5 = \frac{1}{5-1}\sec^{5-2}(x)\tan(x) + \frac{5-2}{5-1}I_{5-2} = \frac{1}{4}\sec^3(x)\tan(x) + \frac{3}{4}I_3$ Now you need $I_3 = \int \sec^3(x) \, dx$. Apply the formula again with $n = 3$: $I_3 = \frac{1}{3-1}\sec^{3-2}(x)\tan(x) + \frac{3-2}{3-1}I_{3-2} = \frac{1}{2}\sec(x)\tan(x) + \frac{1}{2}I_1$ The base case is $I_1 = \int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C$. Substituting back: $I_3 = \frac{1}{2}\sec(x)\tan(x) + \frac{1}{2}\ln|\sec(x) + \tan(x)| + C$ Therefore: $I_5 = \frac{1}{4}\sec^3(x)\tan(x) + \frac{3}{4} \cdot \left[\frac{1}{2}\sec(x)\tan(x) + \frac{1}{2}\ln|\sec(x) + \tan(x)|\right] + C$ $= \frac{1}{4}\sec^3(x)\tan(x) + \frac{3}{8}\sec(x)\tan(x) + \frac{3}{8}\ln|\sec(x) + \tan(x)| + C$ This matches answer C. Answer A stops at the first reduction step without continuing to $I_3$. 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.
13. Evaluate the definite integral $\int_0^{\pi/8} \sec^4(2x) \tan(2x) \, dx$.
    Answer: $\frac{3}{8}$ Let $w = 2x$, so $dw = 2dx$ or $dx = \frac{1}{2}dw$. The integral becomes $\frac{1}{2}\int_{0}^{\pi/4} \sec^4(w) \tan(w) \, dw$. This fits both standard cases. Let's use $u=\sec(w)$, so $du = \sec(w)\tan(w)\,dw$. The integral is $\frac{1}{2}\int \sec^3(w) (\sec(w)\tan(w))\,dw$. In terms of $u$, this is $\frac{1}{2} \int u^3\,du = \frac{1}{2} \frac{u^4}{4} = \frac{1}{8}u^4 = \frac{1}{8}\sec^4(w)$. Evaluating from $w=0$ to $w=\pi/4$: $[\frac{1}{8}\sec^4(w)]_0^{\pi/4} = \frac{1}{8}(\sec^4(\pi/4) - \sec^4(0)) = \frac{1}{8}((\sqrt{2})^4 - 1^4) = \frac{1}{8}(4 - 1) = \frac{3}{8}$.
14. To evaluate $\int \frac{\sec^2(x)}{\sqrt{\tan(x)}} \, dx$, a u-substitution is performed. What is the resulting integral in terms of $u$?
    Answer: $\int u^{-1/2} \, du$ The integrand can be written as $\int (\tan(x))^{-1/2} \sec^2(x) \, dx$. The presence of $\tan(x)$ and its derivative $\sec^2(x)$ suggests the substitution $u = \tan(x)$. The differential is $du = \sec^2(x)\,dx$. Substituting $u$ and $du$ into the integral directly yields $\int u^{-1/2} \, du$.
15. Evaluate the integral $\int \frac{\tan(x)}{\sec(x) - \tan(x)} \, dx$.
    Answer: $\sec(x) + \tan(x) - x + C$ When you encounter integrals involving trigonometric functions like tangent and secant, look for opportunities to use substitution or recognize patterns that simplify through trigonometric identities. To solve $\int \frac{\tan(x)}{\sec(x) - \tan(x)} \, dx$, multiply both numerator and denominator by $(\sec(x) + \tan(x))$: $\frac{\tan(x)}{\sec(x) - \tan(x)} \cdot \frac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)} = \frac{\tan(x)(\sec(x) + \tan(x))}{\sec^2(x) - \tan^2(x)}$ Since $\sec^2(x) - \tan^2(x) = 1$, this becomes: $\tan(x)(\sec(x) + \tan(x)) = \tan(x)\sec(x) + \tan^2(x)$ Now integrate each term: - $\int \tan(x)\sec(x) \, dx = \sec(x) + C_1$ - $\int \tan^2(x) \, dx = \int (\sec^2(x) - 1) \, dx = \tan(x) - x + C_2$ Therefore: $\sec(x) + \tan(x) - x + C$, which is choice C. Choice A incorrectly combines logarithmic and polynomial forms that don't match our integration result. Choice B has the wrong sign on the $x$ term and suggests $\sec(x) - \tan(x)$ instead of $\sec(x) + \tan(x)$. Choice D involves $\ln|\cos(x)|$ and $-\sec(x)$, which would arise from completely different integral setups. Strategy tip: When dealing with rational trigonometric expressions, try multiplying by the conjugate of the denominator. This often eliminates complex denominators using the identity $\sec^2(x) - \tan^2(x) = 1$, making integration much more manageable.
16. Evaluate the indefinite integral $\int \frac{\sec^4(x)}{\tan^2(x)} \, dx$.
    Answer: $-\cot(x) + \tan(x) + C$ 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 $\sec^2(x) = 1 + \tan^2(x)$: $\int \frac{\sec^4(x)}{\tan^2(x)} \, dx = \int \frac{\sec^2(x) \cdot \sec^2(x)}{\tan^2(x)} \, dx = \int \frac{(1 + \tan^2(x)) \sec^2(x)}{\tan^2(x)} \, dx$ Split this into two separate integrals: $\int \frac{\sec^2(x)}{\tan^2(x)} \, dx + \int \frac{\tan^2(x) \sec^2(x)}{\tan^2(x)} \, dx = \int \frac{\sec^2(x)}{\tan^2(x)} \, dx + \int \sec^2(x) \, dx$ The second integral is straightforward: $\int \sec^2(x) \, dx = \tan(x)$ For the first integral, use the identity $\frac{\sec^2(x)}{\tan^2(x)} = \frac{1}{\sin^2(x)} = \csc^2(x)$ Therefore: $\int \csc^2(x) \, dx = -\cot(x)$ Combining both parts: $-\cot(x) + \tan(x) + C$, 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.
17. The integral $\int x \sec(x) \tan(x) \, dx$ is best evaluated using which technique?
    Answer: Integration by parts with $u = x$ and $dv = \sec(x)\tan(x)\,dx$. The integral contains a product of an algebraic function ($x$) and a trigonometric function ($\sec(x)\tan(x)$). This structure suggests integration by parts. The choice of $u$ and $dv$ is guided by the LIATE rule (Log, Inverse, Algebraic, Trig, Exponential). We choose the algebraic term $u=x$, as its derivative simplifies to $du=dx$. The remaining part, $dv = \sec(x)\tan(x)\,dx$, is a standard derivative, and its integral is $v=\sec(x)$. This choice leads to a simpler integral to solve. The alternative choice in B would lead to a more complex integral.
18. Evaluate $\int \frac{\sin^3(x)}{\cos^6(x)} \, dx$.
    Answer: $\frac{1}{5}\sec^5(x) - \frac{1}{3}\sec^3(x) + C$ First, rewrite the integrand in terms of tangent and secant: $\int \frac{\sin^3(x)}{\cos^3(x)} \cdot \frac{1}{\cos^3(x)} \, dx = \int \tan^3(x) \sec^3(x) \, dx$. Since the power of tangent is odd, save a $\sec(x)\tan(x)$ factor and let $u = \sec(x)$. The integral becomes $\int \tan^2(x)\sec^2(x)(\sec(x)\tan(x))\,dx$. Substitute $\tan^2(x) = \sec^2(x) - 1$ to get $\int (\sec^2(x)-1)\sec^2(x)(\sec(x)\tan(x))\,dx$. With $u = \sec(x)$ and $du = \sec(x)\tan(x)\,dx$, this becomes $\int (u^2-1)u^2\,du = \int (u^4 - u^2)\,du = \frac{u^5}{5} - \frac{u^3}{3} + C$. Substituting back gives $\frac{1}{5}\sec^5(x) - \frac{1}{3}\sec^3(x) + C$.
19. If $I_n = \int \sec^n(x) \, dx$ for $n \ge 2$, which of the following gives the reduction formula for $I_n$?
    Answer: $I_n = \frac{1}{n-1}\sec^{n-2}(x)\tan(x) + \frac{n-2}{n-1}I_{n-2}$ The reduction formula is derived using integration by parts on $\int \sec^{n-2}(x) \sec^2(x) \, dx$. Let $u = \sec^{n-2}(x)$ and $dv = \sec^2(x)\,dx$. Then $du = (n-2)\sec^{n-2}(x)\tan(x)\,dx$ and $v = \tan(x)$. Applying the formula $uv - \int v\,du$ gives $I_n = \sec^{n-2}(x)\tan(x) - (n-2)\int \sec^{n-2}(x)\tan^2(x)\,dx$. Using $\tan^2(x) = \sec^2(x)-1$, we get $I_n = \sec^{n-2}(x)\tan(x) - (n-2)\int (\sec^n(x) - \sec^{n-2}(x))\,dx$. This becomes $I_n = \sec^{n-2}(x)\tan(x) - (n-2)I_n + (n-2)I_{n-2}$. Solving for $I_n$ yields $(n-1)I_n = \sec^{n-2}(x)\tan(x) + (n-2)I_{n-2}$, which gives the correct formula.
20. Evaluate the definite integral $$\int_0^{\pi/4} \tan^3(x) \sec^4(x) \, dx$$
    Answer: $\frac{5}{12}$ The power of secant is even, so we save a $\sec^2(x)$ factor and use the identity $\sec^2(x) = 1 + \tan^2(x)$. The integral becomes $\int_0^{\pi/4} \tan^3(x)(1+\tan^2(x))\sec^2(x)\,dx$. Let $u = \tan(x)$, so $du = \sec^2(x)\,dx$. The bounds change: when $x=0, u=\tan(0)=0$; when $x=\pi/4, u=\tan(\pi/4)=1$. The integral in $u$ is $\int_0^1 u^3(1+u^2)\,du = \int_0^1 (u^3+u^5)\,du$. Evaluating this gives $[\frac{u^4}{4} + \frac{u^6}{6}]_0^1 = (\frac{1}{4} + \frac{1}{6}) - 0 = \frac{3+2}{12} = \frac{5}{12}$.

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