Algebraic & Trig Simplification - Calculus 2

1. Before applying partial fraction decomposition, the integrand of $\int \frac{x^3 + 2x}{x^2 - 1} dx$ 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?
   Answer: $x + \frac{3x}{x^2 - 1}$ Performing polynomial long division of $(x^3 + 2x)$ by $(x^2 - 1)$ gives a quotient of $x$ and a remainder of $3x$. Specifically, $x(x^2 - 1) = x^3 - x$, and $(x^3 + 2x) - (x^3 - x) = 3x$. Thus, the expression is rewritten as $x + \frac{3x}{x^2 - 1}$. The other options result from common errors in the long division process, such as incorrect subtraction of terms or miscalculation of the remainder.
2. For the rational function $\frac{2x^3 - x^2 + 3x - 1}{x^4 - x^2}$, what is the correct partial fraction decomposition setup?
   Answer: $\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{x+1}$ 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: $x^4 - x^2 = x^2(x^2 - 1) = x^2(x-1)(x+1)$. Now you can see the complete factorization involves three distinct factors: $x^2$, $(x-1)$, and $(x+1)$. For partial fractions, each factor gets its own term. The repeated factor $x^2$ requires two terms: $\frac{A}{x}$ for the linear factor $x$ and $\frac{B}{x^2}$ for the quadratic factor $x^2$. The distinct linear factors $(x-1)$ and $(x+1)$ each contribute one term with constant numerators: $\frac{C}{x-1}$ and $\frac{D}{x+1}$. This gives us answer choice D. Choice A incorrectly treats $(x^2-1)$ as irreducible, missing that it factors further. Choice B uses $\frac{Bx+C}{x^2}$ 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 $x^2-1$ 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.
3. Consider the integral $\int \frac{x^3 - 8}{x^2 - 4} dx$. Which of the following represents the simplified integrand, valid for $x \neq \pm 2$?
   Answer: $\frac{x^2 + 2x + 4}{x+2}$ The numerator is a difference of cubes, $x^3 - 2^3 = (x-2)(x^2+2x+4)$. The denominator is a difference of squares, $x^2 - 2^2 = (x-2)(x+2)$. The integrand simplifies by canceling the $(x-2)$ term: $\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} = \frac{x^2+2x+4}{x+2}$. Choice B uses an incorrect sign in the factorization of the difference of cubes. Choices C and D result from incorrect factorization or cancellation.
4. To evaluate $\int \sqrt{1+\cos(4\theta)} d\theta$, which simplification of the integrand is the most effective first step, assuming $\cos(2\theta) \ge 0$?
   Answer: $\sqrt{2}\cos(2\theta)$ When you encounter an integral with $\sqrt{1+\cos(4\theta)}$, you should immediately think about trigonometric identities that can simplify expressions under square roots. The key insight is recognizing that this has the form of a double-angle cosine identity. Using the double-angle identity $\cos(2A) = 2\cos^2(A) - 1$, we can rearrange to get $1 + \cos(2A) = 2\cos^2(A)$. Applying this with $A = 2\theta$: $1 + \cos(4\theta) = 2\cos^2(2\theta)$ Therefore: $\sqrt{1+\cos(4\theta)} = \sqrt{2\cos^2(2\theta)} = \sqrt{2}|\cos(2\theta)|$ Since we're given that $\cos(2\theta) \geq 0$, we have $|\cos(2\theta)| = \cos(2\theta)$, so our expression simplifies to $\sqrt{2}\cos(2\theta)$, which is answer C. Let's examine why the other options are incorrect: A) $1+\cos(2\theta)$ is missing the factor of 2 and doesn't account for the square root properly. B) $\sqrt{2}\sin(2\theta)$ would arise if we had started with $1-\cos(4\theta)$ instead, since $1-\cos(2A) = 2\sin^2(A)$. D) This is clearly wrong since we just demonstrated a successful simplification using the double-angle identity. Study tip: Memorize that $1 \pm \cos(2A) = 2\cos^2(A)$ and $1 - \cos(2A) = 2\sin^2(A)$. These identities are crucial for simplifying square root expressions in trigonometric integrals, especially when combined with the condition about sign.
5. To evaluate $\int \frac{1}{1 - \cos x} dx$, 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?
   Answer: $\frac{1+\cos x}{\sin^2 x}$ Multiplying the numerator and denominator by the conjugate, $1+\cos x$, gives $\frac{1+\cos x}{(1-\cos x)(1+\cos x)} = \frac{1+\cos x}{1-\cos^2 x}$. Applying the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we know $1-\cos^2 x = \sin^2 x$. The resulting integrand is $\frac{1+\cos x}{\sin^2 x}$. Choice D shows the correct multiplication but omits the identity application. Choices B and C result from incorrect conjugate multiplication or identity usage.
6. The integral $\int \tan^3(x) \sec^4(x) dx$ is being prepared for a u-substitution. Which rewrite of the integrand is most effective for a substitution with $u=\tan(x)$?
   Answer: $\tan^3(x) (1+\tan^2(x)) \sec^2(x)$ For a substitution $u = \tan x$, we need to reserve a factor of $\sec^2 x$ for $du$. We rewrite $\sec^4 x$ as $\sec^2 x \cdot \sec^2 x$. Then, using the identity $\sec^2 x = 1 + \tan^2 x$, we get $\tan^3(x) (1+\tan^2(x)) \sec^2(x)$. This form is ideal for the substitution. Choice B sets up for $u = \sec x$, which is also possible but often more complex. Choice C is an incomplete rewrite, losing a factor of $\sec^2 x$. Choice D, converting to sine and cosine, is much less efficient.
7. For the integral $\int \frac{dx}{x(1+(\ln x)^2)}$, a substitution is needed. Which substitution and resulting differential correctly transform this integral into a standard arctangent form?
   Answer: $u = \ln x$, so $du = \frac{1}{x} dx$. When you encounter an integral with a complex expression involving natural logarithms, look for substitutions that will simplify the integrand and potentially create a recognizable form like arctangent. The key insight here is recognizing that this integral can be transformed into the standard form $\int \frac{du}{1+u^2} = \arctan(u) + C$ with the right substitution. Let's examine what happens with $u = \ln x$, so $du = \frac{1}{x} dx$. Substituting into the original integral: $\int \frac{dx}{x(1+(\ln x)^2)} = \int \frac{1}{1+(\ln x)^2} \cdot \frac{dx}{x} = \int \frac{du}{1+u^2}$. This is exactly the arctangent form we want, making choice B correct. Let's see why the other options fail. Choice A uses $u = 1+(\ln x)^2$ with $du = \frac{2\ln x}{x} dx$. This substitution doesn't account for the $\frac{1}{x}$ factor properly and won't yield a standard form. Choice D sets $u = (\ln x)^2$ with $du = \frac{2\ln x}{x} dx$. Again, this creates a mismatch because we have $\frac{1}{x} dx$ in our integral, not $\frac{2\ln x}{x} dx$. Choice C, $u = x$, doesn't simplify the logarithmic expression at all. Study tip: When you see integrands containing $\ln x$ terms with $\frac{1}{x}$ factors, try $u = \ln x$ first. The derivative of $\ln x$ is $\frac{1}{x}$, which often creates clean substitutions that reveal standard integral forms.
8. Which algebraic manipulation correctly transforms the integrand of $\int \frac{dx}{x^2 - 6x + 13}$ into a form suitable for an arctangent integration formula?
   Answer: Completing the square to get $\frac{1}{(x-3)^2 + 4}$. To get the integral into the form $\int \frac{du}{u^2+a^2}$, we complete the square on the denominator. $x^2 - 6x + 13 = (x^2 - 6x + 9) - 9 + 13 = (x-3)^2 + 4$. This is the correct transformation. Choice B is mathematically correct but not useful for real-valued integration. Choice C forgets to subtract the 9 that was added to complete the square. Choice D misapplies the procedure for completing the square.
9. To simplify the integral $\int \frac{dx}{\sin(x)\cos(x)}$, which manipulation based on a trigonometric identity is most effective?
   Answer: Multiply the numerator and denominator by 2 to get $\int \frac{2}{2\sin(x)\cos(x)} dx = \int \frac{2}{\sin(2x)} dx$. The most effective simplification is to recognize that the denominator $\sin(x)\cos(x)$ is half of $\sin(2x)$. By multiplying the numerator and denominator by 2, the integrand becomes $\frac{2}{\sin(2x)} = 2\csc(2x)$, which is a standard integral form. The other options either complicate the integral unnecessarily or are based on invalid mathematical operations.
10. To integrate $\int \frac{3x^2 + x + 4}{x^3 + 4x} dx$, the first step is to factor the denominator and set up the partial fraction decomposition. What is the correct setup?
    Answer: $\frac{A}{x} + \frac{Bx+C}{x^2+4}$ When you encounter a rational function like this, partial fraction decomposition requires first factoring the denominator completely. Let's factor $x^3 + 4x = x(x^2 + 4)$. Notice that $x^2 + 4$ cannot be factored further over the real numbers since it has no real roots. The correct setup in option C, $\frac{A}{x} + \frac{Bx+C}{x^2+4}$, follows the fundamental rule: for each linear factor in the denominator, use a constant in the numerator; for each irreducible quadratic factor, use a linear expression in the numerator. Since we have the linear factor $x$ and the irreducible quadratic $x^2 + 4$, we need $\frac{A}{x}$ for the linear part and $\frac{Bx+C}{x^2+4}$ for the quadratic part. Option A incorrectly factors $x^2 + 4$ as $(x+2)(x-2)$, which would only work if the denominator were $x^3 - 4x$. Option B uses only a constant $B$ over the quadratic factor, which is insufficient for integrating terms that will involve $x^2 + 4$. You need the linear numerator $Bx + C$ to handle both the arctangent and logarithmic parts that arise. Option D misrepresents the factorization entirely, treating it as $x^2(x+4)$ instead of $x(x^2+4)$. Remember: irreducible quadratics in partial fractions always require linear numerators. This gives you enough flexibility to match coefficients and integrate successfully using both substitution and arctangent formulas.
11. For the integral $\int \sin^4 x \cos^3 x dx$, which power reduction strategy leads to the most efficient solution?
    Answer: Use $\cos^3 x = \cos x(1 - \sin^2 x)$ and substitute $u = \sin x$ Since the cosine power is odd, factor out one cosine: $\cos^3 x = \cos x \cos^2 x = \cos x(1 - \sin^2 x)$. The integral becomes $\int \sin^4 x \cos x(1 - \sin^2 x) dx = \int \sin^4 x \cos x dx - \int \sin^6 x \cos x dx$. With substitution $u = \sin x$, $du = \cos x dx$, this becomes $\int u^4 du - \int u^6 du$, which is straightforward. Choices A and D involve unnecessarily complex power reduction formulas. Choice B doesn't optimize the factorization.
12. To integrate $\int \frac{\sec x}{\tan x - 1} dx$, it is useful to rewrite the integrand entirely in terms of $\sin x$ and $\cos x$. What is the resulting expression?
    Answer: $\frac{1}{\sin x - \cos x}$ 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 $\frac{\sec x}{\tan x - 1}$ in terms of sine and cosine, recall that $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$. Substituting these definitions: $\frac{\sec x}{\tan x - 1} = \frac{\frac{1}{\cos x}}{\frac{\sin x}{\cos x} - 1}$ To simplify the complex fraction, multiply both numerator and denominator by $\cos x$: $\frac{\frac{1}{\cos x} \cdot \cos x}{(\frac{\sin x}{\cos x} - 1) \cdot \cos x} = \frac{1}{\sin x - \cos x}$ This gives us answer choice B. Let's examine why the other options are incorrect. Choice A, $\frac{\cos x}{\sin x - \cos x}$, has an extra $\cos x$ in the numerator that doesn't belong—this would result from incorrectly leaving $\cos x$ in the numerator after clearing the complex fraction. Choice C, $\frac{1}{\sin x - 1}$, appears if you mistakenly write $\tan x - 1 = \frac{\sin x}{\cos x} - 1 = \frac{\sin x - 1}{\cos x}$, forgetting that subtracting 1 requires a common denominator. Choice D, $\frac{\cos x}{\sin x - 1}$, 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.
13. To evaluate the integral $\int \frac{\sin^3 x}{\cos^2 x} dx$, which of the following is the most effective first step in rewriting the integrand to prepare for a u-substitution?
    Answer: Rewrite $\sin^3 x$ as $\sin x (\sin^2 x)$ and then as $\sin x (1-\cos^2 x)$. The correct strategy is to prepare for a u-substitution with $u = \cos x$. To do this, we need a single $\sin x$ factor left over for $du = -\sin x dx$. By rewriting $\sin^3 x$ as $\sin x (1-\cos^2 x)$, the integrand becomes $\frac{\sin x (1-\cos^2 x)}{\cos^2 x}$, 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.
14. By letting $u = x^2+4$, the integral $\int \frac{x^3}{\sqrt{x^2+4}} dx$ is transformed into an integral in terms of $u$. What is the resulting integral?
    Answer: $\frac{1}{2} \int \frac{u-4}{\sqrt{u}} du$ When you encounter an integral with a radical expression like $\sqrt{x^2+4}$, substitution is often the key to simplification. The goal is to transform both the integrand and the differential into terms of the new variable. Starting with $u = x^2 + 4$, you need to find $du$. Taking the derivative: $du = 2x \, dx$, which means $x \, dx = \frac{1}{2} du$. Now transform each part of the original integral $\int \frac{x^3}{\sqrt{x^2+4}} dx$: - The denominator $\sqrt{x^2+4}$ becomes $\sqrt{u}$ - For the numerator $x^3$, rewrite it as $x^2 \cdot x = (u-4) \cdot x$ since $x^2 = u-4$ - The differential $x \, dx = \frac{1}{2} du$ Putting it together: $\int \frac{(u-4) \cdot x}{\sqrt{u}} dx = \int \frac{u-4}{\sqrt{u}} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{u-4}{\sqrt{u}} du$ This confirms answer D is correct. A is wrong because it raises $(u-4)$ to the $\frac{3}{2}$ power instead of keeping it linear. B misses the $\frac{1}{2}$ factor that comes from the differential transformation. C incorrectly replaces the numerator with $u$ instead of $(u-4)$, forgetting that $x^2 = u-4$, not $u$. Strategy tip: In substitution problems, always work systematically through three steps: find $du$, express all $x$-terms in terms of $u$, and don't forget the factor from the differential transformation.
15. When simplifying $\int \frac{x+2}{\sqrt{4x^2 + 16x + 15}} dx$, what should be done immediately after completing the square in the denominator?
    Answer: Complete the square to get $4x^2 + 16x + 15 = (2x+4)^2 - 1$, then substitute $u = 2x + 4$ Completing the square: $4x^2 + 16x + 15 = (2x)^2 + 2(2x)(4) + 16 - 1 = (2x+4)^2 - 1$. Then substitute $u = 2x + 4$, so $x = \frac{u-4}{2}$ and $dx = \frac{du}{2}$. The integral becomes $\int \frac{\frac{u-4}{2} + 2}{\sqrt{u^2 - 1}} \cdot \frac{du}{2} = \frac{1}{4}\int \frac{u}{\sqrt{u^2 - 1}} du$, which is ready for standard integration techniques.
16. To evaluate $\int \cos^2(3x) \sin^5(3x) dx$, which substitution method is most appropriate after applying trigonometric identities?
    Answer: Use $\sin^5(3x) = \sin(3x)(1-\cos^2(3x))^2$ and substitute $u = \cos(3x)$ 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 $\sin^5(3x)$ has an odd power, you can rewrite it as $\sin^5(3x) = \sin(3x) \cdot \sin^4(3x) = \sin(3x) \cdot (\sin^2(3x))^2$. Using the Pythagorean identity $\sin^2(3x) = 1 - \cos^2(3x)$, this becomes $\sin(3x)(1-\cos^2(3x))^2$. Now if you substitute $u = \cos(3x)$, then $du = -3\sin(3x)dx$, so $\sin(3x)dx = -\frac{1}{3}du$. Your integral transforms into $\int \cos^2(3x) \sin^5(3x) dx = \int u^2(1-u^2)^2 \cdot (-\frac{1}{3})du$, 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 $\cos^2(3x)$ has an even power, substituting $u = \cos(3x)$ won't work cleanly because you'd need $\sin(3x)dx$ for the differential, but you'd have leftover even powers of sine. Strategy tip: For $\int \sin^m(x)\cos^n(x)dx$, if either exponent is odd, use substitution with the function having the even exponent. Save power reduction for when both exponents are even.
17. To prepare the integral $\int \frac{dx}{4x^2+8x+5}$ for an arctangent form, one must complete the square in the denominator. What is the denominator after this process?
    Answer: $4(x+1)^2 + 1$ When you encounter an integral with a quadratic denominator like this, completing the square is essential to transform it into a recognizable arctangent form. The goal is to rewrite the quadratic as a perfect square plus or minus a constant. Starting with $4x^2 + 8x + 5$, you need to factor out the coefficient of $x^2$ first: $4(x^2 + 2x) + 5$. Now complete the square inside the parentheses. Take half of the coefficient of $x$ (which is 2), square it: $(\frac{2}{2})^2 = 1$. Add and subtract this inside the parentheses: $4(x^2 + 2x + 1 - 1) + 5 = 4((x+1)^2 - 1) + 5 = 4(x+1)^2 - 4 + 5 = 4(x+1)^2 + 1$. This confirms that answer B is correct. Looking at the wrong answers: A) $(2x+2)^2 + 5$ expands to $4x^2 + 8x + 4 + 5 = 4x^2 + 8x + 9$, which doesn't match the original denominator. C) $4(x+2)^2 - 11$ expands to $4x^2 + 16x + 5$, showing an incorrect linear term coefficient. D) $(4x+1)^2 + 4$ expands to $16x^2 + 8x + 5$, which has the wrong leading coefficient. Study tip: Always factor out the leading coefficient before completing the square, and remember to distribute it through when adding/subtracting the completing term. Double-check by expanding your final answer to verify it matches the original expression.
18. Which expression represents the correct partial fraction decomposition for the rational function $f(x) = \frac{x^2 + 1}{(x-1)(x^2+4)}$?
    Answer: $\frac{A}{x-1} + \frac{Bx+C}{x^2+4}$ The denominator contains a distinct linear factor $(x-1)$ and an irreducible quadratic factor $(x^2+4)$. 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 $\frac{A}{x-1} + \frac{Bx+C}{x^2+4}$. Choice A incorrectly uses a constant for the quadratic factor. Choice B incorrectly factors $x^2+4$ over the real numbers. Choice D incorrectly uses a linear numerator for the linear factor.
19. To prepare the integral $\int \frac{dx}{\sqrt{8+2x-x^2}}$ for a trigonometric substitution, the expression inside the square root must be rewritten by completing the square. Which expression is the correct transformation?
    Answer: $9 - (x-1)^2$ To complete the square for $8+2x-x^2$, first factor out the negative from the x-terms: $8 - (x^2 - 2x)$. To complete the square for $x^2 - 2x$, we add and subtract 1 inside the parentheses: $8 - (x^2 - 2x + 1 - 1)$. This becomes $8 - ((x-1)^2 - 1)$, which simplifies to $8 - (x-1)^2 + 1 = 9 - (x-1)^2$. The other choices result from sign errors or incorrect application of the completing the square algorithm.
20. To integrate $\int (\sin x + \cos x)^2 dx$, the most effective first step is to expand the binomial. What is the simplified integrand after expansion and application of trigonometric identities?
    Answer: $1 + \sin(2x)$ When you encounter integrands with squared trigonometric expressions like $(\sin x + \cos x)^2$, expanding the binomial and applying fundamental identities is often the key to simplification. Let's expand this binomial systematically: $(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x$ Now apply two crucial trigonometric identities. First, the Pythagorean identity: $\sin^2 x + \cos^2 x = 1$. Second, the double angle identity: $2\sin x \cos x = \sin(2x)$. Substituting these identities gives us: $\sin^2 x + 2\sin x \cos x + \cos^2 x = (\sin^2 x + \cos^2 x) + 2\sin x \cos x = 1 + \sin(2x)$ This matches answer choice D. Now let's examine why the other options fall short. Choice A gives $1 + 2\sin x \cos x$ but fails to apply the double angle identity to simplify $2\sin x \cos x$ into $\sin(2x)$. Choice B shows only the first two terms from the expansion, completely omitting the crucial cross-term $2\sin x \cos x$. Choice C incorrectly suggests that $2\sin x \cos x = 0$, which would only be true at specific values, not generally. Study tip: When you see squared trigonometric expressions, immediately think "Pythagorean identity" for $\sin^2 x + \cos^2 x = 1$ and "double angle identity" for $2\sin x \cos x = \sin(2x)$. These two identities together transform many complex-looking integrands into much simpler forms.

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