Answer: Half-angle identities produce lower-frequency cosine terms that often need to be reduced again, leading to multiple rounds of identity application.

Compare the two strategies: With an odd exponent, one substitution typically collapses the integral into a polynomial. With all even exponents, repeated half-angle identities are needed. Why the correct answer works: Applying $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ or $\cos^2(x) = \frac{1 + \cos(2x)}{2}$ produces terms like $\cos^2(2x)$, which must be reduced again with the same identity, leading to cascading steps. Why distractors fail: Option A is wrong—integration is still performed, not differentiation. Option B is wrong—half-angle identities do simplify even powers. Option D is wrong—even-power trig integrals have exact closed-form antiderivatives.

Answer: $-\cos(x) + \frac{\cos^3(x)}{3} + C$

Apply the odd power strategy: Write $\sin^3(x) = \sin^2(x)\sin(x) = (1 - \cos^2(x))\sin(x)$. Substitute: Let $u = \cos(x)$, $du = -\sin(x)\,dx$. The integral becomes $\int -(1 - u^2)\,du = -u + \frac{u^3}{3} + C = -\cos(x) + \frac{\cos^3(x)}{3} + C$. Why distractors fail: Option B has incorrect signs from mishandling $du = -\sin(x)\,dx$. Option C omits the linear term from expanding $(1 - u^2)$. Option D incorrectly treats $\sin^3(x)$ as a power rule integral without performing substitution.

Answer: Either choice leads to a valid substitution and produces the same antiderivative (up to algebraic equivalence).

Analyze both options: If we strip $\sin(x)$: $u = \cos(x)$, leftover $\sin^2(x) = 1 - \cos^2(x)$ — valid. If we strip $\cos(x)$: $u = \sin(x)$, leftover $\cos^2(x) = 1 - \sin^2(x)$ — also valid. Why the correct answer works: When both exponents are odd, either function can serve as the stripped factor. The resulting antiderivatives differ in form but are algebraically equivalent (they differ by a constant). Why distractors fail: Options A and B are wrong because both approaches are valid—neither is the exclusive path. Option D is wrong because the odd power strategy works perfectly here; half-angle identities are not needed.

Answer: Factor out $\sin(x)$ and rewrite $\sin^2(x)$ as $1 - \cos^2(x)$

Identify the odd exponent: The exponent on $\sin(x)$ is 3, which is odd. The odd power strategy tells us to strip one factor of the function with the odd exponent. Why the correct answer works: Factoring out $\sin(x)$ leaves $\sin^2(x)\cos^2(x)\sin(x)\,dx$. Then $\sin^2(x) = 1 - \cos^2(x)$, giving $(1 - \cos^2(x))\cos^2(x)\sin(x)\,dx$, which is ready for the substitution $u = \cos(x)$. Why distractors fail: Option A is wrong because half-angle identities are used when both exponents are even, not when one is odd. Option C is wrong because we strip the odd-powered factor ($\sin$), not the even-powered one ($\cos$). Option D is wrong because integration by parts is unnecessary here; the odd power strategy is simpler and more direct.

Answer: $u = \sec(x)$

Identify the strategy for odd power of tan: When $n$ is odd in $\int \sec^m(x)\tan^n(x)\,dx$, we save a factor of $\sec(x)\tan(x)$ because $\frac{d}{dx}[\sec(x)] = \sec(x)\tan(x)$. Why the correct answer works: Setting $u = \sec(x)$ gives $du = \sec(x)\tan(x)\,dx$. The remaining even powers of $\tan(x)$ are converted to $\sec(x)$ using $\tan^2(x) = \sec^2(x) - 1$. Why distractors fail: Option A is wrong because $u = \tan(x)$ requires $du = \sec^2(x)\,dx$, which is the strategy for even powers of $\sec$. Option C is irrelevant to sec/tan integrals. Option D is wrong because $\sec(x)\tan(x)$ is the derivative of $\sec(x)$, not a substitution variable itself.

Answer: $u = \tan(x)$, $du = \sec^2(x)\,dx$

Identify the strategy for sec/tan integrals: When $\sec(x)$ has an even exponent, we reserve a factor of $\sec^2(x)$ for $du$ and convert the remaining $\sec^2(x)$ to $1 + \tan^2(x)$. Why the correct answer works: Setting $u = \tan(x)$ gives $du = \sec^2(x)\,dx$. We write $\sec^4(x) = \sec^2(x)\cdot\sec^2(x) = \sec^2(x)(1 + \tan^2(x))$, so the integral becomes $\int (1 + u^2)u^2\,du$. Why distractors fail: Options A and B are wrong because substituting $\sin(x)$ or $\cos(x)$ does not pair naturally with sec/tan expressions. Option D is wrong because $u = \sec(x)\tan(x)$ is not a standard substitution—$\sec(x)\tan(x)$ is the derivative of $\sec(x)$, which is used when $\tan$ has an odd power, not when $\sec$ has an even power.

Answer: $\frac{\sec^3(x)}{3} + C$

Identify the strategy: The exponent on $\tan(x)$ is 1 (odd), so save $\sec(x)\tan(x)$ for $du$ and set $u = \sec(x)$. Compute: Write $\sec^3(x)\tan(x) = \sec^2(x)\cdot\sec(x)\tan(x)$. Let $u = \sec(x)$, $du = \sec(x)\tan(x)\,dx$. Then $\int u^2\,du = \frac{u^3}{3} + C = \frac{\sec^3(x)}{3} + C$. Why distractors fail: Option B incorrectly substitutes with $\tan$ instead of $\sec$. Option C introduces a logarithmic term that does not arise from this integral. Option D has incorrect structure and does not match any valid antiderivative.

Answer: The claim is incorrect because the integral of a power of a function is not the power of the integral of that function.

Identify the logical error: The student is confusing $\int [f(x)]^3\,dx$ with $\left(\int f(x)\,dx\right)^3$. Integration is a linear operator, not a multiplicative one: $\int f^3 \neq (\int f)^3$. Why the correct answer works: In general, $\int [f(x)]^n\,dx \neq \left[\int f(x)\,dx\right]^n$. The correct approach to $\int \sec^3(x)\,dx$ requires integration by parts or a reduction formula. Why distractors fail: Option A is wrong because integration does not distribute over powers. Option C is wrong because the claim is never correct, regardless of parity. Option D is wrong because $\int \sec(x)\,dx = \ln|\sec(x) + \tan(x)| + C$ is a standard and correct result.

Answer: $-\frac{\sin^3(x)\cos(x)}{4} + \frac{3}{4}\int \sin^2(x)\,dx$

Apply the formula with $n = 4$: Substituting $n = 4$: $\int \sin^4(x)\,dx = -\frac{\sin^{4-1}(x)\cos(x)}{4} + \frac{4-1}{4}\int \sin^{4-2}(x)\,dx = -\frac{\sin^3(x)\cos(x)}{4} + \frac{3}{4}\int \sin^2(x)\,dx$. Why the correct answer works: Direct substitution of $n = 4$ into the given formula yields exactly Option A. Why distractors fail: Option B uses $n = 3$ fractions ($\frac{1}{3}$ and $\frac{2}{3}$) instead of $n = 4$. Option C uses $n = 5$ values and has the wrong power on $\sin$. Option D has $\sin^2(x)$ in the leading term instead of $\sin^3(x)$.

Answer: $\int \sin^7(x)\cos^4(x)\,dx$

Check each integral for odd exponents: The odd power strategy requires at least one odd exponent. Option A: 6 and 4, both even. Option B: 4 and 4, both even. Option C: 7 and 4 — 7 is odd. Option D: 6 and 6, both even. Why the correct answer works: In $\int \sin^7(x)\cos^4(x)\,dx$, strip one $\sin(x)$, convert $\sin^6(x) = (1-\cos^2(x))^3$, and substitute $u = \cos(x)$. This reduces to a polynomial integral — no half-angle identities or reduction formulas needed. Why distractors fail: Options A, B, and D all have both exponents even, requiring half-angle identities (even power strategy) and potentially multiple rounds of reduction.

Answer: $\frac{x}{16} - \frac{\sin(4x)}{64} - \frac{\sin^3(2x)}{48} + C$

Set up with half-angle identities: $\sin^2(x)\cos^4(x) = \frac{1-\cos(2x)}{2}\cdot\left(\frac{1+\cos(2x)}{2}\right)^2 = \frac{(1-\cos(2x))(1+\cos(2x))^2}{8}$. Expand: $(1-\cos(2x))(1+\cos(2x))^2 = (1-\cos(2x))(1+2\cos(2x)+\cos^2(2x))$. Expanding: $1 + 2\cos(2x) + \cos^2(2x) - \cos(2x) - 2\cos^2(2x) - \cos^3(2x) = 1 + \cos(2x) - \cos^2(2x) - \cos^3(2x)$. Integrate each term: $\frac{1}{8}\int [1 + \cos(2x) - \cos^2(2x) - \cos^3(2x)]\,dx$. We have: $\int 1\,dx = x$; $\int \cos(2x)\,dx = \frac{\sin(2x)}{2}$; $\int \cos^2(2x)\,dx = \frac{x}{2} + \frac{\sin(4x)}{8}$ (via half-angle); $\int \cos^3(2x)\,dx = \frac{\sin(2x)}{2} - \frac{\sin^3(2x)}{6}$ (via odd power strategy with $u = \sin(2x)$). Combining: $\frac{1}{8}\left[x + \frac{\sin(2x)}{2} - \frac{x}{2} - \frac{\sin(4x)}{8} - \frac{\sin(2x)}{2} + \frac{\sin^3(2x)}{6}\right] = \frac{1}{8}\left[\frac{x}{2} - \frac{\sin(4x)}{8} - \frac{\sin^3(2x)}{6}\right] = \frac{x}{16} - \frac{\sin(4x)}{64} - \frac{\sin^3(2x)}{48} + C$. Why distractors fail: Option A includes a spurious $\frac{\sin(2x)}{64}$ term that should cancel during simplification. Option B has the wrong sign on the $\sin(4x)$ term. Option D corresponds to $\int \sin^2(x)\cos^2(x)\,dx$, which has a simpler integrand.

Answer: $\int \sec^3(x)\,dx$

Check each option: Option A: $\sin$ has odd power → odd power strategy with $u = \cos(x)$. Option B: $\sec^2(x)$ is present → direct $u = \tan(x)$. Option D: $\sin(x)$ is present and $\cos$ is odd → $u = \sin(x)$ or $u = \cos(x)$ both work. Why the correct answer works: $\int \sec^3(x)\,dx$ has $\sec$ with an odd exponent and no $\tan$ factor to work with. Neither the even-$\sec$ strategy nor the odd-$\tan$ strategy applies, so it must be handled via integration by parts or a reduction formula. Why distractors fail: Options A, B, and D all have features (odd power of sin/cos or $\sec^2$/$\sec\tan$ factor present) that allow direct substitution.

Answer: $\frac{\sin^3(x)}{3} - \frac{2\sin^5(x)}{5} + \frac{\sin^7(x)}{7} + C$

Apply the odd power strategy: The exponent on $\cos(x)$ is 5 (odd). Strip one $\cos(x)$: $\cos^5(x)\sin^2(x) = \cos^4(x)\sin^2(x)\cos(x) = (1-\sin^2(x))^2 \sin^2(x)\cos(x)$. Substitute and expand: Let $u = \sin(x)$, $du = \cos(x)\,dx$. The integral becomes $\int (1 - u^2)^2 u^2\,du = \int (1 - 2u^2 + u^4)u^2\,du = \int (u^2 - 2u^4 + u^6)\,du = \frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7} + C$. Why distractors fail: Option B omits the $\frac{\sin^7(x)}{7}$ term from incomplete expansion. Option C uses $\cos$ instead of $\sin$ as the substitution variable. Option D has incorrect signs from expanding $(1-u^2)^2$ incorrectly.

Answer: Either substitution leads to a solvable integral, though one may be algebraically simpler.

Examine both strategies: Even power of $\sec$: reserve $\sec^2(x)$, set $u = \tan(x)$, convert $\sec^2(x) = 1 + \tan^2(x)$. Odd power of $\tan$: reserve $\sec(x)\tan(x)$, set $u = \sec(x)$, convert $\tan^2(x) = \sec^2(x) - 1$. Why the correct answer works: Both strategies are valid. Using $u = \sec(x)$: $\int (u^2 - 1)u^3\,du$. Using $u = \tan(x)$: $\int (1 + u^2)u^3\,du$. Both are polynomial integrals. The algebra differs slightly but both yield equivalent antiderivatives. Why distractors fail: Options A and B each claim only one approach works, which is false. Option D is wrong because a reduction formula is unnecessary when direct substitution is available.

Answer: $\int \frac{1}{8}(1 - \cos(2x))(1 + \cos(2x))^2\,dx$

Apply half-angle identities to each factor: $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ and $\cos^2(x) = \frac{1 + \cos(2x)}{2}$, so $\cos^4(x) = \left(\frac{1 + \cos(2x)}{2}\right)^2$. Combine: $\sin^2(x)\cos^4(x) = \frac{1 - \cos(2x)}{2}\cdot\frac{(1 + \cos(2x))^2}{4} = \frac{1}{8}(1 - \cos(2x))(1 + \cos(2x))^2$. Why distractors fail: Option B simplifies too aggressively and loses the asymmetry between the $\sin^2$ and $\cos^4$ contributions. Option C swaps the roles of $\sin^2$ and $\cos^4$ by reversing the signs. Option D introduces $\sin^2(2x)\cos^2(2x)$, which corresponds to a different integrand.

Answer: The first strips off a $\sin(x)$ factor and uses Pythagorean identities, while the second requires half-angle identities because both exponents are even.

Classify each integral: In $\int \sin^5(x)\cos^2(x)\,dx$, the exponent on $\sin$ is odd (5), so the odd power strategy applies. In $\int \sin^2(x)\cos^2(x)\,dx$, both exponents are even (2 and 2), so the even power strategy applies. Why the correct answer works: The odd power strategy strips one $\sin(x)$ factor and converts $\sin^4(x)$ to $(1-\cos^2(x))^2$ for a $u = \cos(x)$ substitution. The even power strategy uses $\sin^2(x)\cos^2(x) = \frac{1}{4}\sin^2(2x)$ and then the half-angle identity. Why distractors fail: Option A is wrong because neither integral requires integration by parts or partial fractions as the primary method. Option C is wrong because the first integral does not need half-angle identities—it has an odd exponent. Option D is wrong because $\sin^2(x)\cos^2(x)$ cannot be integrated directly without applying identities first.

Answer: Either $u = \sec(x)$ or $u = \tan(x)$ works; using $u = \sec(x)$ gives $\int u^5(u^2-1)^2\,du$ which is a polynomial in $u$.

Check conditions: $\sec$ has an even exponent (6) and $\tan$ has an odd exponent (5). Both standard strategies apply. Strategy with $u = \sec(x)$: Save $\sec(x)\tan(x)$, $du = \sec(x)\tan(x)\,dx$. Remaining: $\sec^5(x)\tan^4(x) = \sec^5(x)(\sec^2(x)-1)^2$. So $\int u^5(u^2-1)^2\,du$. Strategy with $u = \tan(x)$: Save $\sec^2(x)$, $du = \sec^2(x)\,dx$. Remaining: $\sec^4(x)\tan^5(x) = (1+\tan^2(x))^2\tan^5(x)$. So $\int (1+u^2)^2 u^5\,du$. Why the correct answer is best: Both substitutions produce polynomial integrals of similar complexity. Option D correctly recognizes this flexibility. Options B and C each claim only their specific approach, and Option A is completely wrong for sec/tan integrals.

Answer: Yes, this is correct.

Verify step by step: $\sin^2(x)\cos^2(x) = \frac{1}{4}\sin^2(2x) = \frac{1}{4}\cdot\frac{1 - \cos(4x)}{2} = \frac{1}{8}(1 - \cos(4x))$. Integrate: $\int \frac{1}{8}(1 - \cos(4x))\,dx = \frac{1}{8}\left(x - \frac{\sin(4x)}{4}\right) + C = \frac{x}{8} - \frac{\sin(4x)}{32} + C$. Why distractors fail: Option B is wrong because the $\frac{1}{8}$ coefficient arises correctly from $\frac{1}{4} \cdot \frac{1}{2}$. Option C is wrong because $\int \cos(4x)\,dx = \frac{\sin(4x)}{4}$, not $\cos(4x)$. Option D is wrong because the double application of identities naturally produces $\cos(4x)$ and thus $\sin(4x)$ in the answer.

Answer: Integrate $\int \tan^3(x)\sec^2(x)\,dx$ directly using $u = \tan(x)$, and repeat the identity on $\int \tan^3(x)\,dx$.

Analyze the split: The student wrote $\int \tan^5(x)\,dx = \int \tan^3(x)\sec^2(x)\,dx - \int \tan^3(x)\,dx$. Handle each part: For $\int \tan^3(x)\sec^2(x)\,dx$: let $u = \tan(x)$, $du = \sec^2(x)\,dx$, giving $\int u^3\,du = \frac{\tan^4(x)}{4}$. For $\int \tan^3(x)\,dx$: repeat the identity $\tan^3(x) = \tan(x)\sec^2(x) - \tan(x)$ and integrate each piece. Why distractors fail: Option A is wrong because half-angle identities are for even powers of $\sin/\cos$, not $\tan$. Option C is wrong because partial fractions apply to rational functions of polynomials, not trig functions directly. Option D would work but is unnecessarily complicated compared to the direct substitution.

Answer: $\frac{\tan^4(x)}{4} + C$

Identify the substitution: Since $\sec^2(x)$ is the derivative of $\tan(x)$, let $u = \tan(x)$, $du = \sec^2(x)\,dx$. Compute the integral: $\int \sec^2(x)\tan^3(x)\,dx = \int u^3\,du = \frac{u^4}{4} + C = \frac{\tan^4(x)}{4} + C$. Why distractors fail: Option B confuses $\tan$ with $\sec$ in the substitution result. Option C uses the wrong exponent ($u^2$ instead of $u^3$). Option D incorrectly adds an extra term, likely from an unnecessary identity application.

Answer: After stripping one $\sin(x)$, the remaining $\sin^3(x)$ has an odd power that cannot be fully converted to $\cos(x)$.

Attempt the proposed method: Stripping one $\sin(x)$ from $\sin^4(x)$ gives $\sin^3(x)\cdot\sin(x)\,dx$. For the substitution $u = \cos(x)$, $du = -\sin(x)\,dx$, we need the leftover to be entirely in terms of $\cos(x)$. Why the correct answer works: After stripping one $\sin(x)$, we have $\sin^3(x)$. Using $\sin^2(x) = 1 - \cos^2(x)$ converts it to $(1 - \cos^2(x))\sin(x)$, giving $\sin^4(x) = (1 - \cos^2(x))\sin^2(x)$... but we still have $\sin^2(x)$ remaining that we cannot convert away because we already used our stripped factor. The issue is that the original exponent 4 is even, so after stripping one factor, the leftover exponent (3) is odd and cannot be fully expressed in terms of $\cos(x)$ alone. Why distractors fail: Option A is wrong—$\sin^2(x) + \cos^2(x) = 1$ is the Pythagorean identity. Option C is wrong—$u = \cos(x)$ is perfectly valid when $\sin$ has an odd power. Option D is wrong—identities apply regardless of the size of the exponent.

Answer: The substitution in Step 2 should produce $du = -\sin(x)\,dx$, causing a sign error that propagates through Steps 3 and 4.

Check each step: Step 1: $\sin^3(x) = \sin^2(x)\sin(x) = (1-\cos^2(x))\sin(x)$ ✓. Step 2: $u = \cos(x) \Rightarrow du = -\sin(x)\,dx$, so $\sin(x)\,dx = -du$. The student wrote $du = \sin(x)\,dx$, missing the negative sign. Correct result: The integral should be $-\int (1-u^2)u^4\,du = -\frac{u^5}{5} + \frac{u^7}{7} + C = -\frac{\cos^5(x)}{5} + \frac{\cos^7(x)}{7} + C$. Why distractors fail: Option A is wrong because the identity $(1-\cos^2(x))\sin(x)$ is correct. Option C is wrong because the expansion $(1-u^2)u^4 = u^4 - u^6$ is correct (two terms). Option D is wrong because $u = \cos(x)$ is a valid choice when stripping $\sin(x)$.

Answer: $\frac{\pi}{32}$

Apply the half-angle identities: $\sin^4(x)\cos^2(x) = \left(\frac{1-\cos(2x)}{2}\right)^2\cdot\frac{1+\cos(2x)}{2} = \frac{(1-\cos(2x))^2(1+\cos(2x))}{8}$. Expand and integrate: Expanding: $(1-\cos(2x))^2(1+\cos(2x)) = 1 - \cos(2x) - \cos^2(2x) + \cos^3(2x)$. Over $[0, \pi/2]$: $\int_0^{\pi/2}1\,dx = \frac{\pi}{2}$, $\int_0^{\pi/2}\cos(2x)\,dx = 0$, $\int_0^{\pi/2}\cos^2(2x)\,dx = \frac{\pi}{4}$, and $\int_0^{\pi/2}\cos^3(2x)\,dx = 0$. So the result is $\frac{1}{8}\left(\frac{\pi}{2} - \frac{\pi}{4}\right) = \frac{1}{8}\cdot\frac{\pi}{4} = \frac{\pi}{32}$. Why distractors fail: Option B ($\frac{\pi}{16}$) is a common error from miscounting the factor of $\frac{1}{8}$ as $\frac{1}{4}$. Option C ($\frac{\pi}{8}$) is too large. Option D ($\frac{3\pi}{32}$) arises from an arithmetic error in the expansion.

Answer: Approach A is preferable because $\cos$ has an odd exponent, making it a direct polynomial substitution.

Evaluate approach A: Strip $\cos(x)$: $\cos^5(x)\sin^2(x) = \cos^4(x)\sin^2(x)\cos(x) = (1-\sin^2(x))^2\sin^2(x)\cos(x)$. Let $u = \sin(x)$. This becomes a polynomial $\int (1-u^2)^2 u^2\,du$. Evaluate approach B: Approach B would require multiple applications of half-angle identities on all seven combined powers, generating numerous terms with $\cos(2x)$, $\cos(4x)$, etc. — far more algebraic work. Why distractors fail: Option A is wrong because approach A is significantly faster. Option C is wrong because the presence of an odd exponent means half-angle identities are not necessary. Option D is wrong because approach A works perfectly without reduction formulas.

Answer: This approach is valid but less efficient than setting $u = \sec(x)$, which gives $\int u^4\,du$ directly.

Evaluate the proposed approach: With $u = \tan(x)$, $du = \sec^2(x)\,dx$. We would write $\sec^5(x)\tan(x) = \sec^3(x)\cdot\sec^2(x)\cdot\tan(x)$ and need to express $\sec^3(x)\tan(x)$ in terms of $\tan(x)$ — this gets complicated. Compare with the better approach: Since $\tan$ has an odd exponent (1), set $u = \sec(x)$, $du = \sec(x)\tan(x)\,dx$. Then $\sec^5(x)\tan(x) = \sec^4(x)\cdot\sec(x)\tan(x)$, giving $\int u^4\,du = \frac{u^5}{5} + C = \frac{\sec^5(x)}{5} + C$. Why distractors fail: Option A is wrong because while technically workable, it is far more cumbersome. Option C is wrong because $\sec^2(x) = 1 + \tan^2(x)$, so it can be expressed in terms of $\tan$, just not cleanly. Option D is wrong because $\sec^2(x)\,dx$ can be extracted, but the remaining expression is messy.

Answer: Yes, this is correct.

Verify by computing: Strip one $\sin(x)$: $\sin^5(x)\cos^2(x) = (1-\cos^2(x))^2 \cos^2(x)\sin(x)$. Let $u = \cos(x)$, $du = -\sin(x)\,dx$. Expand and integrate: $-\int (1 - 2u^2 + u^4)u^2\,du = -\int (u^2 - 2u^4 + u^6)\,du = -\frac{u^3}{3} + \frac{2u^5}{5} - \frac{u^7}{7} + C$. Why distractors fail: Option A is wrong because the signs match the negative from $du = -\sin(x)\,dx$. Option B is wrong because the coefficients 1/3, 2/5, 1/7 arise correctly from the expansion. Option D is wrong because the substitution $u = \cos(x)$ naturally yields an answer in terms of $\cos$.

Answer: To systematically lower the power of the trigonometric function in successive applications

Definition of a reduction formula: A reduction formula is a recursive relation that expresses an integral involving a parameter $n$ in terms of an integral with a smaller parameter, typically $n-2$. Why the correct answer works: Each application of the reduction formula decreases the exponent by 2, eventually reducing the integral to a base case such as $\int \sin^0(x)\,dx = x + C$ or $\int \sin^1(x)\,dx = -\cos(x) + C$. Why distractors fail: Option A is wrong because the result is still a trigonometric expression, not a polynomial in $x$. Option C is wrong because reduction formulas are typically derived using trigonometric identities and integration by parts. Option D is wrong because reduction formulas apply to both definite and indefinite integrals and do not introduce bounds.

Answer: $\tan(x) + \frac{\tan^3(x)}{3} + C$

Apply the even-sec strategy: Write $\sec^4(x) = \sec^2(x)\cdot\sec^2(x) = (1 + \tan^2(x))\sec^2(x)$. Substitute and integrate: Let $u = \tan(x)$, $du = \sec^2(x)\,dx$. Then $\int (1 + u^2)\,du = u + \frac{u^3}{3} + C = \tan(x) + \frac{\tan^3(x)}{3} + C$. Why distractors fail: Option B incorrectly treats $\sec^4(x)$ like a power rule problem. Option C is missing the $\frac{1}{3}$ coefficient and the standalone $\tan(x)$ term structure is wrong. Option D has coefficient $\frac{2}{3}$ instead of $\frac{1}{3}$.

Answer: $\cos^2(x) = \frac{1 + \cos(2x)}{2}$

Recall the half-angle identity for cosine: The even power strategy uses the identity $\cos^2(x) = \frac{1 + \cos(2x)}{2}$ to rewrite the integrand in a form that can be integrated directly. Why the correct answer works: $\int \cos^2(x)\,dx = \int \frac{1 + \cos(2x)}{2}\,dx = \frac{x}{2} + \frac{\sin(2x)}{4} + C$. Why distractors fail: Option A is the Pythagorean identity, which does not reduce the power—it replaces $\cos^2$ with $1 - \sin^2$. Option C is wrong because $\frac{1 - \cos(2x)}{2} = \sin^2(x)$, not $\cos^2(x)$. Option D is wrong because $\frac{\cos(2x) - 1}{2} = -\sin^2(x)$.

Answer: The double-angle identity $\sin(2x) = 2\sin(x)\cos(x)$

Recognize the even power scenario: Both exponents are even (2 and 2), so the even power strategy applies. We need to reduce powers using identities. Why the correct answer works: Since $\sin(x)\cos(x) = \frac{1}{2}\sin(2x)$, we have $\sin^2(x)\cos^2(x) = \frac{1}{4}\sin^2(2x)$, which can then be integrated using the half-angle identity $\sin^2(u) = \frac{1 - \cos(2u)}{2}$. Why distractors fail: Option A is wrong because the Pythagorean identity alone doesn't reduce even powers—it merely swaps one even power for another. Option C is wrong because the sec/tan identity is for sec/tan integrals, not sin/cos. Option D incorrectly states the sum-to-product formula and would not simplify the integral efficiently.

Answer: $\frac{x}{2} - \frac{\sin(2x)}{4} + C$

Apply the half-angle identity: $\sin^2(x) = \frac{1 - \cos(2x)}{2}$. Integrate: $\int \frac{1 - \cos(2x)}{2}\,dx = \frac{1}{2}\left(x - \frac{\sin(2x)}{2}\right) + C = \frac{x}{2} - \frac{\sin(2x)}{4} + C$. Why distractors fail: Option A has $+\frac{\sin(2x)}{4}$ instead of $-$, which comes from using $\cos^2(x)$ identity instead of $\sin^2(x)$. Option C integrates $\cos(2x)$ incorrectly as $\cos(2x)$ instead of $\frac{\sin(2x)}{2}$. Option D has incorrect coefficients from failing to divide by 2 properly.