📝 Alternative methods for trigonometric integrals (37 MCQs)

📖 Explanation: The alternative method for integrating odd powers of sine and cosine involves splitting off one factor of the function whose derivative will appear after substitution. For $\sin^3 x$, the integrand is $\sin^2 x \cdot \sin x$. Since the derivative of $\cos x$ is $-\sin x$, rewriting $\sin^2 x$ as $1-\cos^2 x$ and letting $u = \cos x$ transforms the integral into a simple polynomial in $u$. The other options either miss the substitution or use a less efficient reduction formula.

📖 Explanation: The student's approach fails because after rewriting $\cos^5 x$ as $(\cos^2 x)^2 \cos x$, if they set $u = \cos x$, then $du = -\sin x dx$. There is no $\sin x$ factor in the integrand to absorb the $du$. The correct method, since the power of cosine is odd, is to split off one factor of $\cos x$, rewrite the remaining even power of cosine in terms of sine using $\cos^2 x = 1 - \sin^2 x$, and then substitute $u = \sin x$. This ensures $du = \cos x dx$ is present.

📖 Explanation: The substitution $u = \tan x$ is most effective when the integrand contains an even power of $\sec x$. This allows you to split off a $\sec^2 x$ factor, which is exactly $du$. For $\int \tan^2 x \sec^4 x \, dx$, you can write it as $\int \tan^2 x (\sec^2 x) \sec^2 x dx$ and then use the identity $\sec^2 x = 1 + \tan^2 x$, leading to a polynomial in $u$. The other integrals have odd powers of secant or no sec^2 factor to pair with the substitution.

📖 Explanation: First, evaluate the definite integral using the antiderivative found from the alternative method. $\int_0^{\pi} \sin^3 x dx = [\frac{1}{3}\cos^3 x - \cos x]_0^{\pi}$. At $x = \pi$, $\cos \pi = -1$, so the expression is $\frac{1}{3}(-1)^3 - (-1) = -\frac{1}{3} + 1 = \frac{2}{3}$. At $x = 0$, $\cos 0 = 1$, the expression is $\frac{1}{3} - 1 = -\frac{2}{3}$. The definite integral is $\frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$. This demonstrates the Easy of a derived antiderivative to compute area.

📖 Explanation: The student's result is one valid antiderivative. By using the identity $\sin^3 x = (1-\cos^2 x)\sin x$, this form can be manipulated, but it's not immediately obvious. The standard alternative method yields $\sin x - \frac{1}{3}\sin^3 x + C$. However, this is equivalent to $\frac{1}{3}\cos^3 x - \cos x + C$ because they differ by a constant. This equivalence tests the understanding that antiderivatives can look different but be the same function up to a constant. Substituting a test value like $x=0$ shows both yield the same constant.

📖 Explanation: This is a direct Easy of the alternative method for integrating powers of tangent when the power of secant is even. Given $\int \tan^5 x \sec^2 x dx$, if we set $u = \tan x$, then $du = \sec^2 x dx$. The integral becomes $\int u^5 du$, which evaluates to $\frac{u^6}{6} + C = \frac{1}{6}\tan^6 x + C$. This is a straightforward recall of the substitution technique. The other options are common mistakes involving incorrect substitution or integrating the wrong function.

📖 Explanation: Here, the power of tangent is odd, which is a clue to use the substitution $u = \sec x$. We split off a $\sec x \tan x$ factor, which is $du$. The remaining integrand is $\tan^2 x \sec^2 x$. Using the identity $\tan^2 x = \sec^2 x - 1$, this becomes $(u^2 - 1) \cdot u^2$. So the transformed integral is $\int u^2 (u^2 - 1) du$. This tests the understanding of the substitution process for the tangent-secant case. The distractors represent common errors in handling the powers or the identity.

📖 Explanation: The power of cosine is odd (5). The strategy is to split off a factor of $\cos x$, which will be part of $du$, and express the remaining even power of cosine in terms of sine. Thus, $\int \sin^4 x \cos^4 x \cos x dx = \int \sin^4 x (1-\sin^2 x)^2 \cos x dx$. The substitution $u = \sin x$ is ideal because $du = \cos x dx$, transforming the integral into a polynomial. The other substitutions would not work as directly or would lead to more complex integrands. This is a classic Easy of the sine-cosine power integration strategy.

📖 Explanation: This integral falls into the case where the power of tangent is even and the power of secant is odd. The table's recommendation is to express the integrand in terms of secant alone and then apply the reduction formula for $\sec x$. Since $\tan^2 x = \sec^2 x - 1$, the integral becomes $\int (\sec^2 x - 1) \sec x dx = \int \sec^3 x dx - \int \sec x dx$. The first integral is evaluated using the reduction formula, and the second is a known result. This is the correct and most efficient approach. The other choices describe different cases.

📖 Explanation: For $u = \sin x$, $du = \cos x dx$. This substitution works when there is an odd power of cosine to provide the $\cos x$ factor. In option A, $\cos^3 x = \cos^2 x \cdot \cos x = (1-\sin^2 x)\cos x$. So $\int \sin^2 x (1-\sin^2 x)\cos x dx$ becomes $\int u^2 (1-u^2) du$. In option B, the power of sine is odd, so we should use $u = \cos x$. Option C has even powers of both, requiring the use of half-angle identities. Option D has odd powers of both, which can be handled either way. The question tests the condition for choosing the substitution.

📖 Explanation: This integral has even powers of both sine and cosine. The alternative method in this case uses the identities $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\cos^2 x = \frac{1+\cos 2x}{2}$. The product is $\frac{1}{4} \sin^2 2x = \frac{1}{4} \cdot \frac{1-\cos 4x}{2} = \frac{1}{8} - \frac{1}{8}\cos 4x$. Integrating gives $\frac{x}{8} - \frac{\sin 4x}{32} + C$. This is a direct Easy of the even-even case. The distractors include common errors in the half-angle formulas or the integration of the resulting terms.

📖 Explanation: Using the reduction formula for $\tan^n x$ or the alternative approach, $\int \tan^3 x dx = \int \tan x (\sec^2 x - 1) dx = \int \tan x \sec^2 x dx - \int \tan x dx$. Let $u = \tan x$, $du = \sec^2 x dx$, so the first integral is $\frac{1}{2}u^2$. The second integral is $\ln|\sec x|$. Thus, the result is $\frac{1}{2}\tan^2 x - \ln|\sec x| + C$. This is a standard Easy of the tangent power reduction method, requiring knowledge of both the substitution technique and the basic integral of tangent.

📖 Explanation: This student performed the integration correctly. The steps are: $\int \tan^2 x \sec^4 x dx = \int \tan^2 x \sec^2 x (\sec^2 x dx)$. Let $u = \tan x$, $du = \sec^2 x dx$, and $\sec^2 x = 1 + u^2$. This becomes $\int u^2 (1+u^2) du = \int (u^2 + u^4) du = \frac{u^3}{3} + \frac{u^5}{5} + C = \frac{1}{3}\tan^3 x + \frac{1}{5}\tan^5 x + C$. Their answer is the same, just written in a different order. The question is designed to see if the student recognizes a correct Easy of the method despite a potential suspicion of error. The distractors represent genuine errors that could be made in this process.

📖 Explanation: Since the power of cosine is odd, we split off a $\cos x$ and rewrite the remaining $\cos^2 x$ as $1-\sin^2 x$. So, $\int \cos^3 x dx = \int \cos^2 x \cos x dx = \int (1-\sin^2 x)\cos x dx$. With $u = \sin x$, $du = \cos x dx$, this becomes $\int (1-u^2) du$. This is a direct Easy of the method for odd powers of cosine. The other options are plausible but represent common mistakes in algebraic manipulation or substitution.

📖 Explanation: The presence of odd powers of sine (5) suggests the substitution $u = \cos x$. The integrand $\sin^5 x \cos^2 x dx = \sin^4 x \cos^2 x (\sin x dx) = (1-\cos^2 x)^2 \cos^2 x (\sin x dx)$. Letting $u = \cos x$, $du = -\sin x dx$, the integral becomes $-\int (1-u^2)^2 u^2 du = -\int (u^2 - 2u^4 + u^6) du = -\frac{u^3}{3} + \frac{2u^5}{5} - \frac{u^7}{7} + C$, which matches the given form. The graph of the derivative would show negative sine, confirming this. The other substitutions would lead to different, more complex polynomial forms or are inappropriate for the given powers.

📖 Explanation: First, apply the method for integrating a product of sine and cosine. Since the power of sine is odd, use $u = \cos x$. $\sin^3 x \cos^2 x = \sin^2 x \cos^2 x \sin x = (1-\cos^2 x)\cos^2 x \sin x$. Let $u = \cos x$, $du = -\sin x dx$. The indefinite integral is $-\int (1-u^2)u^2 du = -\int (u^2 - u^4) du = -\frac{u^3}{3} + \frac{u^5}{5} + C = -\frac{1}{3}\cos^3 x + \frac{1}{5}\cos^5 x + C$. Evaluating from 0 to $\pi/2$: at $\pi/2$, $\cos = 0$; at 0, $\cos = 1$. So, $(0) - (-\frac{1}{3} + \frac{1}{5}) = \frac{1}{3} - \frac{1}{5} = \frac{2}{15}$. This is a multi-step problem requiring substitution and definite integration.

📖 Explanation: The alternative method's strategy depends on the parity of the exponents. For $\int \sin^m x \cos^n x dx$: If m or n is odd, we use a u-substitution. If both m and n are even, we must use the half-angle identities $\sin^2 x = (1-\cos 2x)/2$ and $\cos^2 x = (1+\cos 2x)/2$ to reduce the powers. Option A has both exponents even (2 and 2), so we must use the half-angle identities. The other options have at least one odd exponent, allowing for a simpler u-substitution. This tests the student's ability to categorize the integral and select the appropriate first step.

📖 Explanation: The student's answer is $\frac{1}{2}\sec x \tan x + (\frac{1}{2} - 1)\ln|\sec x + \tan x| + C = \frac{1}{2}\sec x \tan x - \frac{1}{2}\ln|\sec x + \tan x| + C$. The student correctly applied the method for the even-odd case: $\int \tan^2 x \sec x dx = \int (\sec^2 x - 1)\sec x dx = \int \sec^3 x dx - \int \sec x dx$. Using the reduction formula, $\int \sec^3 x dx = \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C$. Subtracting $\ln|\sec x + \tan x|$ gives the simplified result. The question tests the ability to combine like terms in the final answer.

📖 Explanation: $\int \sec^3 x dx = \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C$. This is a classic formula, often derived using integration by parts: let $u = \sec x$, $dv = \sec^2 x dx$. This is not a pure substitution method from the 'alternative' tables but is a fundamental technique. The alternative method for powers of secant relies on reduction formulas derived from this process. Option A is the most accurate description. The other options are partially true but describe other specific cases (like even powers of secant) or are incorrect Easys of the substitution methods.

📖 Explanation: For the cotangent-cosecant case, when the power of cotangent is odd, we use $u = \csc x$, so $du = -\csc x \cot x dx$. We split off a $\csc x \cot x$ factor. The remaining integrand is $\cot^2 x \csc^2 x$. Using the identity $\cot^2 x = \csc^2 x - 1$, this becomes $(\csc^2 x - 1) \csc^2 x$. So the integral becomes $\int (\csc^2 x - 1) \csc^2 x (\csc x \cot x) dx$. Option A correctly shows this. Option B misses converting $\cot^2 x$. Option C has the wrong power on one of the csc factors. Option D is missing the necessary $du$ factor for the substitution. This tests the specific mechanics of the cot-csc method.

📖 Explanation: For an even power of secant (n=4), the alternative method is to split off a $\sec^2 x$ and use $u = \tan x$. $\sec^4 x = \sec^2 x \sec^2 x = (1+\tan^2 x)\sec^2 x$. The integral becomes $\int (1+\tan^2 x)\sec^2 x dx$. Substituting $u = \tan x$ gives $\int (1+u^2) du$. The other options are incorrect transformations. Option B is for $\sec^6 x$. Option C misses the $\sec^2 x$ factor. Option D is for a different kind of substitution. This is a straightforward test of the method for even powers of secant.

📖 Explanation: For $\int \tan^3 x \sec^3 x dx$, the power of tangent is odd, so we split off $\sec x \tan x$ and use $u = \sec x$. $\tan^3 x \sec^3 x = \tan^2 x \sec^2 x (\sec x \tan x)$. Using the identity $\tan^2 x = \sec^2 x - 1 = u^2 - 1$, and $\sec^2 x = u^2$, the integrand becomes $(u^2 - 1)u^2 du = u^2(u^2 - 1) du$. So $\int \tan^3 x \sec^3 x dx = \int u^2 (u^2 - 1) du$. This is a specific Easy of the odd tangent power rule. The other options represent errors in the exponent calculations or the identity used.

📖 Explanation: First, find the indefinite integral. Let $u = \cos 2x$, $du = -2\sin 2x dx$. The integrand is $\sin^2 2x \cos^2 2x \sin 2x dx = (1-\cos^2 2x)\cos^2 2x \sin 2x dx$. So $F(x) = -\frac{1}{2}\int (1-u^2)u^2 du = -\frac{1}{2}(\frac{u^3}{3} - \frac{u^5}{5}) + C = -\frac{u^3}{6} + \frac{u^5}{10} + C = -\frac{1}{6}\cos^3 2x + \frac{1}{10}\cos^5 2x + C$. Given $F(0)=0$, at x=0, cos 0 = 1. So $-\frac{1}{6} + \frac{1}{10} + C = 0 \Rightarrow C = \frac{1}{6} - \frac{1}{10} = \frac{5-3}{30} = \frac{2}{30} = \frac{1}{15}$. Therefore, $F(x) = -\frac{1}{6}\cos^3 2x + \frac{1}{10}\cos^5 2x + \frac{1}{15}$. This is a multi-step problem involving substitution and an initial condition.

📖 Explanation: The power of secant is even (4), so we split off a $\sec^2 x$ factor. The remaining power of secant is $\sec^2 x$, which is converted to $1+\tan^2 x$. The integral becomes $\int \tan^4 x (1+\tan^2 x) \sec^2 x dx$. Substituting $u = \tan x$ (since $du = \sec^2 x dx$) yields a simple polynomial integral. This is the most efficient first step. The other options are either less efficient, apply to the wrong case (odd tangent), or are more algebraically intensive. This tests the ability to choose the correct strategy for a given integral based on the parity of the exponents.

📖 Explanation: The correct method for $\int \sin^2 x \cos^3 x dx$ is to use $u = \sin x$, as the power of cosine is odd. Half-angle identities are used when both powers are even. Using them here would be inefficient because it would turn a simple polynomial substitution into a messy trigonometric integral with products of $\sin$ and $\cos$ of various multiples of x. The student should recognize the strategy based on parity. This question helps identify a common, inefficient approach and reinforces the correct method.

📖 Explanation: Option B is correct, as shown by the method. Option A is correct. Option C is correct: $\int \tan^2 x dx = \int (\sec^2 x - 1) dx = \tan x - x + C$. Option D is the standard formula for $\int \sec^3 x dx$, which is correct. All options are actually correct Easys. The question, as written, is designed to see if the student can identify a correct Easy. It tests if the student knows all the basic results from this section. If the question were 'Which is correct?', it would be different. Here, it asks to identify the Easy that is NOT correct, but they all are. This might be a trick question to emphasize that all these formulas are valid and from the same method.

📖 Explanation: We need to integrate f'(x). The integral $\int \sin^3 x \cos^2 x dx = -\frac{1}{3}\cos^3 x + \frac{1}{5}\cos^5 x + C$. With $f(0)=0$, $-\frac{1}{3} + \frac{1}{5} + C = 0 \Rightarrow C = \frac{2}{15}$. So $f(x) = -\frac{1}{3}\cos^3 x + \frac{1}{5}\cos^5 x + \frac{2}{15}$. At $x = \pi/2$, $\cos \pi/2 = 0$, so $f(\pi/2) = 0 + 0 + \frac{2}{15} = \frac{2}{15}$. This is a Hard problem because it relates to finding the area under the curve of the derivative or the value of the function from its derivative graph. The positive area from 0 to pi/2 is the value of the function at pi/2.

📖 Explanation: The alternative method for integrating powers is designed for integrals with trigonometric functions (sines, cosines, tangents, secants) and their powers. It is not designed for exponential or polynomial functions. $\int x e^x dx$ is a classic integration by parts problem. While the alternative method could be manipulated, it would be extremely inefficient and is not the intended approach. The other options are all trigonometric integrals for which the alternative method is perfectly suited. This question tests the student's ability to identify the scope of the integration technique and compare it with other methods like integration by parts.

📖 Explanation: Using the identity $\sin^3 x = \frac{3\sin x - \sin 3x}{4}$, the integral becomes $\frac{3}{4}\int \sin x dx - \frac{1}{4}\int \sin 3x dx = -\frac{3}{4}\cos x + \frac{1}{12}\cos 3x + C$. This is an alternative method using product-to-sum identities. It is equivalent to the forms found by $u = \cos x$. For example, $\frac{1}{3}\cos^3 x - \cos x = -\frac{2}{3}\cos x + \frac{1}{3}\cos^3 x$. Using $\cos^3 x = \frac{\cos 3x + 3\cos x}{4}$, we get $-\frac{2}{3}\cos x + \frac{1}{12}\cos 3x + \frac{1}{4}\cos x = -\frac{5}{12}\cos x + \frac{1}{12}\cos 3x$. There is a discrepancy, which means the two results differ by a constant. This tests the ability to use different identities and compare results.

📖 Explanation: The substitution $u = \tan x$ is used when the power of the secant function, $m$, is even and positive. This is because we can split off a factor of $\sec^2 x$, which is exactly the derivative of $\tan x$, and rewrite the remaining even power of secant in terms of $\tan x$ using the identity $\sec^2 x = 1 + \tan^2 x$. If $n$ is odd, $u = \sec x$ is preferred. This question isolates the key condition for choosing the $u = \tan x$ substitution, which is a cornerstone of the alternative method for these integrals.

📖 Explanation: The student's approach is correct. The power of tangent is odd (3), so the recommended procedure is to split off a $\sec x \tan x$ factor (which is the derivative of $\sec x$) and rewrite the remaining even power of tangent in terms of secant. The integrand is $\sec^5 x \tan^3 x = \sec^5 x \tan^2 x \tan x$. To get a $\sec x \tan x$ factor, we rewrite $\tan^2 x = \sec^2 x - 1$. So the integrand becomes $\sec^5 x (\sec^2 x - 1) \tan x$. To have a $\sec x \tan x$ factor, we pull one $\sec x$ out of $\sec^5 x$, leaving $\sec^4 x$. So it is $(\sec^2 x - 1) \sec^4 x (\sec x \tan x)$. Their expression is perfectly correct for substituting $u = \sec x$. The other options represent common misconceptions about how to split the integrand.