📝 Integrals of sin^n x cos^m x (36 MCQs)

📖 Explanation: The primary goal of reduction formulas like $\int \sin^n x \, dx = -\frac{1}{n}\sin^{n-1}x\cos x + \frac{n-1}{n}\int \sin^{n-2}x \, dx$ is to systematically lower the exponent by 2. This process is repeated until the exponent becomes 0 (if even) or 1 (if odd), at which point standard integrals like $\int 1 \, dx = x$ or $\int \sin x \, dx = -\cos x$ can be applied, making the integral solvable.

📖 Explanation: Since the exponent of cosine is odd (5), the standard strategy is to save one factor of $\cos x$ for $du$ and convert the remaining even power of cosine into sine. This gives $\int \sin^4 x \cos^4 x \cos x \, dx$. Using $\cos^2 x = 1 - \sin^2 x$, the integrand becomes $\int \sin^4 x (1 - \sin^2 x)^2 \cos x \, dx$, and the substitution $u = \sin x$, $du = \cos x \, dx$ simplifies the integral to $\int u^4 (1-u^2)^2 \, du$, which is a straightforward polynomial integration.

📖 Explanation: The correct reduction formula is $\int \sin^3 x \, dx = -\frac{1}{3}\sin^2 x\cos x + \frac{2}{3}\int \sin x \, dx$. The student omitted the integral $\frac{2}{3}\int \sin x \, dx$, which evaluates to $-\frac{2}{3}\cos x$. Thus, the complete answer is $-\frac{1}{3}\sin^2 x\cos x - \frac{2}{3}\cos x + C$. Omitting this term leads to an incorrect antiderivative, as differentiating the student's answer would not yield $\sin^3 x$. This is a common error when students do not fully apply the reduction formula.

📖 Explanation: Squaring the identity gives $\sin^4 x = \left(\frac{1}{2}(1-\cos 2x)\right)^2 = \frac{1}{4}(1 - 2\cos 2x + \cos^2 2x)$. This method is necessary to reduce the power of sine. A common mistake is to forget to square the entire expression or to incorrectly handle the constant factor. Expanding and applying the double-angle identity again to $\cos^2 2x$ will allow the integral to be evaluated in terms of $x$, $\sin 2x$, and $\sin 4x$, as seen in the reduction formula result for $\sin^4 x$.

📖 Explanation: The expression $\sin^2 x \cos^2 x = \frac{1}{4}\sin^2 2x$ is correct. However, integrating $\sin^2 2x$ requires using the identity $\sin^2 2x = \frac{1}{2}(1-\cos 4x)$, not simply treating it as $\sin 2x$. The student's result $-\frac{1}{8}\cos 2x$ would be the integral of $\frac{1}{4}\sin 2x$, not $\frac{1}{4}\sin^2 2x$. This demonstrates a failure to recognize that the power of the sine function must be reduced. The correct answer involves $x$, $\sin 4x$, and $\cos 4x$.

📖 Explanation: When the exponent of sine ($m$) is even and cosine ($n$) is odd, we save one factor of $\cos x$ for $du$. This allows us to express the remaining even power of cosine in terms of sine using $\cos^2 x = 1 - \sin^2 x$. The substitution $u = \sin x$ is then effective. This is a critical distinction; saving the derivative of the substitution function is the key to this technique. If $m$ were odd and $n$ even, we would choose $u = \cos x$.

📖 Explanation: The third procedure in the standard integration strategy for $\sin^m x \cos^n x$ is used when both $m$ and $n$ are even. In this case, the half-angle identities $\sin^2 x = \frac{1}{2}(1-\cos 2x)$ and $\cos^2 x = \frac{1}{2}(1+\cos 2x)$ are applied. The integral $\int \sin^2 x \cos^2 x \, dx$ falls into this category. The other options have at least one odd exponent, which allows for a $u$-substitution and does not necessitate the use of half-angle identities in the same way.

📖 Explanation: By symmetry, the graph of $\sin^4 x$ over $[0,\pi]$ is the reflection of $\cos^4 x$ over the interval $[0,\pi]$. Specifically, $\sin^4 x = \cos^4(\pi/2 - x)$. Since the substitution $u = \pi/2 - x$ maps the interval $[0,\pi]$ to itself, the areas under both curves are identical. This demonstrates a strong Hard of function symmetry and integral properties. Both integrals evaluate to $\frac{3\pi}{8}$, a result that can be verified using reduction formulas.

📖 Explanation: When $m$ is odd, we save a factor of $\sin x$ for $du$, as $du = -\sin x \, dx$. Then $\sin^5 x = \sin^4 x \cdot \sin x = (\sin^2 x)^2 \sin x = (1 - \cos^2 x)^2 \sin x$. With $u = \cos x$, $\sin^5 x \, dx = (1-u^2)^2 \sin x \, dx = -(1-u^2)^2 \, du$. Option D correctly identifies the polynomial in $u$ that remains after the substitution is fully applied. It is crucial to account for the $\sin x$ factor that appears in $du$ and to correctly substitute $\sin^2 x$ with $1-u^2$.

📖 Explanation: The reduction formula is specifically designed to handle integrals of powers of sine or cosine in a systematic way. By setting $n=6$, the formula yields $\int \cos^6 x \, dx = \frac{1}{6}\cos^5 x \sin x + \frac{5}{6}\int \cos^4 x \, dx$. This reduces the problem from a power of 6 to a power of 4, and the process can be repeated until the integral becomes $\int \cos^0 x \, dx = \int 1 \, dx = x$. While other methods exist, the direct Easy of the reduction formula is the most straightforward Easy of this theorem.

📖 Explanation: The integration of products of sine and cosine, i.e., $\int \sin^m x \cos^n x \, dx$, requires a careful analysis of whether $m$, $n$, or both are odd or even. This determines whether a $u$-substitution (with $u = \sin x$ or $u = \cos x$) or the half-angle identities (for the case where both are even) is the most efficient method. Integrating a single power, like $\sin^m x$, usually only requires one reduction formula. The parity analysis is the critical first step in tackling product integrals.

📖 Explanation: The function $f(x) = \sin^7 x \cos^3 x$ is odd about $x = \pi$ because $f(2\pi - x) = -\sin^7 x \cos^3 x = -f(x)$. The interval $[0, 2\pi]$ is symmetric around $\pi$, so the integral evaluates to 0. A common and significant error is to perform the full integration and obtain a complicated expression, then substitute the limits and get a non-zero result, missing the symmetry property. Recognizing symmetry can dramatically simplify the problem and avoid lengthy calculations.

📖 Explanation: Since both exponents are even, the standard strategy is to apply the half-angle identities. The integral becomes $\int \frac{1}{2}(1-\cos 2x) \cdot \frac{1}{4}(1+\cos 2x)^2 \, dx$. This can be expanded and simplified into a sum of integrals involving $\cos 2x$ and $\cos^2 2x$, etc. The half-angle identity approach is systematic and directly reduces the power of the trigonometric functions. Using product-to-sum would also work but is more complex for products with different powers.

📖 Explanation: Using the reduction formula: $\int_0^{\pi/2} \sin^3 x \, dx = [-\frac{1}{3}\sin^2 x \cos x]_0^{\pi/2} + \frac{2}{3} \int_0^{\pi/2} \sin x \, dx$. The first term evaluates to 0. The remaining integral is $\frac{2}{3} [-\cos x]_0^{\pi/2} = \frac{2}{3}(0 + 1) = \frac{2}{3}$. This is a standard result in calculus and a direct Easy of the reduction formula. Understanding these standard values can help in quickly verifying more complex calculations.

📖 Explanation: Using $\sin^2 x = \frac{1}{2}(1 - \cos 2x)$, the integral is $\frac{x}{2} - \frac{\sin 2x}{4} + C$. Using the identity $\sin 2x = 2\sin x \cos x$, this becomes $\frac{x}{2} - \frac{\sin x \cos x}{2} + C$. This demonstrates how different valid forms of the answer can be equivalent. Recognizing both forms is important for comparing answers from different methods or verifying results against CAS outputs. This is a good example of how trigonometric identities allow for multiple representations of the same result.

📖 Explanation: Since the exponent of cosine is odd (3), we save one factor of $\cos x$ for $du$. The remaining factor $\cos^2 x$ is replaced by $1 - \sin^2 x$. This gives $\int \sin^4 x (1 - \sin^2 x) \cos x \, dx$. Option B is incorrect because it uses $1 - \cos^2 x$ which is not the identity for $\cos^2 x$. The substitution $u = \sin x$ then makes this a straightforward polynomial integral. This highlights the essential strategy for integrals with an odd power of cosine.

📖 Explanation: The Wallis sine formula states $\int_0^{\pi/2} \sin^n x \, dx = \frac{\pi}{2} \cdot \frac{1 \cdot 3 \cdot 5 \cdots (n-1)}{2 \cdot 4 \cdot 6 \cdots n}$ for even $n$. Thus, for $n=6$, the integral from 0 to $\pi/2$ is $\frac{\pi}{2} \cdot \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} = \frac{\pi}{2} \cdot \frac{15}{48} = \frac{15\pi}{96} = \frac{5\pi}{32}$. By symmetry, $\int_0^{\pi} \sin^6 x \, dx = 2 \int_0^{\pi/2} \sin^6 x \, dx = 2 \cdot \frac{5\pi}{32} = \frac{5\pi}{16}$. This is a direct Easy of a standard formula and showcases the use of symmetry to extend the interval.

📖 Explanation: Since the exponent of sine is odd, we save a factor of $\sin x$ for $du$. With $u = \cos x$, $du = -\sin x \, dx$. The integrand becomes $\sin^2 x \cos^2 x \sin x \, dx = (1 - \cos^2 x) \cos^2 x \sin x \, dx = (1-u^2)u^2 (\sin x \, dx) = u^2(1-u^2)(-du) = (-u^2 + u^4) du$. This simplifies to $u^4 - u^2$ after multiplying by -1 and reordering. The correct expression is $-u^2 + u^4$, which can be integrated directly as a polynomial. This demonstrates a correct Easy of a $u$-substitution and the relevant trig identity.

📖 Explanation: The product-to-sum identities like $\sin A \cos B = \frac{1}{2}[\sin(A-B) + \sin(A+B)]$ are most effective when the integral is a product of sines and cosines with different arguments, such as $\int \sin mx \cos nx \, dx$. This method avoids repeated use of half-angle identities and is more direct. For $\int \sin 2x \cos 4x \, dx$, the identity immediately gives $\frac{1}{2} \int (\sin(-2x) + \sin(6x)) \, dx$, which is trivial. For products with powers, other strategies like $u$-substitution or half-angle are preferred.

📖 Explanation: Since $du = \cos x \, dx$, the $\cos x$ factor in the integrand is exactly $du$. The $\sin^2 x$ becomes $u^2$. Thus, the integral becomes $\int u^2 \, du$. This is a classic example of the integration by substitution method. The student correctly identifies the derivative of the substitution, which is crucial for simplifying the integral. This is a fundamental skill in calculus and a building block for more complex integrals like those with higher powers.

📖 Explanation: The reduction formula for $\cos^n x$ is derived by letting $u = \cos^{n-1} x$ and $dv = \cos x \, dx$. Then $v = \sin x$ and $du = (n-1)\cos^{n-2} x (-\sin x) dx = -(n-1)\cos^{n-2} x \sin x \, dx$. The integration by parts formula gives $\cos^{n-1} x \sin x + (n-1) \int \sin^2 x \cos^{n-2} x \, dx$. Replacing $\sin^2 x$ with $1 - \cos^2 x$ and solving for the integral yields the formula. The derivative of the power of cosine is a crucial part of the integration by parts process. This question tests the understanding of the derivation of the formula, not just its Easy.

📖 Explanation: Using the disk method, the volume is $V = \pi \int_0^\pi \sin^4 x \, dx$. Using the reduction formula or half-angle identities, $\int_0^\pi \sin^4 x \, dx = \frac{3\pi}{8}$. Therefore, $V = \pi \cdot \frac{3\pi}{8} = \frac{3\pi^2}{8}$. This is a classic Easy of integration techniques to find the volume of a solid of revolution. It combines the disk method with the techniques for integrating powers of sine, demonstrating a multi-step problem that tests both Hard of volume and computational skill with trigonometric integrals.

📖 Explanation: The identity $\sin^3 x = \frac{3}{4}\sin x - \frac{1}{4}\sin 3x$ is correct. Integrating term by term yields $-\frac{3}{4}\cos x + \frac{1}{12}\cos 3x + C$. Using $\cos 3x = 4\cos^3 x - 3\cos x$, the result becomes $-\frac{3}{4}\cos x + \frac{1}{12}(4\cos^3 x - 3\cos x) + C = -\frac{3}{4}\cos x + \frac{1}{3}\cos^3 x - \frac{1}{4}\cos x + C = \frac{1}{3}\cos^3 x - \cos x + C$. This is the known result for $\int \sin^3 x \, dx$. This confirms that the student's answer, while in a different form, is correct. Recognizing equivalent forms is a key skill in calculus.

📖 Explanation: The reduction formula $\int \sin^n x \, dx = -\frac{1}{n}\sin^{n-1} x \cos x + \frac{n-1}{n} \int \sin^{n-2} x \, dx$ reduces the exponent by 2 each time. For $n=12$ (even), it takes 6 Easys to reach $\int \sin^0 x \, dx = \int 1 \, dx = x$. For $n=11$ (odd), it takes 5 Easys to reach $\int \sin^1 x \, dx = \int \sin x \, dx = -\cos x$. This understanding is crucial for efficiently evaluating higher powers and planning the integration path. The number of steps is directly determined by the parity of the exponent.

📖 Explanation: For any positive even integer $n$, $\sin^n x$ has a maximum value of 1 at $x = \pi/2$ because $\sin(\pi/2) = 1$. As $n$ increases, the function becomes 'flatter' near the peaks and drops off more steeply near the edges of the interval, so the area under the curve decreases. This is because values of $\sin x$ less than 1 become smaller when raised to a higher power. This has significant implications for probability and physics, as it describes the shape of distributions. The integral $\int_0^\pi \sin^n x \, dx$ is often used to compute volumes and probabilities.

📖 Explanation: Applying the half-angle identities gives $\sin^4 x = \left( \frac{1}{2}(1-\cos 2x) \right)^2 = \frac{1}{4}(1-\cos 2x)^2$ and $\cos^2 x = \frac{1}{2}(1+\cos 2x)$. Multiplying these gives $\frac{1}{8} (1-\cos 2x)^2 (1+\cos 2x)$. Expanding this expression will produce a sum of powers of $\cos 2x$, which can then be further reduced using the half-angle identity again. This question tests the student's ability to correctly apply the identities and combine constants, a common source of errors.

📖 Explanation: The integral $\int \sin^4 x \cos^3 x \, dx$ has an odd exponent on cosine, so we can factor out one cosine, convert the rest to sine, and use $u$-substitution. The integral $\int \sin^4 x \cos^2 x \, dx$ has both exponents even, requiring the more complex half-angle identities. This demonstrates a critical point in integration strategy: recognizing the parity of the exponents is the key to choosing the most efficient method. The odd exponent makes the integral a straightforward polynomial after substitution, whereas the all-even case is more involved.

📖 Explanation: Since $\sin x \cos x = \frac{1}{2} \sin 2x$, we have $(\sin x \cos x)^4 = \left( \frac{1}{2} \sin 2x \right)^4 = \frac{1}{16} \sin^4 2x$. This is a clever simplification that turns a product of even powers of sine and cosine into a single power of a sine function. This is a very useful trick for this specific form. The student must be careful with the constant factor, as $(1/2)^4 = 1/16$, not $1/8$ or $1/4$. This approach is far more efficient than expanding using half-angle identities multiple times.

📖 Explanation: Since the exponent of sine is odd, we save a factor of $\sin x$ for $du$, so $du = \cos x \, dx$. The integrand becomes $\sin^4 x \cos^2 x \sin x \, dx = (\sin^2 x)^2 \cos^2 x (\sin x \, dx) = (1-\cos^2 x)^2 \cos^2 x (\sin x \, dx)$. With $u = \cos x$, the integrand becomes $-(1-u^2)^2 u^2 \, du$. Option D correctly represents the integrand with the negative sign that comes from $\sin x \, dx = -du$. This is a common point of confusion; the sign must be carefully tracked to avoid an incorrect final answer. The polynomial in $u$ is $-(u^2 - 2u^4 + u^6)$.

📖 Explanation: The function $f(x) = \sin^n x \cos^m x$ has a maximum value that depends on $n$ and $m$. This maximum is not always 1 because the product of two functions less than 1 is less than 1 unless both are 1, which is impossible except at a single point. For example, for $n=m=2$, $f(x) = \sin^2 x \cos^2 x = \frac{1}{4}\sin^2 2x$, which has a maximum of $1/4$ at $x = \pi/4$. This highlights the effect of powers on the shape and magnitude of the function, which has Easys in probability and physics. The maximum value of $\sin^n x \cos^m x$ can be found using calculus and is often used in optimization problems.

📖 Explanation: The integral $\int \sin^2 x \cos^2 x \, dx$ can be rewritten as $\frac{1}{4} \int \sin^2 2x \, dx$. Using the identity $\sin^2 2x = \frac{1}{2}(1-\cos 4x)$, the integral becomes $\frac{1}{8} \int (1-\cos 4x) \, dx = \frac{x}{8} - \frac{\sin 4x}{32} + C$. The student's answer is correct. This question tests the student's ability to recognize a correct result and to understand the steps involved in deriving it. It also highlights the importance of carefully handling constants when applying double-angle identities. The most common errors here are in the constant coefficients, so this is a good way to check for that.

📖 Explanation: The substitution $u = \tan x$ is typically not the first choice for integrals of the form $\int \sin^m x \cos^n x \, dx$ unless $m$ and $n$ are both even or one is negative. For $\int \sin^3 x \cos^4 x \, dx$, the standard $u$-substitution ($u = \cos x$) would be more direct. However, if we were to force a tangent substitution, we would rewrite the integrand in terms of $\tan x$ and $\sec x$, which is not a standard first approach for simple sine-cosine products. This question is designed to test the student's ability to identify the most appropriate method for a given integral, not just follow a rote procedure. The presence of odd and even powers guides the strategy.

📖 Explanation: Using the half-angle identities and the properties of definite integrals, $\int_0^\pi \sin^4 x \cos^2 x \, dx = \frac{1}{8} \int_0^\pi (1-\cos 2x)^2(1+\cos 2x) \, dx$. Expanding and integrating term by term using the fact that $\int_0^\pi \cos kx \, dx = 0$ for integer $k$, the constant term becomes $\frac{1}{8} \int_0^\pi 1 \, dx = \frac{\pi}{8}$, but the average value of the sine and cosine powers over a full period is $\frac{1}{4}$, leading to $\frac{\pi}{16}$. This problem requires careful algebraic manipulation and knowledge of definite integrals of even functions. It also tests the ability to integrate powers of cosine. A full solution would involve expanding the integrand and using the reduction formula or half-angle identities.

📖 Explanation: Since both $m$ and $n$ are even, the standard strategy is to use half-angle identities. The substitution $u = \sin x$ would be less efficient because it would require expressing $\cos^{10} x$ in terms of $u$ and $du$, which is not a simple replacement. The half-angle identities reduce the powers of sine and cosine, leading to a sum of integrals of $\cos kx$, which are straightforward. The substitution method is generally only useful when one exponent is odd. This question tests the student's ability to choose the most appropriate method based on the parity of the exponents.

📖 Explanation: This is a simple $u$-substitution. Let $u = \sin x$, then $du = \cos x \, dx$. The integral becomes $\int u^5 \, du = \frac{u^6}{6} + C = \frac{1}{6}\sin^6 x + C$. The given result for $\int \sin^5 x \, dx$ is a distractor. This question tests the student's ability to recognize that the presence of the derivative of the substitution makes the problem trivial, regardless of the complexity of the reduction formula for the sine-only integral. It highlights the importance of pattern recognition in integration.