| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the indefinite integral of the function \( 7 \sin (1-7 x) \), we will use the substitution method. Let's proceed step by step. 1. **Substitution**: Let \( u = 1 - 7x \). Then, the differential \( du \) is given by: \[ du = -7 dx \implies dx = -\frac{1}{7} du \] 2. **Rewrite the integral**: Substitute \( u \) and \( dx \) into the integral: \[ \int 7 \sin (1-7 x) \, dx = \int 7 \sin u \cdot -\frac{1}{7} \, du \] 3. **Simplify the integral**: The constants \( 7 \) and \( -\frac{1}{7} \) cancel each other out: \[ \int 7 \sin u \cdot -\frac{1}{7} \, du = -\int \sin u \, du \] 4. **Integrate**: The integral of \( \sin u \) is \( -\cos u \): \[ -\int \sin u \, du = -\left( -\cos u \right) = \cos u \] 5. **Substitute back \( u \)**: Recall that \( u = 1 - 7x \): \[ \cos u = \cos (1 - 7x) \] 6. **Add the constant of integration**: The indefinite integral includes an arbitrary constant \( C \): \[ \int 7 \sin (1-7 x) \, dx = \cos (1 - 7x) + C \] Thus, the final answer is: \[ \boxed{\cos (1 - 7x) + C} \] |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| deepseek-r1 | To find the indefinite integral of \(7 \sin(1 - 7x)\), we use the substitution method. Let \(u = 1 - 7x\). Then, \(du = -7 \, dx\), which implies \(dx = -\frac{du}{7}\). Substituting into the integral, we get: \[ \int 7 \sin(1 - 7x) \, dx = \int 7 \sin(u) \left(-\frac{du}{7}\right) = -\int \sin(u) \, du \] The integral of \(\sin(u)\) is \(-\cos(u) + C\). Therefore, \[ -\int \sin(u) \, du = -(-\cos(u)) + C = \cos(u) + C \] Substituting back \(u = 1 - 7x\), we obtain: \[ \cos(1 - 7x) + C \] To verify, we differentiate \(\cos(1 - 7x) + C\): \[ \frac{d}{dx} [\cos(1 - 7x) + C] = -\sin(1 - 7x) \cdot (-7) = 7 \sin(1 - 7x) \] This confirms the result is correct. Therefore, the indefinite integral is \[ \boxed{\cos(1 - 7x) + C} \] |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| gpto1 | To find the indefinite integral of the function \(7 \sin(1 - 7x)\), we first consider the integral: \[ \int 7 \sin(1 - 7x) \, dx. \] We can factor out the constant 7: \[ 7 \int \sin(1 - 7x) \, dx. \] Next, we perform a substitution to simplify the integral. Let: \[ u = 1 - 7x \implies du = -7 \, dx \implies dx = -\frac{1}{7} du. \] Substituting back into the integral: \[ 7 \int \sin(u) \left(-\frac{1}{7} du\right) = -\int \sin(u) \, du. \] Integrating \(-\sin(u)\) with respect to \(u\): \[ -\int \sin(u) \, du = -(-\cos(u)) + C = \cos(u) + C. \] Finally, substituting back \(u = 1 - 7x\): \[ \cos(1 - 7x) + C. \] Therefore, the indefinite integral of \(7 \sin(1 - 7x)\) is: \[ \boxed{\cos(1 - 7x) + C}. \] \boxed{\cos(1\,-\,7x)\,+\,C} |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| gpt-4o-2024-08-06 | To find the indefinite integral of the function \( 7 \sin(1 - 7x) \), we start by considering the integral: \[ \int 7 \sin(1 - 7x) \, dx \] We can factor out the constant 7 from the integral: \[ = 7 \int \sin(1 - 7x) \, dx \] To solve this integral, we use a substitution method. Let: \[ u = 1 - 7x \] Then, the derivative of \( u \) with respect to \( x \) is: \[ \frac{du}{dx} = -7 \] This implies: \[ dx = -\frac{1}{7} \, du \] Substituting \( u \) and \( dx \) into the integral, we have: \[ 7 \int \sin(u) \left(-\frac{1}{7}\right) \, du = -\int \sin(u) \, du \] The integral of \(\sin(u)\) is: \[ -\cos(u) + C \] Thus, the integral becomes: \[ -\left(-\cos(u)\right) + C = \cos(u) + C \] Substituting back \( u = 1 - 7x \), we get: \[ \cos(1 - 7x) + C \] Therefore, the indefinite integral of \( 7 \sin(1 - 7x) \) is: \[ \boxed{\cos(1 - 7x) + C} \] |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| DeepSeek-Coder-V2-Lite-Instruct | To find the indefinite integral of the function \( 7 \sin (1-7 x) \), we will use the method of substitution. Let's proceed step-by-step: 1. **Substitution**: Let \( u = 1 - 7x \). Then, we need to find \( du \) in terms of \( dx \): \[ du = -7 \, dx \quad \Rightarrow \quad dx = -\frac{1}{7} \, du \] 2. **Rewrite the integral**: Substitute \( u \) and \( dx \) into the integral: \[ \int 7 \sin (1-7 x) \, dx = \int 7 \sin u \left( -\frac{1}{7} \, du \right) \] 3. **Simplify the integral**: The constant 7 in the numerator and denominator cancels out: \[ \int 7 \sin u \left( -\frac{1}{7} \, du \right) = -\int \sin u \, du \] 4. **Integrate**: The integral of \( -\sin u \) is: \[ -\int \sin u \, du = -\cos u + C \] 5. **Substitute back**: Replace \( u \) with \( 1 - 7x \): \[ -\cos u + C = -\cos (1 - 7x) + C \] Thus, the indefinite integral of \( 7 \sin (1-7 x) \) is: \[ -\cos (1 - 7x) + C \] Therefore, the final answer is: \[ \boxed{-\cos (1 - 7x) + C} \] |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the integral to be solved The integral we need to solve is $\int 7 \sin (1-7 x) dx$. ## Step 2: Recall the general formula for the integral of sine The general formula for the integral of $\sin(ax)$ is $-\frac{1}{a} \cos(ax) + C$, where $a$ is a constant and $C$ is the constant of integration. ## Step 3: Apply the formula to our specific integral In our case, the integral is $\sin(1-7x)$. To apply the formula, we recognize that $a = -7$ because the argument of the sine function is $1-7x$, which we can treat as $-7(x - \frac{1}{7})$. However, since the derivative of $1-7x$ with respect to $x$ is $-7$, we directly use this as our $a$ for the purpose of integration, but we must also account for the fact that we're integrating $\sin$ of a function of $x$ rather than just $\sin(ax)$. The integral then becomes $-\frac{1}{-7} \cos(1-7x) + C$. ## Step 4: Simplify the integral Simplifying the integral gives us $\frac{1}{7} \cos(1-7x) + C$. ## Step 5: Account for the coefficient outside the sine function Since the original function is $7 \sin (1-7 x)$, we must multiply our integral by 7 to account for the coefficient, resulting in $7 \cdot \frac{1}{7} \cos(1-7x) + C$. ## Step 6: Simplify the expression Simplifying the expression gives us $\cos(1-7x) + C$. The final answer is: $\boxed{\cos(1-7x) + C}$ |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the integral to be solved The given function to integrate is $7 \sin (1-7 x)$. ## Step 2: Recall the integral formula for sine function The integral of $\sin(ax+b)$ with respect to $x$ is given by $-\frac{1}{a} \cos(ax+b) + C$, where $a$ and $b$ are constants and $C$ is the constant of integration. ## Step 3: Apply the integral formula to the given function Here, $a = -7$ and $b = 1$. Applying the formula, we get the integral of $7 \sin (1-7 x)$ as $7 \times \left( -\frac{1}{-7} \cos(1-7x) \right) + C$. ## Step 4: Simplify the integral expression Simplifying the expression gives $\cos(1-7x) + C$. The final answer is: $\boxed{\cos(1-7x) + C}$ |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| Qwen2-72B-Instruct | To find the indefinite integral of \(7 \sin(1-7x)\), we use substitution. Let's set \(u = 1 - 7x\). Then, \(du = -7dx\), so \(dx = -\frac{1}{7}du\). Substituting these into the integral gives: \[ \int 7 \sin(1-7x) dx = \int 7 \sin(u) \left(-\frac{1}{7}\right) du = -\int \sin(u) du. \] The antiderivative of \(\sin(u)\) is \(-\cos(u)\), so: \[ -\int \sin(u) du = -(-\cos(u)) + C = \cos(u) + C. \] Substituting back for \(u\) gives: \[ \cos(1-7x) + C. \] Therefore, the indefinite integral of \(7 \sin(1-7x)\) is: \[ \boxed{\cos(1-7x) + C}. \] |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| baichuan4 | To find the indefinite integral of the function $7 \sin(1-7x)$, we can use the standard integration technique for sine functions. The general rule for integrating sine functions is: $$\int \sin(ax+b) \, dx = -\frac{1}{a} \cos(ax+b) + C$$ where $a$ and $b$ are constants and $C$ is the constant of integration. In our case, we have $a = -7$ and $b = 1$. Therefore, we can apply the rule to find the integral: $$\int 7 \sin(1-7x) \, dx = 7 \int \sin(1-7x) \, dx$$ Using the rule, we get: $$7 \int \sin(1-7x) \, dx = 7 \left( -\frac{1}{-7} \cos(1-7x) \right) + C$$ Simplifying this expression, we obtain: $$7 \left( \frac{1}{7} \cos(1-7x) \right) + C = \cos(1-7x) + C$$ Thus, the indefinite integral of the function $7 \sin(1-7x)$ is: $$\boxed{\cos(1-7x) + C}$$ where $C$ is the constant of integration. |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| doubao | Let $u=1-7x$, then $du=-7dx$. So the integral becomes: $\int 7\sin(1-7x)dx=\int 7\sin u \cdot (-\frac{1}{7}du)=-\int \sin u du=-\cos u+C=-\cos(1-7x)+C$. So the answer is $\boxed{-\cos(1-7x)+C}$. |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| gpto1 | To find the indefinite integral of the function \(7 \sin(1 - 7x)\), we first consider the integral: \[ \int 7 \sin(1 - 7x) \, dx. \] We can factor out the constant 7: \[ 7 \int \sin(1 - 7x) \, dx. \] Next, we perform a substitution to simplify the integral. Let: \[ u = 1 - 7x \implies du = -7 \, dx \implies dx = -\frac{1}{7} du. \] Substituting back into the integral: \[ 7 \int \sin(u) \left(-\frac{1}{7} du\right) = -\int \sin(u) \, du. \] Integrating \(-\sin(u)\) with respect to \(u\): \[ -\int \sin(u) \, du = -(-\cos(u)) + C = \cos(u) + C. \] Finally, substituting back \(u = 1 - 7x\): \[ \cos(1 - 7x) + C. \] Therefore, the indefinite integral of \(7 \sin(1 - 7x)\) is: \[ \boxed{\cos(1 - 7x) + C}. \] \boxed{\cos(1\,-\,7x)\,+\,C} |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| kimi | 1. **Identify the function to integrate**: We need to find the indefinite integral of the function \(7 \sin(1 - 7x)\). 2. **Rewrite the function using trigonometric identities**: Recall the trigonometric identity for the sine of a difference: \[ \sin(a - b) = \sin a \cos b - \cos a \sin b \] Here, \(a = 1\) and \(b = 7x\). Thus, \[ \sin(1 - 7x) = \sin 1 \cos(7x) - \cos 1 \sin(7x) \] 3. **Substitute this into the original function**: \[ 7 \sin(1 - 7x) = 7(\sin 1 \cos(7x) - \cos 1 \sin(7x)) \] Simplify the expression: \[ 7 \sin(1 - 7x) = 7 \sin 1 \cos(7x) - 7 \cos 1 \sin(7x) \] 4. **Integrate term by term**: We now integrate each term separately. - For the first term, \(7 \sin 1 \cos(7x)\): \[ \int 7 \sin 1 \cos(7x) \, dx \] Since \(\sin 1\) is a constant, we can factor it out: \[ 7 \sin 1 \int \cos(7x) \, dx \] The integral of \(\cos(7x)\) is: \[ \int \cos(7x) \, dx = \frac{1}{7} \sin(7x) \] Thus, \[ 7 \sin 1 \int \cos(7x) \, dx = 7 \sin 1 \cdot \frac{1}{7} \sin(7x) = \sin 1 \sin(7x) \] - For the second term, \(-7 \cos 1 \sin(7x)\): \[ \int -7 \cos 1 \sin(7x) \, dx \] Since \(\cos 1\) is a constant, we can factor it out: \[ -7 \cos 1 \int \sin(7x) \, dx \] The integral of \(\sin(7x)\) is: \[ \int \sin(7x) \, dx = -\frac{1}{7} \cos(7x) \] Thus, \[ -7 \cos 1 \int \sin(7x) \, dx = -7 \cos 1 \left(-\frac{1}{7} \cos(7x)\right) = \cos 1 \cos(7x) \] 5. **Combine the results**: \[ \int 7 \sin(1 - 7x) \, dx = \sin 1 \sin(7x) + \cos 1 \cos(7x) \] 6. **Add the constant of integration**: Since we are finding an indefinite integral, we add the constant \(C\): \[ \int 7 \sin(1 - 7x) \, dx = \sin 1 \sin(7x) + \cos 1 \cos(7x) + C \] ### Conclusion: \[ \boxed{\sin 1 \sin(7x) + \cos 1 \cos(7x) + C} \] |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| minimax | To find the indefinite integral of the function $7 \sin(1 - 7x)$, we can use the substitution method. Let's set $u = 1 - 7x$. Then, we differentiate both sides with respect to $x$ to find $\frac{du}{dx}$: $$\frac{du}{dx} = -7$$ Now, we can solve for $dx$: $$dx = -\frac{du}{7}$$ Substituting $u$ and $dx$ into the integral, we get: $$\int 7 \sin(1 - 7x) \, dx = \int 7 \sin(u) \left(-\frac{du}{7}\right)$$ Simplify the integral: $$= -\int \sin(u) \, du$$ The integral of $\sin(u)$ is $-\cos(u)$. So we have: $$= -(-\cos(u)) + C$$ $$= \cos(u) + C$$ Now, we substitute back the original variable $x$ for $u$: $$u = 1 - 7x$$ So the final answer is: $$\boxed{\cos(1 - |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| wenxinyiyan | 首先，我们需要找到函数$7 \sin (1-7 x)$的一个原函数。 我们可以利用换元积分法，令$u = 1 - 7x$，则$du = -7 dx$。 将$7 \sin (1-7 x)$写为$- \sin u du$，然后对其进行积分，得到： $\int{- \sin u du} = \cos u + C$其中$C$是积分常数。 将$u = 1 - 7x$代回，得到： $\cos (1 - 7x) + C$所以，函数$7 \sin (1-7 x)$的一个不定积分为$\cos (1 - 7x) + C$。 最终答案为：$\boxed{\cos (1 - 7x) + C}$。 |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| zhipu | To find the indefinite integral of the function \( 7 \sin (1-7x) \), we can use the method of substitution. 1. **Substitution:** Let \( u = 1 - 7x \). Then, the differential \( du \) is given by: \[ du = -7 \, dx \quad \Rightarrow \quad dx = -\frac{1}{7} \, du \] 2. **Rewrite the Integral:** Substitute \( u \) and \( dx \) into the integral: \[ \int 7 \sin(1 - 7x) \, dx = \int 7 \sin(u) \left( -\frac{1}{7} \, du \right) \] 3. **Simplify the Integral:** The \( 7 \) and \( -\frac{1}{7} \) will cancel each other out: \[ \int 7 \sin(u) \left( -\frac{1}{7} \, du \right) = -\int \sin(u) \, du \] 4. **Integrate:** The integral of \( \sin(u) \) is \( -\cos(u) \): \[ -\int \sin(u) \, du = -(-\cos(u)) = \cos(u) \] 5. **Substitute Back:** Replace \( u \) with the original expression \( 1 - 7x \): \[ \cos(u) = \cos(1 - 7x) \] 6. **Include the Constant of Integration:** Don't forget to add the constant of integration \( C \): \[ \int 7 \sin(1 - 7x) \, dx = \cos(1 - 7x) + C \] Therefore, the indefinite integral of \( 7 \sin(1 - 7x) \) is: \[ \boxed{\cos(1 - 7x) + C} \] |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |
| deepseek-v3 | To find the indefinite integral of the function \( 7 \sin(1 - 7x) \), follow these steps: 1. **Identify the Integral:** \[ \int 7 \sin(1 - 7x) \, dx \] 2. **Factor Out the Constant:** \[ 7 \int \sin(1 - 7x) \, dx \] 3. **Use Substitution:** Let \( u = 1 - 7x \). Then, \( du = -7 \, dx \), which implies \( dx = -\frac{1}{7} du \). 4. **Substitute and Integrate:** \[ 7 \int \sin(u) \left(-\frac{1}{7}\right) du = - \int \sin(u) \, du \] The integral of \( \sin(u) \) is \( -\cos(u) \): \[ - \int \sin(u) \, du = -(-\cos(u)) + C = \cos(u) + C \] 5. **Substitute Back for \( u \):** \[ \cos(1 - 7x) + C \] 6. **Final Answer:** \[ \boxed{\cos(1 - 7x) + C} \] |
7 \left(\frac{1}{7} \sin (1) \sin (7 x)+\frac{1}{7} \cos (1) \cos (7 x)\right) |