| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the indefinite integral of the function \( 5 \sec^2(5x + 1) - 8 \sin(7 - 8x) \), we will use the following integration rules: 1. \(\int \sec^2(ax + b) \, dx = \frac{1}{a} \tan(ax + b) + C\) 2. \(\int \sin(ax + b) \, dx = -\frac{1}{a} \cos(ax + b) + C\) Let's break down the given function into two parts and integrate each part separately. ### Part 1: Integral of \( 5 \sec^2(5x + 1) \) Using the first integration rule: \[ \int 5 \sec^2(5x + 1) \, dx = 5 \int \sec^2(5x + 1) \, dx \] The integral of \(\sec^2(5x + 1)\) is \(\frac{1}{5} \tan(5x + 1)\): \[ 5 \int \sec^2(5x + 1) \, dx = 5 \cdot \frac{1}{5} \tan(5x + 1) = \tan(5x + 1) \] ### Part 2: Integral of \( -8 \sin(7 - 8x) \) Using the second integration rule: \[ \int -8 \sin(7 - 8x) \, dx = -8 \int \sin(7 - 8x) \, dx \] The integral of \(\sin(7 - 8x)\) is \(-\frac{1}{8} \cos(7 - 8x)\): \[ -8 \int \sin(7 - 8x) \, dx = -8 \cdot -\frac{1}{8} \cos(7 - 8x) = \cos(7 - 8x) \] ### Combining the Results Combining the results from both parts, we get: \[ \int \left( 5 \sec^2(5x + 1) - 8 \sin(7 - 8x) \right) \, dx = \tan(5x + 1) + \cos(7 - 8x) + C \] Thus, the indefinite integral is: \[ \boxed{\tan(5x + 1) + \cos(7 - 8x) + C} \] |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| deepseek-r1 | To find the indefinite integral of the function \( 5 \sec^2(5x + 1) - 8 \sin(7 - 8x) \), we can break it into two separate integrals: \[ \int \left[ 5 \sec^2(5x + 1) - 8 \sin(7 - 8x) \right] dx = \int 5 \sec^2(5x + 1) dx - \int 8 \sin(7 - 8x) dx \] For the first integral, \( \int 5 \sec^2(5x + 1) dx \), we use the substitution \( u = 5x + 1 \). Then, \( du = 5 dx \) or \( dx = \frac{du}{5} \): \[ \int 5 \sec^2(u) \cdot \frac{du}{5} = \int \sec^2(u) du = \tan(u) + C = \tan(5x + 1) + C \] For the second integral, \( - \int 8 \sin(7 - 8x) dx \), we use the substitution \( v = 7 - 8x \). Then, \( dv = -8 dx \) or \( dx = -\frac{dv}{8} \): \[ - \int 8 \sin(v) \cdot \left( -\frac{dv}{8} \right) = - \left( - \int \sin(v) dv \right) = - \left( -\cos(v) \right) + C = \cos(v) + C = \cos(7 - 8x) + C \] However, we need to correct the sign here. The integral of \( 8 \sin(7 - 8x) dx \) is \( -\cos(7 - 8x) + C \), so the negative of this integral becomes \( \cos(7 - 8x) + C \). But upon re-evaluating the substitution steps carefully, we find: \[ - \int 8 \sin(7 - 8x) dx = - \left( -\cos(7 - 8x) \right) + C = \cos(7 - 8x) + C \] But differentiating our result shows the correct sign. The integral of \( 8 \sin(7 - 8x) dx \) is \( -\cos(7 - 8x) + C \), so the negative of this integral is \( \cos(7 - 8x) + C \). However, differentiating the combined result shows that the correct integral should be: \[ \tan(5x + 1) - \cos(7 - 8x) + C \] Combining both results, we get the indefinite integral: \[ \boxed{\tan(5x + 1) - \cos(7 - 8x) + C} \] |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| gpto1 | To find the indefinite integral of the function \( f(x) = 5 \sec^2(5x+1) -8 \sin(7 -8 x) \), we will integrate each term separately. **First term**: \[ \int 5 \sec^2(5x+1) \, dx \] Recall that the derivative of \(\tan(u)\) is \(\sec^2(u) \cdot u'\). Therefore, the integral of \(\sec^2(u)\) is \(\tan(u)\) divided by the derivative of \(u\): \[ \int \sec^2(au+b) \, dx = \frac{1}{a} \tan(au + b) + C \] Applying this to our function: \[ \int 5 \sec^2(5x+1) \, dx = \frac{5}{5} \tan(5x + 1) + C = \tan(5x + 1) + C \] **Second term**: \[ -8 \int \sin(7 -8 x) \, dx \] The integral of \(\sin(u)\) is \(-\cos(u)\) divided by the derivative of \(u\): \[ \int \sin(au + b) \, dx = -\frac{1}{a} \cos(au + b) + C \] So we have: \[ -8 \int \sin(7 -8 x) \, dx = -8 \left( -\frac{1}{8} \cos(7 -8 x) \right) + C = -\cos(7 -8 x) + C \] **Combining the results**: \[ \int f(x) \, dx = \tan(5x + 1) - \cos(7 -8 x) + C \] **Final Answer**: \[ \boxed{\,\tan(5x+1)\ -\ \cos(7\ -\ 8x)\ +\ C\,} \] \boxed{\,\tan(5x+1)\ -\ \cos(7\ -\ 8x)\ +\ C\,} |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| gpt-4o-2024-08-06 | To find the indefinite integral of the function \(5 \sec^2(5x+1) - 8 \sin(7-8x)\), we will integrate each term separately. 1. **Integrating \(5 \sec^2(5x+1)\):** The integral of \(\sec^2(u)\) with respect to \(u\) is \(\tan(u) + C\). Here, \(u = 5x + 1\). Therefore, we have: \[ \int 5 \sec^2(5x+1) \, dx = 5 \int \sec^2(5x+1) \, dx \] Let \(u = 5x + 1\), then \(du = 5 \, dx\) or \(dx = \frac{1}{5} \, du\). Substituting, we get: \[ 5 \int \sec^2(u) \cdot \frac{1}{5} \, du = \int \sec^2(u) \, du = \tan(u) + C = \tan(5x+1) + C \] 2. **Integrating \(-8 \sin(7-8x)\):** The integral of \(\sin(v)\) with respect to \(v\) is \(-\cos(v) + C\). Here, \(v = 7 - 8x\). Therefore, we have: \[ \int -8 \sin(7-8x) \, dx = -8 \int \sin(7-8x) \, dx \] Let \(v = 7 - 8x\), then \(dv = -8 \, dx\) or \(dx = -\frac{1}{8} \, dv\). Substituting, we get: \[ -8 \int \sin(v) \cdot -\frac{1}{8} \, dv = \int \sin(v) \, dv = -\cos(v) + C = -\cos(7-8x) + C \] Therefore, the integral becomes: \[ -8(-\cos(7-8x)) = 8 \cos(7-8x) \] Combining both results, the indefinite integral of the given function is: \[ \int \left(5 \sec^2(5x+1) - 8 \sin(7-8x)\right) \, dx = \tan(5x+1) + 8 \cos(7-8x) + C \] Thus, the final answer is: \[ \boxed{\tan(5x+1) + 8 \cos(7-8x) + C} \] |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| DeepSeek-Coder-V2-Lite-Instruct | To find the indefinite integral of the function \( 5 \sec^2(5x+1) - 8 \sin(7-8x) \), we will integrate each term separately. 1. **Integrate \( 5 \sec^2(5x+1) \):** The integral of \( \sec^2(u) \) is \( \tan(u) \), where \( u \) is a function of \( x \). Here, \( u = 5x + 1 \). \[ \int 5 \sec^2(5x+1) \, dx = 5 \int \sec^2(5x+1) \, dx = 5 \tan(5x+1) \] 2. **Integrate \( -8 \sin(7-8x) \):** The integral of \( \sin(u) \) is \( -\cos(u) \), where \( u \) is a function of \( x \). Here, \( u = 7 - 8x \). \[ \int -8 \sin(7-8x) \, dx = -8 \int \sin(7-8x) \, dx = -8 \left( -\frac{1}{8} \cos(7-8x) \right) = 8 \cos(7-8x) \] Combining these results, we get: \[ \int (5 \sec^2(5x+1) - 8 \sin(7-8x)) \, dx = 5 \tan(5x+1) + 8 \cos(7-8x) + C \] Thus, the final answer is: \[ \boxed{5 \tan(5x+1) + 8 \cos(7-8x) + C} \] |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the integral to be solved The given function to integrate is $5 \sec ^2(5 x+1)-8 \sin (7-8 x)$. ## Step 2: Break down the integral into manageable parts The integral can be broken down into two separate integrals: $\int 5 \sec ^2(5 x+1) dx - \int 8 \sin (7-8 x) dx$. ## Step 3: Solve the first integral The first integral is $\int 5 \sec ^2(5 x+1) dx$. This is a standard integral that results in $\tan(5x+1)$, considering the coefficient $5$ and the chain rule due to the $5x+1$ inside the $\sec^2$ function. So, the solution to the first integral is $5 \cdot \frac{1}{5} \tan(5x+1) = \tan(5x+1)$. ## Step 4: Solve the second integral The second integral is $\int -8 \sin (7-8 x) dx$. This can be solved by recognizing it as a standard integral of the sine function, which results in the cosine function. Considering the coefficient $-8$ and the chain rule due to the $7-8x$ inside the sine function, the solution involves a $-\frac{1}{8}$ factor due to the derivative of $7-8x$ being $-8$. So, the solution to the second integral is $-8 \cdot -\frac{1}{8} \cos(7-8x) = \cos(7-8x)$. ## Step 5: Combine the solutions of the two integrals Combining the solutions from step 3 and step 4, the indefinite integral of the given function is $\tan(5x+1) + \cos(7-8x) + C$, where $C$ is the constant of integration. The final answer is: $\boxed{\tan(5x+1) + \cos(7-8x)}$ |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the components of the function that need to be integrated. The function given is $5 \sec ^2(5 x+1)-8 \sin (7-8 x)$. This function has two components that need to be integrated separately: $5 \sec ^2(5 x+1)$ and $-8 \sin (7-8 x)$. ## Step 2: Integrate the first component, $5 \sec ^2(5 x+1)$. The integral of $\sec ^2(u)$ is $\tan(u)$. Here, $u = 5x + 1$, so we need to adjust the integral accordingly. The derivative of $u$ with respect to $x$ is $5$, so we will have $\frac{1}{5} \tan(5x + 1)$ as part of the integral, but since the coefficient of $\sec ^2(5 x+1)$ is $5$, the integral of this part becomes $5 \cdot \frac{1}{5} \tan(5x + 1) = \tan(5x + 1)$. ## Step 3: Integrate the second component, $-8 \sin (7-8 x)$. The integral of $\sin(u)$ is $-\cos(u)$. Here, $u = 7 - 8x$, so we adjust the integral accordingly. The derivative of $u$ with respect to $x$ is $-8$, so we will have $-\frac{1}{-8} \cos(7 - 8x)$ as part of the integral, but since the coefficient of $\sin (7-8 x)$ is $-8$, the integral of this part becomes $-8 \cdot -\frac{1}{-8} \cos(7 - 8x) = \cos(7 - 8x)$. ## Step 4: Combine the results of the integrals of the two components. The integral of the entire function is the sum of the integrals of its components: $\tan(5x + 1) + \cos(7 - 8x)$. ## Step 5: Add a constant of integration to the result. Since this is an indefinite integral, we must add a constant of integration, $C$, to the result: $\tan(5x + 1) + \cos(7 - 8x) + C$. The final answer is: $\boxed{\tan(5x + 1) + \cos(7 - 8x) + C}$ |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| Qwen2-72B-Instruct | To find the indefinite integral of the given function, we need to integrate each term separately. The first term is $5\sec^2(5x + 1)$, and its integral is straightforward because the derivative of $\tan(x)$ is $\sec^2(x)$. However, we have a chain rule factor here due to the argument $5x + 1$, so we need to adjust for that. The integral of this term will be: $$\int 5\sec^2(5x + 1) dx = \frac{5}{5}\tan(5x + 1) + C_1 = \tan(5x + 1) + C_1$$ The second term is $-8\sin(7 - 8x)$. This term also requires us to account for the chain rule factor due to the argument $7 - 8x$. The integral of this term will be: $$\int -8\sin(7 - 8x) dx = \frac{-8}{-8}\cos(7 - 8x) + C_2 = \cos(7 - 8x) + C_2$$ Combining both integrals, we get: $$\int (5\sec^2(5x + 1) - 8\sin(7 - 8x)) dx = \tan(5x + 1) + \cos(7 - 8x) + C$$ where $C = C_1 + C_2$ is the combined constant of integration. Therefore, the indefinite integral of the given function is: $$\boxed{\tan(5x + 1) + \cos(7 - 8x) + C}$$ |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| baichuan4 | To find the indefinite integral of the given function $5 \sec^2(5x+1) - 8 \sin(7-8x)$, we can split the problem into two separate integrals and solve each one individually. First, we know that the derivative of $\tan(u)$ is $\sec^2(u)$, where $u$ is any differentiable function. Therefore, the integral of $\sec^2(u)$ is $\tan(u) + C$, where $C$ is the constant of integration. For the first part of the integral, $5 \sec^2(5x+1)$, we can set $u = 5x+1$. Then, the derivative of $u$ is $du = 5dx$, and $dx = \frac{du}{5}$. We can substitute this into the integral to get: $$ \int 5 \sec^2(5x+1) dx = \int 5 \sec^2(u) \frac{du}{5} = \int \sec^2(u) du = \tan(u) + C = \tan(5x+1) + C_1 $$ where $C_1$ is the constant of integration for this part of the integral. Next, we deal with the second part of the integral, $-8 \sin(7-8x)$. We know that the derivative of $\cos(u)$ is $-\sin(u)$, so the integral of $-\sin(u)$ is $\cos(u) + C$. For this part, we can set $v = 7-8x$. Then, the derivative of $v$ is $dv = -8dx$, and $dx = -\frac{dv}{8}$. Substituting this into the integral gives: $$ \int -8 \sin(7-8x) dx = \int -8 \sin(v) \left(-\frac{dv}{8}\right) = \int \sin(v) dv = -\cos(v) + C_2 = -\cos(7-8x) +_2 _2 $$ where $C_2$ is the constant of integration for this part of the integral. Finally, we combine the two parts of the integral to get the complete solution: $$ \int \left(5 \sec^2(5x+1) - 8 \sin(7-8x)\right) dx = \tan(5x+1) - \cos(7-8x) + C $$ where $C = C_1 + C_2$ is the overall constant of integration. Therefore, the indefinite integral of the given function is: $$ \boxed{\tan(5x+1) - \cos(7-8x) + C} $$ |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| doubao | The integral of $5\sec^2(5x+1)$ is $\tan(5x+1)$ (using the fact that the integral of $\sec^2 u$ is $\tan u$). The integral of $-8\sin(7-8x)$ is $8\cos(7-8x)\times\frac{1}{8}=\cos(7-8x)$ (using the integration rule for $\sin u$). So the indefinite integral is $\tan(5x+1)+\cos(7-8x)+C$ (where $C$ is the constant of integration), and the final answer is $\boxed{\tan(5x+1)+\cos(7-8x)+C}$. |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| gpto1 | To find the indefinite integral of the function \( f(x) = 5 \sec^2(5x+1) -8 \sin(7 -8 x) \), we will integrate each term separately. **First term**: \[ \int 5 \sec^2(5x+1) \, dx \] Recall that the derivative of \(\tan(u)\) is \(\sec^2(u) \cdot u'\). Therefore, the integral of \(\sec^2(u)\) is \(\tan(u)\) divided by the derivative of \(u\): \[ \int \sec^2(au+b) \, dx = \frac{1}{a} \tan(au + b) + C \] Applying this to our function: \[ \int 5 \sec^2(5x+1) \, dx = \frac{5}{5} \tan(5x + 1) + C = \tan(5x + 1) + C \] **Second term**: \[ -8 \int \sin(7 -8 x) \, dx \] The integral of \(\sin(u)\) is \(-\cos(u)\) divided by the derivative of \(u\): \[ \int \sin(au + b) \, dx = -\frac{1}{a} \cos(au + b) + C \] So we have: \[ -8 \int \sin(7 -8 x) \, dx = -8 \left( -\frac{1}{8} \cos(7 -8 x) \right) + C = -\cos(7 -8 x) + C \] **Combining the results**: \[ \int f(x) \, dx = \tan(5x + 1) - \cos(7 -8 x) + C \] **Final Answer**: \[ \boxed{\,\tan(5x+1)\ -\ \cos(7\ -\ 8x)\ +\ C\,} \] \boxed{\,\tan(5x+1)\ -\ \cos(7\ -\ 8x)\ +\ C\,} |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| kimi | To find the indefinite integral of the function \(5 \sec^2(5x + 1) - 8 \sin(7 - 8x)\), we will integrate each term separately. First, consider the term \(5 \sec^2(5x + 1)\). 1. **Integrate \(5 \sec^2(5x + 1)\):** - Recall that the derivative of \(\tan(u)\) with respect to \(u\) is \(\sec^2(u)\). - Let \(u = 5x + 1\). Then, \(du = 5 \, dx\), so \(dx = \frac{1}{5} \, du\). - Substitute \(u\) and \(dx\) into the integral: \[ \int 5 \sec^2(5x + 1) \, dx = \int 5 \sec^2(u) \cdot \frac{1}{5} \, du = \int \sec^2(u) \, du \] - The integral of \(\sec^2(u)\) is \(\tan(u)\): \[ \int \sec^2(u) \, du = \tan(u) + C \] - Substitute back \(u = 5x + 1\): \[ \tan(5x + 1) + C \] Next, consider the term \(-8 \sin(7 - 8x)\). 2. **Integrate \(-8 \sin(7 - 8x)\):** - Recall that the derivative of \(\cos(u)\) with respect to \(u\) is \(-\sin(u)\). - Let \(u = 7 - 8x\). Then, \(du = -8 \, dx\), so \(dx = -\frac{1}{8} \, du\). - Substitute \(u\) and \(dx\) into the integral: \[ \int -8 \sin(7 - 8x) \, dx = \int -8 \sin(u) \cdot -\frac{1}{8} \, du = \int \sin(u) \, du \] - The integral of \(\sin(u)\) is \(-\cos(u)\): \[ \int \sin(u) \, du = -\cos(u) + C \] - Substitute back \(u = 7 - 8x\): \[ -\cos(7 - 8x) + C \] Now, combine the results of the two integrals: \[ \int (5 \sec^2(5x + 1) - 8 \sin(7 - 8x)) \, dx = \tan(5x + 1) - \cos(7 - 8x) + C \] Thus, the indefinite integral is: \[ \boxed{\tan(5x + 1) - \cos(7 - 8x) + C} \] |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| minimax | To find the indefinite integral of the function $f(x) = 5 \sec^2(5x+1) - 8 \sin(7-8x)$, we integrate each term separately. The first term is a standard integral of the form $\int \sec^2(ax) \, dx$, which is $\frac{1}{a} \tan(ax) + C_1$, where $C_1$ is the constant of integration for the first term. The second term is a standard integral of the form $\int \sin(bx) \, dx$, which is $-\frac{1}{b} \cos(bx) + C_2$, where $b$ is a constant and $C_2$ is the constant of integration for the second term. Let's integrate each term: 1. For the first term $5 \sec^2(5x+1)$: $$\int 5 \sec^2(5x+1) \, dx = \frac{5}{5} \int \sec^2(5x+1) \, d(5x+1) = \tan(5x+1) + C_1$$ 2. For the second term |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| wenxinyiyan | 首先，我们分别考虑两个部分的积分。 对于 $5 \sec ^2(5 x+1)$，我们可以使用积分公式 $\int \sec ^2(u) du = \tan(u) + C$，其中 $u$ 是某个函数。 令 $u = 5x + 1$，则 $du = 5 dx$。 因此，$\int 5 \sec ^2(5 x+1) dx = \tan(5x + 1) + C_1$，其中 $C_1$ 是积分常数。 对于 $-8 \sin (7-8 x)$，我们可以使用积分公式 $\int \sin(u) du = -\cos(u) + C$，其中 $u$ 是某个函数。 令 $u = 7 - 8x$，则 $du = -8 dx$。 因此，$\int -8 \sin (7-8 x) dx = \cos(7 - 8x) + C_2$，其中 $C_2$ 是积分常数。 综合以上两部分，我们得到： $\int \left( 5 \sec ^2(5 x+1)-8 \sin (7-8 x) \right) dx = \tan(5x + 1) + \cos(7 - 8x) + C$ 其中 $C = C_1 + C_2$ 是积分常数。 所以，最终答案是： $\boxed{\tan(5x + 1) + \cos(7 - 8x) + C}$ |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| zhipu | To find the indefinite integral of the function \( 5 \sec^2(5x + 1) - 8 \sin(7 - 8x) \), we can integrate each term separately. 1. **Integrate the first term \( 5 \sec^2(5x + 1) \):** We know that the integral of \( \sec^2(u) \) with respect to \( u \) is \( \tan(u) \). Here, \( u = 5x + 1 \), so we need to use the chain rule. \[ \int 5 \sec^2(5x + 1) \, dx = 5 \int \sec^2(5x + 1) \, dx \] Let \( u = 5x + 1 \). Then, \( du = 5 \, dx \) or \( dx = \frac{du}{5} \). \[ 5 \int \sec^2(u) \cdot \frac{du}{5} = \int \sec^2(u) \, du = \tan(u) = \tan(5x + 1) \] 2. **Integrate the second term \( -8 \sin(7 - 8x) \):** We know that the integral of \( \sin(u) \) with respect to \( u \) is \( -\cos(u) \). Here, \( u = 7 - 8x \), so we need to use the chain rule. \[ \int -8 \sin(7 - 8x) \, dx = -8 \int \sin(7 - 8x) \, dx \] Let \( u = 7 - 8x \). Then, \( du = -8 \, dx \) or \( dx = \frac{du}{-8} \). \[ -8 \int \sin(u) \cdot \frac{du}{-8} = \int \sin(u) \, du = -\cos(u) = -\cos(7 - 8x) \] 3. **Combine the results:** Adding the integrals of both terms, we get: \[ \int \left( 5 \sec^2(5x + 1) - 8 \sin(7 - 8x) \right) \, dx = \tan(5x + 1) + \cos(7 - 8x) + C \] where \( C \) is the constant of integration. Thus, the indefinite integral of the given function is: \[ \boxed{\tan(5x + 1) + \cos(7 - 8x) + C} \] |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |
| deepseek-v3 | To find the indefinite integral of the function \( 5 \sec^2(5x + 1) - 8 \sin(7 - 8x) \), we'll integrate each term separately. 1. **Integrate \( 5 \sec^2(5x + 1) \):** Recall that the integral of \( \sec^2(u) \) is \( \tan(u) \). Let \( u = 5x + 1 \), then \( du = 5 \, dx \). \[ \int 5 \sec^2(5x + 1) \, dx = \int \sec^2(u) \, du = \tan(u) + C_1 = \tan(5x + 1) + C_1 \] 2. **Integrate \( -8 \sin(7 - 8x) \):** Recall that the integral of \( \sin(u) \) is \( -\cos(u) \). Let \( u = 7 - 8x \), then \( du = -8 \, dx \). \[ \int -8 \sin(7 - 8x) \, dx = \int \sin(u) \, du = -\cos(u) + C_2 = -\cos(7 - 8x) + C_2 \] 3. **Combine the results:** Add the two integrals together and combine the constants \( C_1 \) and \( C_2 \) into a single constant \( C \). \[ \int \left(5 \sec^2(5x + 1) - 8 \sin(7 - 8x)\right) \, dx = \tan(5x + 1) - \cos(7 - 8x) + C \] **Final Answer:** \[ \boxed{ \tan(5x + 1) - \cos(7 - 8x) + C } \] |
-\sin (7) \sin (8 x)-\cos (7) \cos (8 x)+\tan (5 x+1) |