Question

$\int \frac{1}{7} \sin\left(\frac{x}{7} + 10\right)\, dx$ is equal to:

• A. $\frac{1}{7} \cos\left(\frac{x}{7} + 10\right) + C$
• B. $-\frac{1}{7} \cos\left(\frac{x}{7} + 10\right) + C$
• C. $-\cos\left(\frac{x}{7} + 10\right) + C$
• D. $-7 \cos\left(\frac{x}{7} + 10\right) + C$
• E. $\cos(x + 70) + C$

Hint:

When integrating, always divide the final result by the coefficient of $x$. In this case, dividing by $1/7$ is equivalent to multiplying by $7$, which perfectly cancels the $1/7$ already present in the question.

Answer 1

The Correct Option is C

Concept: The integral of $\sin(ax + b)$ is given by $-\frac{1}{a} \cos(ax + b) + C$. In this problem, the constant coefficient outside the integral is managed alongside the internal coefficient of $x$.

Step 1: Identify the components of the integrand.
The integrand is $\frac{1}{7} \sin\left(\frac{x}{7} + 10\right)$.
The coefficient of $x$ is $a = \frac{1}{7}$.

Step 2: Apply the integration rule.
The general rule is $\int \sin(ax + b) dx = -\frac{1}{a} \cos(ax + b) + C$. Including the constant $\frac{1}{7}$ from the problem:
\[ I = \frac{1}{7} \left[ -\frac{1}{1/7} \cos\left(\frac{x}{7} + 10\right) \right] + C \]

Step 3: Simplify the expression.
The term $\frac{1}{1/7}$ simplifies to $7$: \[ I = \frac{1}{7} \left[ -7 \cos\left(\frac{x}{7} + 10\right) \right] + C \] Multiplying the terms: \[ I = -\cos\left(\frac{x}{7} + 10\right) + C \]