Question

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

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

Answer 1

The correct option is C $-\cos (\frac{x}{7}+10)+C$

Let $I=\int \frac{1}{7}\sin (\frac{x}{7}+10)dx$
$=\frac{1}{7}\int \sin (\frac{x}{7}+10)=\frac{1}{7}\frac{-\cos (\frac{x}{7}+10)}{\frac{1}{7}}$
$=-\cos (\frac{x}{7}+10)+C$