pressions matching the value of indefinite int (-sin(4t))dt 0.25cos(4t)+C (cos4t)/(4)+C -cost+C -4cos(4t)+C

The correct answer is $\frac {cos4t}{4}+C$.<br /><br />To solve the integral $\int (-sin(4t))dt$, we can use the substitution method. Let $u = 4t$, then $du = 4dt$ or $dt = \frac{du}{4}$.<br /><br />Substituting these values into the integral, we get:<br /><br />$\int (-sin(4t))dt = \int (-sin(u) \cdot \frac{du}{4}) = -\frac{1}{4} \int sin(u) du$<br /><br />Now, we can integrate $\int sin(u) du$ which is equal to $-\cos(u) + C$. Substituting back $u = 4t$, we get:<br /><br />$-\frac{1}{4} \int sin(u) du = -\frac{1}{4}(-\cos(4t) + C) = \frac{\cos(4t)}{4} + C$<br /><br />Therefore, the correct answer is $\frac {cos4t}{4}+C$.