Question

evaluate the integral
int _ { 0 } ^ { pi } ( - 1 sin ( t ) + 5 cos ( t ) ) d t
integral =

Explanation:

Step1: Integrate term - by - term

$$\int_{0}^{\pi}(-\sin(t)+5\cos(t))dt=\int_{0}^{\pi}(-\sin(t))dt+\int_{0}^{\pi}5\cos(t)dt$$
Using the integral formulas $\int\sin(t)dt = -\cos(t)+C$ and $\int\cos(t)dt=\sin(t)+C$.
For $\int_{0}^{\pi}(-\sin(t))dt$, we have $[\cos(t)]_{0}^{\pi}$.
For $\int_{0}^{\pi}5\cos(t)dt$, we have $[5\sin(t)]_{0}^{\pi}$.

Step2: Evaluate the definite integrals

For $[\cos(t)]_{0}^{\pi}$, substitute the upper and lower limits: $\cos(\pi)-\cos(0)=(- 1)-1=-2$.
For $[5\sin(t)]_{0}^{\pi}$, substitute the upper and lower limits: $5\sin(\pi)-5\sin(0)=5\times0 - 5\times0 = 0$.

Answer:

$-2$