Solve ∫ (from 0 to pi) of (2sin(t)+8cos(t))dt

$$\int _ { 0 } ^ { \pi } ( 2 \sin ( t ) + 8 \cos ( t ) ) d t$$

Evaluate

$$4$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2\left(\sin(t)+4\cos(t)\right)\mathrm{d}t$$

Integrate the sum term by term.

$$\int 2\sin(t)\mathrm{d}t+\int 8\cos(t)\mathrm{d}t$$

Factor out the constant in each of the terms.

$$2\left(\int \sin(t)\mathrm{d}t+4\int \cos(t)\mathrm{d}t\right)$$

Use \int \sin(t)\mathrm{d}t=-\cos(t) from the table of common integrals to obtain the result. Multiply 2 times -\cos(t).

$$2\left(-\cos(t)+4\int \cos(t)\mathrm{d}t\right)$$

Use \int \cos(t)\mathrm{d}t=\sin(t) from the table of common integrals to obtain the result.

$$2\left(-\cos(t)+4\sin(t)\right)$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$-2\cos(\pi )+8\sin(\pi )-\left(-2\cos(0)+8\sin(0)\right)$$

Simplify.

$$4$$