Solve ∫ (from 0 to pi) of (4sin(t)-8cos(t))dt

$$\int _ { 0 } ^ { \pi } ( 4 \sin ( t ) - 8 \cos ( t ) ) d t$$

Evaluate

$$8$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

$$\int 4\sin(t)\mathrm{d}t+\int -8\cos(t)\mathrm{d}t$$

Factor out the constant in each of the terms.

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

Use \int \sin(t)\mathrm{d}t=-\cos(t) from the table of common integrals to obtain the result. Multiply 4 times -\cos(t).

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

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

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

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.

$$-4\cos(\pi )-8\sin(\pi )-\left(-4\cos(0)-8\sin(0)\right)$$

Simplify.

$$8$$