Solve ∫ (from 0 to pi) of (2cosx+sinx) wrt x

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

Evaluate

$$2$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

$$2\sin(x)+\int \sin(x)\mathrm{d}x$$

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

$$2\sin(x)-\cos(x)$$

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.

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

Simplify.

$$2$$