Solve ∫ _{pi/6}^3(2cosx-5sinx) wrt x

$$\int _ { \frac { \pi } { 6 } } ^ { 3 } ( 2 \cos x - 5 \sin x ) d x$$

Evaluate

$$\frac{4\sin(3)+10\cos(3)-5\sqrt{3}-2}{2}\approx -9.997849486$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2\cos(x)-5\sin(x)\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2\cos(x)\mathrm{d}x+\int -5\sin(x)\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int \cos(x)\mathrm{d}x-5\int \sin(x)\mathrm{d}x$$

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

$$2\sin(x)-5\int \sin(x)\mathrm{d}x$$

Use \int \sin(x)\mathrm{d}x=-\cos(x) from the table of common integrals to obtain the result. Multiply -5 times -\cos(x).

$$2\sin(x)+5\cos(x)$$

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.

$$2\sin(3)+5\cos(3)-\left(2\sin(\frac{1}{6}\pi )+5\cos(\frac{1}{6}\pi )\right)$$

Simplify.

$$2\sin(3)+5\cos(3)-1-\frac{5\sqrt{3}}{2}$$