Solve ∫ (from 1 to 2) of (1/x-3cosx)

$$\int _ { 1 } ^ { 2 } ( \frac { 1 } { x } - 3 \cos x )$$

Evaluate

$$\ln(2)+3\sin(1)-3\sin(2)\approx 0.489667855$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{1}{x}-3\cos(x)\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{1}{x}\mathrm{d}x+\int -3\cos(x)\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int \frac{1}{x}\mathrm{d}x-3\int \cos(x)\mathrm{d}x$$

Use \int \frac{1}{x}\mathrm{d}x=\ln(|x|) from the table of common integrals to obtain the result.

$$\ln(|x|)-3\int \cos(x)\mathrm{d}x$$

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

$$\ln(|x|)-3\sin(x)$$

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.

$$\ln(|2|)-3\sin(2)-\left(\ln(|1|)-3\sin(1)\right)$$

Simplify.

$$\ln(2)-3\sin(2)+3\sin(1)$$