Solve ∫ (from 0 to frac { pi) of {2}}(2cosx+sinx-1) wrt x

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } ( 2 \cos x + \sin x - 1 ) d x$$

Evaluate

$$-\frac{\pi }{2}+3\approx 1.429203673$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2\cos(x)+\sin(x)-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2\cos(x)\mathrm{d}x+\int \sin(x)\mathrm{d}x+\int -1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int \cos(x)\mathrm{d}x+\int \sin(x)\mathrm{d}x+\int -1\mathrm{d}x$$

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

$$2\sin(x)+\int \sin(x)\mathrm{d}x+\int -1\mathrm{d}x$$

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

$$2\sin(x)-\cos(x)+\int -1\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$2\sin(\frac{1}{2}\pi )-\cos(\frac{1}{2}\pi )-\frac{\pi }{2}-\left(2\sin(0)-\cos(0)-0\right)$$

Simplify.

$$3-\frac{\pi }{2}$$