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

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

Evaluate

$$0$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

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.

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

Simplify.

$$0$$