Solve ∫ 2(sinx-cosx+x) wrt x

$$\int 2 ( \sin x - \cos x + x ) d x$$

Evaluate

$$-2\sin(x)-2\cos(x)+x^{2}+С$$

Solution Steps

Use the distributive property to multiply 2 by \sin(x)-\cos(x)+x.

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

Use the distributive property to multiply 2 by \sin(x)-\cos(x).

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Use \int \sin(x)\mathrm{d}x=-\cos(x) from the table of common integrals to obtain the result. Multiply 2 times -\cos(x).

$$2\left(-\cos(x)-\int \cos(x)\mathrm{d}x+\int x\mathrm{d}x\right)$$

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

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 2 times \frac{x^{2}}{2}.

$$-2\cos(x)-2\sin(x)+x^{2}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$-2\cos(x)-2\sin(x)+x^{2}+С$$

Differentiate w.r.t. x

$$2\left(\sin(x)-\cos(x)+x\right)$$