Solve ∫ (2cosx-6sinx) wrt x

$$\int ( 2 \cos x - 6 \sin x ) d x$$

Evaluate

$$2\sin(x)+6\cos(x)+С$$

Solution Steps

Integrate the sum term by term.

$$\int 2\cos(x)\mathrm{d}x+\int -6\sin(x)\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\left(\int \cos(x)\mathrm{d}x-3\int \sin(x)\mathrm{d}x\right)$$

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

$$2\left(\sin(x)-3\int \sin(x)\mathrm{d}x\right)$$

Use \int \sin(x)\mathrm{d}x=-\cos(x) from the table of common integrals to obtain the result. Multiply -6 times -\cos(x).

$$2\left(\sin(x)+3\cos(x)\right)$$

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.

$$2\sin(x)+6\cos(x)+С$$

Differentiate w.r.t. x

$$2\left(\cos(x)-3\sin(x)\right)$$