Solve ∫ 361-19sinx wrt x

$$\int 361 - 19 \sin x d x$$

Evaluate

$$19\cos(x)+361x+С$$

Solution Steps

Integrate the sum term by term.

$$\int 361\mathrm{d}x+\int -19\sin(x)\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 361\mathrm{d}x-19\int \sin(x)\mathrm{d}x$$

Find the integral of 361 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$19\left(19x-\int \sin(x)\mathrm{d}x\right)$$

Use \int \sin(x)\mathrm{d}x=-\cos(x) from the table of common integrals to obtain the result. Multiply -19 times -\cos(x).

$$19\left(19x+\cos(x)\right)$$

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.

$$361x+19\cos(x)+С$$

Differentiate w.r.t. x

$$19\left(-\sin(x)+19\right)$$