Why $C=1$ in indefinite integral $\int{\sin x\,\mathrm{d}x}$

Question

I am reading Introduction to Calculus and Analysis by Richard Courant. In Section 3.15, g.The Dirichlet Integral, it said $\int{\sin x \,\mathrm{d}x}=1-\cos x$,why $C=1$ here?

Answer 1

We have, for any constant $C$, $$ \int \sin x \:dx=-\cos x+C $$ since $$ (-\cos x+C)'=\sin x. $$ Then one is allowed to take $C=1$.

In some cases, taking a specific value of the constant of integration is very useful. For example, when integrating by parts in the following evaluation, $$ \int_{0}^M{\frac{\sin x}{x} \,\mathrm{d}x}=\left.\frac{1-\cos x}{x}\right|_0^M+\int_{0}^M{\frac{1-\cos x}{x^2} \,\mathrm{d}x} $$ the first term on the right hand side admits a finite limit as $x \to 0$ because of the choice of $1 -\cos x$ instead of the usual $-\cos x$ for the antiderivative of $\sin x$. Then one may easily conclude to the convergence of $\displaystyle \int_{0}^M{\frac{\sin x}{x} \,\mathrm{d}x}$ as $M \to \infty$.