Integrals and derivatives of trigonometric functions can be tricky. You may have been told in your math class that $\dfrac{d}{dx} \sin(x) = \cos(x)$, but not given a good explanation for it. In this article, we will show you the derivatives and integrals of all the trigonometric functions while also proving that they are true.

### Derivatives of \(\sin(x)\) and \(\cos(x)\)

To find the derivatives of $\sin x$ and $\cos x$, we need two identities: $\displaystyle \lim_{x\to 0} \frac{\sin x}{x} = 1$, and $\displaystyle \lim_{x\to 0} \frac{1-\cos x}{x} = 0$.

Now, we can use the definition of a derivative on $\sin x$: $\dfrac{d}{dx} \sin x = \displaystyle \lim_{h\to 0} \frac{\sin(x+h) – \sin(x)}{h} = \frac{\sin x \cos h-\cos x \sin h-\sin x}{h} = \cos x(\lim_{h\to0} \frac{\sin h}{h}) + \sin x(\lim_{h\to0} \frac{1-\cos h}{h}) = \cos x$.

We can do the same for $\cos x$: $\dfrac{d}{dx} \cos x = \displaystyle \lim_{h\to 0} \frac{\cos(x+h) – \cos(x)}{h} = \frac{\cos x \cos h-\sin x \sin h-\cos x}{h} = \cos x(\lim_{h\to0} \frac{\cos h-1}{h}) – \sin x(\lim_{h\to0} \frac{\sin h}{h}) = -\sin x$.

### Integrals of \(\sin(x)\) and \(\cos(x)\)

Once we have the derivatives, the integrals will be quite easy to do. For the integral of $\sin x$, we simply take integral of both sides of the following: $\dfrac{d}{dx} \cos x = -\sin x \to \cos x + C =\displaystyle \int -\sin x\,dx \to -\cos x + C =\displaystyle \int \sin x\,dx$. It is the same for $\cos x$: $\dfrac{d}{dx} \sin x = \cos x \to \sin x + C = \displaystyle\int \cos x\,dx$.

### Conclusion

You now know how to take the derivatives and integrals of $\sin x$ and $\cos x$, and, even better, you know the underlying logic, so if you ever forget the derivatives or integrals, you will remember how to find them again.