Integrals of trigonometric functions can be difficult to find. You may have been told in your math class that \(\displaystyle\int \cos x \, dx = \sin x + C\), but not given a good explanation for it. In this article, we will explain how to find the integral of –cos x and show that the integral of -cos x is -sin x.
Derivative of \(\sin x\)
It is much easier to show that the derivative of \(\sin x\) is \(\cos x\) and then use the fact that integrals are antiderivatives, than to try to find the integral of \(\cos x\) directly. To find the derivative of \(\sin x\), we will use the limits \(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1\) and \(\displaystyle \lim_{x \to 0} \frac{1-\cos x}{x} = 0\).
We will find the derivative of \(\sin x\) using the limit definition of a derivative and sine angle sum formula:
\[ \begin{align*} \frac{d}{dx} \sin x &= \lim_{h \to 0} \frac{\sin(x+h) – \sin x}{h}\\ &= \lim_{h \to 0}\frac{\sin x \cos h + \cos x \sin h – \sin x}{h}\\ &= \cos x \left( \lim_{h \to 0} \frac{\sin h}{h} \right) + \sin x \left( \lim_{h \to 0} \frac{1 – \cos h}{h} \right)\\ &= \cos x. \end{align*} \]
Integral of -cos x
Since integral is another name for antiderivative, if the derivative of \(\sin x\) is \(\cos x\), the integral of \(\cos x\) must be \(\sin x\) (plus a constant), so we have the answer to the original question:
\[ \begin{align*} \int(-\cos x)\,dx = -\sin x + C. \end{align*} \]
Conclusion
Knowing that $\dfrac{d}{dx}\sin x = \cos x$ will help you remember that \(\displaystyle\int\cos x\,dx = \sin x + C\). Furthermore, you now know how to prove that $\dfrac{d}{dx}\sin x = \cos x$. You now also know the importance of knowing the limits \(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1\) and \(\displaystyle \lim_{x \to 0} \frac{1-\cos x}{x} = 0\). Good luck with your future math adventures!
