### Introduction

Hello everyone! In this blog post, we will discuss how to take the integral of $\sin 2x$. This is important to know, because it touches on two important concepts – integrals of trigonometric functions and $u$-substitution.

### Integrals of Trig Functions

We know that the derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$. If you want a proof of these concepts, you can find one ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/.._MIT18_01SCF10_Ses8d.pdf here and https://www.albert.io/blog/derivative-of-cosx-proof-review/ here. Clearly, because the derivative of $\cos x$ is $-\sin x$, our integral will have something to do with $\cos x$, but to figure out exactly what it is, we will need a tactic called $u$-substitution.

### $u$-substitution

To do the tactic of $u$-substitution, we will take our integral $\displaystyle \int \sin 2x dx$ and express it in terms of some variable $u$, so that it becomes $\displaystyle \int g(u) dx * \dfrac{du}{dx}$ for some function $g(u)$. In this case, it is convenient to take $u = 2x$ so that the function becomes $\sin u$. In this case, we get the integral to be $\int \sin u dx * \dfrac{du}{dx} = \int 2\sin u du = -2\cos x$, which is our final answer.

### Conclusion

In this article, we went over how to take the integral of $\sin 2x$, which touches on some very important subjects – $u$-substitution and the integrals of trig functions. It is absolutely essential that you understand these for your calculus course. Good luck on your future math adventures!