Anti-Derivatives of Sine and Cosine Functions

Hey everyone!

Today, we are going to do some cool things about integrals of trigonometric functions.

Assuming you all all familiar with $\sin(x)$ and $\cos(x)$, some strange things will happen when you take the integral of either of them. Here is what happens:

$$\int \sin(x) \, dx = -\cos(x) + C$$ $$\int \cos(x) \, dx = \sin(x) + C$$ $$\int -\sin(x) \, dx = \cos(x) + C$$ $$\int -\cos(x) \, dx = -\sin(x) + C$$

(Here, $C$ is the constant of integration!)

So, we can easily find that the integrals of these two trig functions tend to be periodic. But why do we get that?

If we look at the graph of $\sin(x)$ or $\cos(x)$, these two functions are both like a curve bouncing back and forth around the $x$-axis. Because the slope functions would decrease when the acceleration of the function decrease, and the same thing happens if the acceleration of the function increases, $\cos(x)$, which is the derivative of $\sin(x)$, seems to move leftward by $\pi/2$.

These are just for sine and cosine functions. When it comes to functions like $\sec(x)$ or $\cot(x)$, it gets more complex, and we will discover more about that in our next exercise. Hope you enjoy it so far!

Eric

At this point you likely know (or can easily learn!) the following two very common derivatives thanks to this:

$\frac{d}{dx}(\sin(x)) = \cos(x)$

$\frac{d}{dx}(\cos(x)) = -\sin(x)$

These equations can be integrated to get two equally common antiderivative statements:

$\sin(x) + C = \int{\cos(x)dx}$

$\cos(x) + C = \int{-\sin(x)dx}$

$C$ represents a constant. This must be included as there are multiple antiderivatives of sine and cosine, all of which only differ by a constant. If the equations are re-differentiated, the constants become zero (the derivative of a constant is always zero).

Now let's find the anti-derivatives of more trig functions using the anti-derivatives of $\sin x$ and $\cos x$.

Now, let's enjoy the show of substitution!

$$\begin{align*} \int \tan x dx &=\int \frac{\sin x}{\cos x} dx\\ &=\int \frac{du}{-u}\\ &=-\ln |u| +C=-\ln|\cos x|+C \end{align*}$$

Let's find $\int \sec x dx$ together!

So far, we could do nothing on the expression $\frac{1}{\cos x}dx$ since it's now hard to find some other functions related to this.

If you got the question above right, you're excellent!

Now it's time for substitution. Let $u=\sin x$.

Finally, bring the $\sin x$ back! We get the final result: $$\begin{align*} \int \sec x dx&=\frac{1}{2}\ln \left|\frac{u+1}{1-u}\right|+C\\ &=\frac{1}{2}\ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\\ &=\frac{1}{2}\ln \frac{(1+\sin x)^2}{1-\sin^2 x}+C\\ &=\frac{1}{2}\ln \left( \frac{1+\sin x}{\cos x}\right)^2+C\\ &=\ln \left|\frac{1+\sin x}{\cos x}\right| +C\\ &=\ln |\sec x+\tan x|+C. \end{align*}$$ Mission accomplished!

If you want to see more tricks, please click 'continue'.

Since $\frac{\sec x +\tan x}{\sec x +\tan x}=1$, we multiply the integral $\int \sec x dx$ with this. Then, the incredible thing happens: The numerator $\sec^2 x +\tan x \sec x$ turns out to be the derivative of the denominator $\sec x +\tan x$. Therefore: $$\begin{align*} \int \sec x dx&=\int \frac{\sec^2 x +\tan x \sec x}{\sec x +\tan x} dx\\ &=\int \frac{d(\sec x +\tan x )}{\sec x +\tan x}=\ln |\sec x +\tan x|+C \end{align*}$$.

You must be totally amazed.