Further Integrals

Polynomial functions are part of a group of integrands with common antiderivatives. These indefinite integrals are sometimes called immediate integrals or elementary integrals because they immediately follow from the derivatives you already know. In this topic, you will take a look at the rules of integration for other functions with common antiderivatives besides constants and powers, including exponential functions and a few trigonometric functions.

Integral of exponential functions

An exponential function follows the form

$$f(x)=a^{kx}$$In order to integrate this function, let's recall its derivative

$$\frac{d}{dx} a^{kx} = k \cdot a^{kx} \cdot \ln(a)$$Rearranging factors,

$$a^{kx} = \frac{1}{k \cdot \ln(a)} \cdot \frac{d}{dx}(a^{kx})$$Integrating both sides, we find that

$$\int a^{kx}dx = \frac{a^{kx}}{k \cdot \ln(a)} + C$$As an example, let's consider the case when $a = e \approx 2.71828$ and $k=1$. For these values, this function becomes the natural exponential function:

$$f(x)=e^x$$You can see how

$$\int e^x dx = e^x + C$$Which is a direct reminder of how the derivative of the natural exponential function is still the natural exponential function!

Integral of sine and cosine functions

The trigonometric functions sine and cosine share a neat cyclical relationship when it comes to differentiation

$$\begin{align*} &\frac{d}{dx} \sin(x) = \cos(x) \\ &\frac{d}{dx} \cos(x) = - \sin(x) \\ &\frac{d}{dx} [-\sin(x)] = -\cos(x) \\ &\frac{d}{dx} [-\cos(x)] = \sin(x) \end{align*}$$As one function acts as the antiderivative for the other, you can see how this cyclical nature is maintained when it comes to integration

$$\begin{align*} &\frac{d}{dx} [-\cos(x)] = \sin(x) &\implies& \int \sin(x) \ dx = - \cos(x) + C \\ &\frac{d}{dx} [-\sin(x)] = -\cos(x) &\implies& \int - \cos(x) \ dx = - \sin(x) + C \\ &\frac{d}{dx} \cos(x) = - \sin(x) &\implies& \int -\sin(x) \ dx = \cos(x) + C \\ &\frac{d}{dx} \sin(x) = \cos(x) &\implies& \int \cos(x) \ dx = \sin(x) + C \end{align*}$$Another way of looking at this phenomenon is by taking a look at the plots of these functions

You can observe how the differentiation process is equivalent to shift the argument of the function by $\pi/2$ in one direction, while the integration process shifts the same argument in the opposite direction.

Integral of secant and cosecant squared functions

Let's consider the following indefinite integrals:

$$\int \sec^2(x) dx \quad \text{and} \quad \int \csc^2(x) dx$$They might appear intricate at first, but you can see that these two also have immediate antiderivatives.

First, let's recall the following derivatives:

$$\frac{d}{dx} \tan(x) = \frac{d}{dx} \left( \frac{\sin(x)}{\cos(x)} \right) = \cos(x) \cdot \frac{1}{\cos(x)} + \sin(x) \cdot \frac{\sin(x)}{\cos^2(x)} = 1 + \tan^2(x)$$and

$$\frac{d}{dx} \cot(x) = \frac{d}{dx} \left( \frac{\cos(x)}{\sin(x)} \right) = -\sin(x) \cdot \frac{1}{\sin(x)} - \cos(x) \cdot \frac{\cos(x)}{\sin^2(x)} = -1 - \cot^2(x)$$Then, let's recall the fundamental trigonometric identity

$$\sin^2(x) + \cos^2(x) = 1$$Multiplying both sides by $1/\cos^2(x)$, you get

$$\frac{\sin^2(x)}{\cos^2(x)} + \frac{\cos^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)} \implies \tan^2(x) + 1 = \sec^2(x)$$Or, if instead you multiply both sides by $1/\sin^2(x)$, you obtain

$$\frac{\sin^2(x)}{\sin^2(x)} + \frac{\cos^2(x)}{\sin^2(x)} = \frac{1}{\sin^2(x)} \implies 1 + \cot^2(x) = \csc^2(x)$$Thus,

$$\frac{d}{dx} \tan(x) = \sec^2(x) \implies \int \sec^2(x) dx = \tan(x) + C$$and

$$\frac{d}{dx} \cot(x) = -\csc^2(x) \implies \int \csc^2(x) dx = -\cot(x) + C$$

Conclusion

In this topic, you've learned the rules to determine antiderivatives for the following functions:

• Exponential functions:

$$f(x) = a^{kx} \implies \int a^{kx}dx = \frac{a^{kx}}{k \cdot \ln(a)} + C$$

• Sine and Cosine functions:

$$\begin{align*} f(x) = \sin(x) &\implies \int \sin(x) \ dx = - \cos(x) + C \\ f(x) = \cos(x) &\implies \int \cos(x) \ dx = \sin(x) +C \end{align*}$$

• Secant squared and Cosecant squared functions:

$$\begin{align*} f(x) = \sec^2(x) &\implies\int \sec^2(x) dx = \tan(x) + C \\ f(x) = \csc^2(x) &\implies \int \csc^2(x) \ dx = -\cot(x) +C \end{align*}$$