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The Fundamental Trigonometric Integrals

April 6th, 2009 |

Author: Merrin

In our studies of trigonometric integrals, it is useful to have knowledge of what we can call the fundamental trigonometric integrals. These integrals are just the first power of each trigonometric functions. Luckily, these integrals can be computed in terms of elementary functions. A good starting point is to recall our knowledge of antiderivatives involving trigonometric functions.

$\begin{array}{l} {\cfrac{d}{dx} \text{sin}\,\, x=+\text{cos}\,\, x} \\ {\cfrac{d}{dx} \text{cos}\,\, x=-\text{sin}\,\, x} \\ {\cfrac{d}{dx} \text{tan}\,\, x=+\text{sec}\,\, ^{2} x} \\ {\cfrac{d}{dx} \text{cot}\,\, x=-\text{csc}\,\, ^{2} x} \\ {\cfrac{d}{dx} \text{sec}\,\, x=+\text{sec}\,\, x\text{tan}\,\, x} \\ {\cfrac{d}{dx} \text{csc}\,\, x=-\text{csc}\,\, x\text{cot}\,\, x} \end{array}$

Perhaps some of these equations will help us in our task

Example 1. Find

$\displaystyle \int _{}^{}\text{sin}\,\, xdx\quad \text{and}\quad \displaystyle \int _{}^{}\text{cos}\,\, xdx$

Solution 1. We spot from our table of derivatives two equations that will be useful

$\cfrac{d}{dx} \text{sin}\,\, x=\text{cos}\,\, x\quad \quad \quad \, \cfrac{d}{dx} \text{cos}\,\, x=-\text{sin}\,\, x$

So we can just integrate both sides of the equation and get

$\int \left(\cfrac{d}{dx} \text{sin}\,\, x\right) dx=\displaystyle \int \text{cos}\,\, xdx } \\ {\displaystyle \int \text{cos}\,\, xdx =\text{sin}\,\, x+C$

We can do the same for the second equation and we have after moving the minus sign across that

${\cfrac{d}{dx} \text{cos}\,\, x=-\text{sin}\,\, x} \\ {\displaystyle \int (\text{cos}\,\, x)'dx =-\displaystyle \int \text{sin}\,\, xdx } \\ {\displaystyle \int \text{sin}\,\, xdx=-\text{cos}\,\, x+C }$

If this makes no sense remember that we are just using the fundamental theorem of calculus. The antiderivative F(x) obeys

$\cfrac{d}{dx} F(x)=\cfrac{d}{dx} \displaystyle \int f(x)dx=f(x)$

Or written another way

$\displaystyle \int \cfrac{df}{dx} dx=f(x)+C$

Example 2. Evaluate the fundamental trigonometric integrals

$\displaystyle \int _{}^{}\text{tan}\,\, xdx\quad \text{and}\quad \displaystyle \int _{}^{}\text{cot}\,\, xdx$

Solution 2. One method to solve these integrals is to use a \textit{u} substitution.

$\displaystyle \int \text{tan}\,\, xdx=\displaystyle \int \cfrac{\text{sin}\,\, x}{\text{cos}\,\, x} dx=-\displaystyle \int \cfrac{du}{u} =-\ln \left|u\right|+C \\ \displaystyle \int \text{tan}\,\, xdx=-\ln \left|\text{cos}\,\, x\right|+C \\ \displaystyle \int \text{tan}\,\, xdx=+\ln \left|\text{sec}\,\, x\right|+C$

A similar calculation can be done for cotangent x

$\displaystyle \int \text{cot}\,\, xdx=\displaystyle \int \cfrac{\text{cos}\,\, x}{\text{sin}\,\, x} dx=\displaystyle \int \cfrac{du}{u} =\ln \left|u\right|+C \\ \displaystyle \int \text{cot}\,\, xdx= +\ln \left|\text{sin}\,\, x\right|+C \\ \displaystyle \int \text{cot}\,\, xdx= -\ln \left|\text{csc}\,\, x\right|+C$

Example 3. Evaluate the fundamental trigonometric integrals

$\displaystyle \int \text{sec}\,\, xdx\qquad \qquad \displaystyle \int\text{csc}\,\, xdx$

Solution 3. These integrals are a little trickier than the previous ones we calculated. Working with secant is more common so we will calculate that integral first.

$\displaystyle \int \cfrac{dx}{\text{cos}\,\, x}$

We need some more meat in the integral to use our techniques of integration. One way to proceed is as follows.

$\cfrac{1}{\text{cos}\,\, x} =\cfrac{\text{cos}\,\, x}{\text{cos}\,\, ^{2} x} =\cfrac{\text{cos}\,\, x}{1-\text{sin}\,\, ^{2} x}$

We can now do a u substitution because the derivative of sine is cosine.

$\displaystyle \int \cfrac{dx}{\text{cos}\,\, x} =\displaystyle \int \cfrac{du}{1-u^{2}}$

Now you should recognize the substituted integrand as the derivative of the hyperbolic arctangent of $u$ from your table of derivatives. So our first form of the solution is

$\displaystyle \int \text{sec}\,\, xdx=\text{arctanh}\,(\text{sin}\,\, x)+C$

This is not the usual form that it is written in but it is a legitimate answer. Instead we use partial fractions to get the more common expression.

$\cfrac{1}{1-u^{2} } =\cfrac{1}{2(1-u)} +\cfrac{1}{2(1+u)}$

$\displaystyle \int \text{sec}\,\, xdx=\cfrac{1}{2} \ln \left|1+u\right|-\cfrac{1}{2} \ln \left|1-u\right| +C=\cfrac{1}{2} \ln \left|\cfrac{1+\text{sin}\,\, x}{1-\text{sin}\,\, x} \right|+C$

Doing some more algebra tricks we can write

$\cfrac{1}{2} \ln \left|\cfrac{1+\text{sin}\,\, x}{1-\text{sin}\,\, x} \right|=\cfrac{1}{2} \ln \left|\cfrac{(1+\text{sin}\,\, x)(1+\text{sin}\,\, x)}{(1-\text{sin}\,\, x)(1+\text{sin}\,\, x)} \right|=\cfrac{1}{2} \ln \left|\cfrac{(1+\text{sin}\,\, x)^{2} }{\text{cos}\,\, ^{2} x} \right|$

$\displaystyle \int{\text{sec}\,\, x}dx=\ln \left|\cfrac{1}{\text{cos}\,\, x} +\cfrac{\text{sin}\,\, x}{\text{cos}\,\, x} \right|+C$

In the usual form, we can write

$\displaystyle \int \text{sec}\,\, xdx=\ln \left|\text{sec}\,\, x+\text{tan}\,\, x\right| +C$

A similar calculation can be done for cosecant.

$\int \cfrac{dx}{\text{sin}\,\, x} }={\displaystyle \int \cfrac{dx\text{sin}\,\, x}{1-\text{cos}\,\, ^{2} x} =-\displaystyle \int \cfrac{du}{1-u^{2} } =-\text{arctanh}\,(\text{cos}\,\, x)+C$

We jump to the end, because all the calculations are just same as the previous integral with sine and cosine reversed.

$\displaystyle \int \text{csc}\,\, x dx = -\ln\left|\text{csc}\,\, x\ +\text{cot}\,\, x \right| +C$

We have learned the integrals of the fundamental trigonometric derivatives. It turns out you can solve the integrals of trigonometric functions to any positive integer powers. These formula are derived in a later section, but be aware

that these solutions exist so we can refer to them. The formulas which give these integrals are called reduction formulas because they solve the integrals in terms of integrals of reduced powers. This business works for trigonometric as well as hyperbolic functions. We will see later how these functions are related, but in the next section we catch up our theory of elementary function integrals using the hyperbolics.

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