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A lecture on trigonometric integrals and trigonometric substitution

(Feature)

Trigonometric Integrals

First, we must recall a few trigonometric identities:

$\displaystyle \sin^2x+\cos^2x$	$\displaystyle =$	$\displaystyle 1$	(1)
$\displaystyle \sec^2x$	$\displaystyle =$	$\displaystyle 1 + \tan^2 x$	(2)
$\displaystyle \sin^2x$	$\displaystyle =$	$\displaystyle \frac{1-\cos (2x)}{2}$	(3)
$\displaystyle \cos^2x$	$\displaystyle =$	$\displaystyle \frac{1+\cos(2x)}{2}$	(4)
$\displaystyle \sin(2x)$	$\displaystyle =$	$\displaystyle 2\sin x \cos x$	(5)
$\displaystyle \cos(2x)$	$\displaystyle =$	$\displaystyle \cos^2x - \sin^2 x.$	(6)

The most usual integrals which involve trigonometric functions can be solved using the identities above.

Example 1.1 $\int \sin x dx = -\cos x +C$ and $\int \cos x dx = \sin x +C$ are immediate integrals.

Example 1.2 For $\int \sin^2x dx,\ \int \cos^2x dx$ we use formulas (3) and (4) respectively, e.g.

$\displaystyle \int \sin^2x dx =\int \frac{1-\cos (2x)}{2} dx = \frac{1}{2} \int (1 - \cos(2x)) dx = \frac{1}{2}\left(x-\frac{\sin(2x)}{2}\right)+C.$

Example 1.3 For integrals of the form $\int \cos^m x\sin x\ dx$ or $\int \sin^m x \cos x\ dx$ we use substitution with $u=\cos x$ or $u=\sin x$ respectively, e.g.

$\displaystyle \int \cos^2x \sin x dx = \int -u^2 du = -\frac{u^3}{3} + C= -\frac{\cos^3 x}{3} +C.\ [u=\cos x, du=-\sin x dx]$

In the following examples, we use equations (1) in the forms $\sin^2x=1-\cos^2x$ or $\cos^2x=1-\sin^2x$ to transform the integral into one of the type described in Example 1.3.

Example 1.4

$\displaystyle \int \sin^3 x dx$	$\displaystyle =$	$\displaystyle \int \sin^2x\sin x dx=\int(1-\cos^2 x)\sin x dx$
	$\displaystyle =$	$\displaystyle \int \sin x dx - \int \cos^2 x \sin x dx$
	$\displaystyle =$	$\displaystyle -\cos x + \frac{\cos^3 x}{3} + C.$

Similarly one can solve $\int \cos^3 x dx$ .

Example 1.5

$\displaystyle \int \cos^3x\sin^2 x dx$	$\displaystyle =$	$\displaystyle \int \cos^2 x \cos x \sin^2x dx = \int (1-\sin^2x)\cos x \sin^2 x dx$
	$\displaystyle =$	$\displaystyle \int \cos x \sin^2 x dx - \int \cos x \sin^4 x dx$
	$\displaystyle =$	$\displaystyle \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C.$

Example 1.6 In order to solve $\int \cos^5 x \sin^3 x dx$ we express it first as $\int \cos^5x \sin^2 x \sin x=\int \cos^5 x (1-\cos^2x)\sin x dx$ and then proceed as in the previous example.

One can use similar tricks to solve integrals which involve products of powers of $\sec x$ and $\tan x$ , by using Equation (2). Also, recall that the derivative of $\tan x$ is $\sec^2 x$ while the derivative of $\sec x$ is $\sec x \tan x$ .

Example 1.7

$\displaystyle \int \tan^5x\sec^4 x dx$	$\displaystyle =$	$\displaystyle \int \tan^5 x \sec^2 x \sec^2x dx = \int \tan^5 x(1+\tan^2x)\sec^2 x dx$
	$\displaystyle =$	$\displaystyle \int \tan^5 x \sec^2 x dx + \int \tan^7 x \sec^2 x dx$
	$\displaystyle =$	$\displaystyle \frac{\tan^6 x}{6} + \frac{\tan^8 x}{8} + C.$

Example 1.8

$\displaystyle \int \tan^3x\sec^4 x dx$	$\displaystyle =$	$\displaystyle \int \tan x \tan^2 x \sec^4x dx = \int \tan x(\sec^2x-1)\sec^4 x dx$
	$\displaystyle =$	$\displaystyle \int \tan x \sec x \sec^5 x dx - \int \tan x \sec x\sec^3 x dx$
	$\displaystyle =$	$\displaystyle \frac{\sec^6 x}{6} - \frac{\sec^4 x}{4} + C.$

Trigonometric Substitutions

One can easily deduce that $\int_0^1 \sqrt{1-x^2}dx$ has value $\frac{\pi}{4}$ . Why? Simply because the graph of the function $y=\sqrt{1-x^2}$ is half a circumference of radius (because if you square both sides of $y=\sqrt{1-x^2}$ you obtain which is the equation of a circle or radius ). Therefore, the area under the graph is a quarter of the area of a circle.

How does one compute $\int_0^1 \sqrt{1-x^2} dx$ without using the geometry of the problem? This is the prototype of integral where a trigonometric substitution will work very nicely. Notice that neither substitution nor integration by parts will work appropriately.

Example 2.1 Suppose we want to solve $\int_0^1 \sqrt{1-x^2}dx$ with analytic methods. We will use a substitution $x=\sin \theta$ (so $\theta$ will be our new variable of integration), because, as we know from Equation (1), $\sqrt{1-x^2}=\sqrt{1-\sin^2 \theta}=\cos \theta$ , thus getting rid of the pesky square root. Notice that $dx=\cos\theta d\theta$ . We also need to find the new limits of integration with respect to the new variable of integration, namely $\theta$ . When $x=0=\sin\theta$ we must have $\theta=0$ . Similarly, when $x=1=\sin\theta$ one has $\theta=\pi/2$ . We are now ready to integrate:

$\displaystyle \int_0^1 \sqrt{1-x^2}dx$	$\displaystyle =$	$\displaystyle \int_0^{\pi/2} (\cos \theta)\cos \theta d\theta=\int_0^{\pi/2} \cos^2 \theta d\theta$
	$\displaystyle =$	$\displaystyle \int_0^{\pi/2} \frac{1+\cos(2\theta)}{2}d\theta = \frac{1}{2}\left(\theta+\frac{\sin(2\theta)}{2}\right)_0^{\pi/2}=\pi/4.$

Notice that we made use of Equation (4) in the second line.

Example 2.2 Similarly, one can solve $\int_0^r\sqrt{r^2 - x^2}dx$ by using a substitution $x=r\sin\theta$ . Indeed, $\sqrt{r^2-x^2}=\sqrt{r^2-r^2\sin^2 \theta}=r\cos \theta$ and $dx=r\cos\theta d\theta$ . The limits of integration with respect to $\theta$ are again $\theta=0$ to $\theta=\pi/2$ (check this!). Thus:

$\displaystyle \int_0^r \sqrt{r^2-x^2}dx$	$\displaystyle =$	$\displaystyle \int_0^{\pi/2} r^2(\cos \theta)\cos \theta d\theta=r^2\int_0^{\pi/2} \cos^2 \theta d\theta$
	$\displaystyle =$	$\displaystyle r^2\int_0^{\pi/2} \frac{1+\cos(2\theta)}{2}d\theta = \frac{r^2}{2}\left(\theta+\frac{\sin(2\theta)}{2}\right)_0^{\pi/2}=r^2\pi/4.$

Thus, we have proved that a quarter of a circle of radius

has area $r^2\pi/4$ which implies that the area of such a circle is $\pi r^2$ , as usual.

The trigonometric substitutions usually work when expressions like $\sqrt{r^2-x^2}$ , $\sqrt{r^2+x^2}$ , $\sqrt{x^2-r^2}$ appear in the integral at hand, for some real number . Here is a table of the suggested change of variables in each particular case:

If you see...	try this...	because...
$\sqrt{1-x^2}$	$x=\sin \theta$	$\sqrt{1-\sin^2\theta}=\cos \theta$
$\sqrt{r^2-x^2}$	$x=r\sin\theta$	$\sqrt{r^2-\sin^2\theta}=r\cos\theta$
$\sqrt{1+x^2}$	$x=\tan \theta$	$\sqrt{1+\tan^2\theta}=\sec \theta$
$\sqrt{r^2+x^2}$	$x=r\tan\theta$	$\sqrt{r^2+\tan^2\theta}=r\sec\theta$
$\sqrt{x^2-1}$	$x=\sec \theta$	$\sqrt{\sec^2\theta-1}=\tan \theta$
$\sqrt{x^2-r^2}$	$x=r\sin\theta$	$\sqrt{\sec^2\theta-1}=r\tan\theta$

Remark 2.3 The above are “suggested” substitutions, they may not be the most ideal choice! For example, for the integral $\int 2x\sqrt{1-x^2}dx$ , the change will work much better than $x=\sin \theta$ .

Example 2.4 We would like to find the value of

$\displaystyle \int_{\sqrt{2}}^2 \frac{1}{x^3\sqrt{x^2-1}} dx.$

Since neither a

-substitution nor integration by parts seem appropriate, we try $x=\sec \theta$ , $dx=\sec\theta \tan \theta d\theta$ . When $x=\sqrt{2}=\sec\theta$ one has $\theta=\pi/4$ while

implies $\theta=\pi/3$ . Hence:

$\displaystyle \int_{\sqrt{2}}^2 \frac{1}{x^3\sqrt{x^2-1}} dx$

$\displaystyle =$

$\displaystyle \int_{\pi/4}^{\pi/3} \frac{\sec\theta \tan\theta}{\sec^3\theta \t... ...\pi/3} \frac{1}{\sec^2\theta}d\theta=\int_{\pi/4}^{\pi/3} \cos^2 \theta d\theta$

and the last integral is easy to compute using Equation (4).

"A lecture on trigonometric integrals and trigonometric substitution" is owned by alozano.

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Attachments:: integration of rational function of sine and cosine (Topic) by pahio

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Cross-references: ideal, real number, expressions, implies, line, limits, square root, variable, analytic, integration by parts, geometry, area, circle, sides, square, radius, circumference, function, graph, derivative, powers, products, similar, order, type, Transform, equations, identities, trigonometric functions, integrals, trigonometric identities
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This is version 1 of A lecture on trigonometric integrals and trigonometric substitution, born on 2006-01-31.
Object id is 7576, canonical name is ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution.
Accessed 7697 times total.

Classification:

AMS MSC:

26A36 (Real functions :: Functions of one variable :: Antidifferentiation)

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