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First, we must recall a few trigonometric identities: \begin{eqnarray} \sin^2x+\cos^2x &=& 1\\ \sec^2x &=& 1 + \tan^2 x\\ \sin^2x &=& \frac{1-\cos (2x)}{2}\\ \cos^2x &=& \frac{1+\cos(2x)}{2}\\ \sin(2x) &=& 2\sin x \cos x\\ \cos(2x) &=& \cos^2x - \sin^2 x. \end{eqnarray} 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. $$\int \sin^2x dx =\int \frac{1-\cos (2x)}{2} dx = \frac{1}{2} \int (1 - \cos(2x)) dx = \frac{1}{2}\left(x-\frac{\sin(2x)}{2}\right)+C.$$
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. $$\int \cos^2x \sin x dx = \int -u^2 du = -\frac{u^3}{3} + C= -\frac{\cos^3 x}{3} +C.\ [u=\cos x,\ du=-\sin x dx]$$
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 \begin{eqnarray*} \int \sin^3 x dx &=& \int \sin^2x\sin x dx=\int(1-\cos^2 x)\sin x dx\\ &=& \int \sin x dx - \int \cos^2 x \sin x dx\\ &=& -\cos x + \frac{\cos^3 x}{3} + C. \end{eqnarray*}Similarly one can solve $\int \cos^3 x dx$ .
Example 1.5 \begin{eqnarray*} \int \cos^3x\sin^2 x dx &=& \int \cos^2 x \cos x \sin^2x dx = \int (1-\sin^2x)\cos x \sin^2 x dx\\ &=& \int \cos x \sin^2 x dx - \int \cos x \sin^4 x dx\\ &=& \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C. \end{eqnarray*}
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 \begin{eqnarray*} \int \tan^5x\sec^4 x dx &=& \int \tan^5 x \sec^2 x \sec^2x dx = \int \tan^5 x(1+\tan^2x)\sec^2 x dx\\ &=& \int \tan^5 x \sec^2 x dx + \int \tan^7 x \sec^2 x dx\\ &=& \frac{\tan^6 x}{6} + \frac{\tan^8 x}{8} + C. \end{eqnarray*}
Example 1.8 \begin{eqnarray*} \int \tan^3x\sec^4 x dx &=& \int \tan x \tan^2 x \sec^4x dx = \int \tan x(\sec^2x-1)\sec^4 x dx\\ &=& \int \tan x \sec x \sec^5 x dx - \int \tan x \sec x\sec^3 x dx\\ &=& \frac{\sec^6 x}{6} - \frac{\sec^4 x}{4} + C. \end{eqnarray*}
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 $r=1$ (because if you square both sides of $y=\sqrt{1-x^2}$ you obtain $x^2+y^2=1$ which is the equation of a circle or radius $r=1$ ). 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: \begin{eqnarray*} \int_0^1 \sqrt{1-x^2}dx &=& \int_0^{\pi/2} (\cos \theta)\cos \theta d\theta=\int_0^{\pi/2} \cos^2 \theta d\theta\\ &=& \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. \end{eqnarray*}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: \begin{eqnarray*} \int_0^r \sqrt{r^2-x^2}dx &=& \int_0^{\pi/2} r^2(\cos \theta)\cos \theta d\theta=r^2\int_0^{\pi/2} \cos^2 \theta d\theta\\ &=& 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. \end{eqnarray*}Thus, we have proved that a quarter of a circle of radius $r$ 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 $r$ . 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 $u=1-x^2$ will work much better than $x=\sin \theta$ .
Example 2.4 We would like to find the value of $$\int_{\sqrt{2}}^2 \frac{1}{x^3\sqrt{x^2-1}} dx.$$ Since neither a $u$ -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 $x=2$ implies $\theta=\pi/3$ . Hence: \begin{eqnarray*}\int_{\sqrt{2}}^2 \frac{1}{x^3\sqrt{x^2-1}} dx &=& \int_{\pi/4}^{\pi/3} \frac{\sec\theta \tan\theta}{\sec^3\theta \tan\theta} d\theta = \int_{\pi/4}^{\pi/3} \frac{1}{\sec^2\theta}d\theta=\int_{\pi/4}^{\pi/3} \cos^2 \theta d\theta \end{eqnarray*}and the last integral is easy to compute using Equation (4).
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