]> IB HL2 - Chapter 07

Integration Techniques

7.1: Basic Integration Rules

The Rules

Rules, rules, rules…always rules…

Well, there isn't a simple answer. You've got to know the techniques, and you've got to know when to apply the techniques. You can start by looking at the rules listed on pages 484 and 485 of the textbook.

However, there is no substitute for just doing a bunch of them. Take a shot at a few of them!

Examples

[1.] Find $\int (tan x) \cdot ln (cos x) d x$ .

Well, what to do, what to do…specifically, what should we use for $u$ ? Not the cosine; not the tangent…how about the log of cosine? $u = ln (cos x)$ ⇒ $d u = \frac{1}{cos x} (- sin x) d x$ ⇒ $d u = - tan x \cdot d x$ . So the integral becomes $- \int u \cdot d u = - \frac{1}{2} {(ln (cos x))}^{2} + C$ .

Well, that wasn't so bad…

[2.] Find $\int \frac{1 + cos x}{sin x} d x$ .

Hmmm…what to use for $u$ ? Not sine (the 1 prevents the replacement of $d u$ ); not cosine (that would put $d u$ in the denominator)…perhaps this one needs some work before starting. There are at least two options that I see; the simplest is to split the fraction. That results in $\int \frac{1 + cos x}{sin x} d x = \int (\frac{1}{sin x} + \frac{cos x}{sin x}) d x = \int (csc x + cot x) d x$ . This can be done! $\int (csc x + cot x) d x = - ln | csc x + cot x | + ln | sin x | + C$ .

[3.] Find $\int \frac{1}{(x - 1) \sqrt{4 x^{2} - 8 x + 3}} d x$ .

Oh dear. Using the radicand as $u$ won't work…the squared term in the radicand (which itself is in the denominator) makes me think that this might be the pattern for an inverse trig function. However, to make that work, the radicand must be a sum or difference of squares. So…complete the square?

$4 x^{2} - 8 x + 3 = 4 (x^{2} - 2 x) + 3 = 4 (x^{2} - 2 x + 1) + 3 - 4 = {(2 (x - 1))}^{2} - 1$

Well, this looks promising. Let $u = 2 (x - 1)$ . Then $d u = 2 \cdot d x$ . Start replacing!

$\int \frac{1}{(x - 1) \sqrt{4 x^{2} - 8 x + 3}} d x = 2 \int \frac{1}{2 (x - 1) \sqrt{{(2 (x - 1))}^{2} - 1}} d x = 2 \int \frac{1}{u \sqrt{u^{2} - 1}} \frac{d u}{2} = \int \frac{1}{u \sqrt{u^{2} - 1}} d u$

This is now in the pattern for arcsecant.

$\int \frac{1}{u \sqrt{u^{2} - 1}} d u = {sec}^{- 1} | 2 (x - 1) | + C$

7.2: Integration by Parts

The Idea

We've managed to "undo" (through integration) the sum, difference and chain rules…but we've not yet taken a stab at the product rule. Remember that it says if $y = f (x) \cdot g (x)$ , then the derivative is $\frac{dy}{d x} = f (x) \cdot g^{'} (x) + f^{'} (x) \cdot g (x)$ .

The Method

Turn the product rule around…first, differential form: $dy = [f (x) \cdot g^{'} (x) + f^{'} (x) \cdot g (x)] d x$ . Now, integrate both sides: $\int dy = \int f (x) \cdot g^{'} (x) d x + \int f^{'} (x) \cdot g (x) d x$ results in $y = f (x) \cdot g (x) = \int f (x) \cdot g^{'} (x) d x + \int f^{'} (x) \cdot g (x) d x$ .

"So what?" you ask. Well, if you have an integral that can be written as $\int f (x) \cdot g^{'} (x) d x$ , and the integral $\int f^{'} (x) \cdot g (x) d x$ is much easier (or at least possible), then you can use this equation to turn things around, since $\int f (x) \cdot g^{'} (x) d x = f (x) \cdot g (x) - \int f^{'} (x) \cdot g (x) d x$ .

This method—called integration by parts—is usually written this way: $\int u \cdot d v = u v - \int v \cdot d u$ . The real trick is in determining what should become part of $u$ , and what should become $d v$ . The textbook has some hints about this on page 493.

Examples

[4.] Find $\int x \cdot ln (x) d x$ .

Well, this is clearly a candidate for integration by parts. Once that's decided, then you'll always let the log part become part of $u$ : let $u = ln (x)$ and let $d v = x \cdot d x$ . Then $d u = \frac{1}{x} d x$ and $v = \frac{1}{2} x^{2}$ (remember that there is no need for a constant of integration until you get to the end of the whole problem). So $\int x \cdot ln (x) d x = (ln (x)) (\frac{1}{2} x^{2}) - \int (\frac{1}{2} x^{2}) (\frac{1}{x} d x)$ . Simplify a bit—you'll find that the integral part of that is quite manageable now: $- \frac{1}{2} \int x \cdot d x = - \frac{1}{4} x^{2} + C$ . So the answer is $\frac{1}{2} ln (x) (x^{2}) - \frac{1}{4} x^{2} + C$ .

[5.] Find $\int \frac{ln (2 x)}{x^{2}} d x$ .

OK, you know the drill. $u = ln (2 x)$ and $d v = \frac{1}{x^{2}} d x = x^{- 2} d x$ . Then $d u = \frac{1}{2 x} 2 \cdot d x = \frac{1}{x} d x$ and $v = - \frac{1}{x}$ . So $\int \frac{ln (2 x)}{x^{2}} d x = ln (2 x) (- \frac{1}{x}) - \int (- \frac{1}{x}) (\frac{1}{x} d x)$ . Once again, the integral is much easier now! $- \int (- \frac{1}{x}) (\frac{1}{x} d x) = \int \frac{1}{x^{2}} d x$ …which is exactly what we just integrated for $d v$ ! So the answer is $\int \frac{ln (2 x)}{x^{2}} d x = - \frac{ln (2 x)}{x} - \frac{1}{x} + C$ .

[6.] Find $\int x^{2} cos x \cdot d x$ .

There are actually four options here…perhaps the two most obvious ones involve $x^{2}$ and $cos x$ . The question is: which should be $u$ , and which should be $d v$ ? Let's try $u = x^{2}$ and $d v = cos x d x$ . That makes $d u = 2 x d x$ and $v = sin x$ . That gives us $\int x^{2} cos x \cdot d x = (x^{2}) (sin x) - \int (sin x) 2 x d x$ .

Uh-oh. Now we've got a new problem with the new integral part…perhaps we should have switched $u$ and $d v$ …

However, I recommend patience. Sometimes (especially when the integral involves those functions that lead back to themselves when integrated repeatedly) you have to do it twice to make everything work out! So, let's tackle that new integral, letting $u = 2 x$ and $d v = sinx d x$ . That makes $d u = 2 d x$ and $v = - cos x$ . So the problem now becomes . Hey, that did it! $\int (- cos x) (2 d x) = - 2 \int cos x d x = - 2 sin x + C$ , so we get our answer: $x^{2} sin x + 2 x cos x - 2 sin x + C$ .

7.3: Trigonometric Integrals

Sines and Cosines

Note that integrating ${sin}^{2} x$ is a lot harder than integrating $sin (2 x)$ …we have some identities that can convert squares of sines and cosines into sines and cosines of multiple angles. Specifically… ${sin}^{2} x = \frac{1 - cos (2 x)}{2}$ and ${cos}^{2} x = \frac{1 + cos (2 x)}{2}$ . This can be done repeatedly, if necessary.

What if the integral involves both sine and cosine? Well, part of one can becomes $u$ , and the rest can become $d u$ …there is a handy chart on page 497 that details this procedure.

Secants and Tangents

Because of the derivatives of secant and tangent, and because one form of the Pythagorean identity equates tangent and secant, integrals involving the two can be solved much like integrals involving sines and cosines. There is a handy chart on page 500.

Mismatched Angles

Dealing with something like $\int (sin (5 x)) (sin (4 x)) d x$ presents its own challenges. However, we know the sum and difference identities for sines and cosines, which can be used to create integrals of the type first mentioned in this section. There are general results of this process on page 502. You could also use repeated integration by parts…

Examples

[7.] Find $\int ({cos}^{3} x) ({sin}^{4} x) d x$ .

Let's peel off one cosine to act as $d u$ , and convert everything else into sines (I pick the odd power since the squares of sines and cosines can be interchanged easily; splitting the odd power gives me the $d u$ and leaves everything else in an easy-to-convert form): $({cos}^{2} x) ({sin}^{4} x) (cos x) = (1 - {sin}^{2} x) ({sin}^{4} x) (cos x) = {sin}^{4} x cos x - {sin}^{6} x cos x$ .

Split this into two integrals and substitute $u = sin x$ in each: $\int ({cos}^{3} x) ({sin}^{4} x) d x = \int ({sin}^{4} x) (cos x) d x - \int ({sin}^{6} x) (cos x) d x = \int u^{4} d u - \int u^{6} d u$ . This is easy! $\int u^{4} d u - \int u^{6} d u = \frac{1}{5} u^{5} - \frac{1}{7} u^{7} + C$ . Back substitute to get the answer: $\frac{1}{5} {sin}^{5} x - \frac{1}{7} {sin}^{7} x + C$ .

[8.] Find $\int ({tan}^{3} 2 x) ({sec}^{3} 2 x) d x$ .

Well, either we should peel off a ${sec}^{2} x$ to act as $d u$ , or one each of tangent and secant. Pulling out ${sec}^{2} x$ leaves us with only one more secant, which can't be (easily) exchanged for a tangent…thus, we should pull out a tangent and a secant. That means that the $u$ must be secant, so convert everything else into secant. $({tan}^{2} 2 x) ({sec}^{2} 2 x) (tan 2 x \cdot sec 2 x) = ({sec}^{2} 2 x - 1) ({sec}^{2} 2 x) (tan 2 x \cdot sec 2 x) = ({sec}^{4} 2 x - {sec}^{2} 2 x) (tan 2 x \cdot sec 2 x)$ . Split this into two integrals, and watch for that pesky 2 running around in there!

$\int ({sec}^{4} 2 x) (tan 2 x \cdot sec 2 x) d x - \int ({sec}^{2} 2 x) (tan 2 x \cdot sec 2 x) d x = \frac{1}{2} \int u^{4} d u - \frac{1}{2} \int u^{2} d u$

Keep going… $\frac{1}{2} \int u^{4} d u - \frac{1}{2} \int u^{2} d u = \frac{1}{10} u^{5} - \frac{1}{6} u^{3} + C$ .

Re-substitute: $\frac{1}{10} {sec}^{5} 2 x - \frac{1}{6} {sec}^{3} 2 x + C$ .

[9.] Find $\int (sin (4 x)) (cos (3 x)) d x$ .

The identity substitution is easy…I'll attack this with repeated integration by parts. First, let $u = sin (4 x)$ and $d v = cos (3 x) d x$ . That makes $d u = 4 cos (4 x) d x$ and $v = \frac{1}{3} sin (3 x)$ . So the integral becomes $(sin (4 x)) (\frac{1}{3} sin (3 x)) - \int (\frac{1}{3} sin (3 x)) (4 cos (4 x) d x)$ . Clean that up a bit, and do it again: $\frac{1}{3} (sin (4 x)) (sin (3 x)) - \frac{4}{3} \int (sin (3 x)) (cos (4 x)) d x$ . This time, let $u = cos (4 x)$ and $d v = sin (3 x) d x$ . That makes $d u = - 4 sin (4 x) d x$ and $v = - \frac{1}{3} cos (3 x)$ . The integral part becomes $\int (sin (3 x) (cos (4 x)) d x = (cos (4 x)) (- \frac{1}{3} cos (3 x)) - \int (- \frac{1}{3} cos (3 x)) (- 4 sin (4 x) d x)$ . The whole thing is $\int (sin (4 x) (cos (3 x)) d x = \frac{1}{3} (sin (4 x)) (sin (3 x)) + \frac{4}{3} (cos (4 x)) (\frac{1}{3} cos (3 x)) + \frac{16}{9} \int (cos (3 x)) (sin (4 x) d x)$ .

Notice that the beginning and end of that are the same thing! This sometimes happens when integrating by parts—part of your answer is the question again! This is not a bad thing. Move that part back to the other side!

$\int (sin (4 x) (cos (3 x)) d x - \frac{16}{9} \int (cos (3 x)) (sin (4 x) d x) = \frac{1}{3} (sin (4 x)) (sin (3 x)) + \frac{4}{3} (cos (4 x)) (\frac{1}{3} cos (3 x))$

$- \frac{7}{9} \int (cos (3 x)) (sin (4 x) d x) = \frac{1}{3} (sin (4 x)) (sin (3 x)) + \frac{4}{3} (cos (4 x)) (\frac{1}{3} cos (3 x))$

$\int (cos (3 x)) (sin (4 x) d x) = (- \frac{9}{7}) (\frac{1}{3} (sin (4 x)) (sin (3 x)) + \frac{4}{3} (cos (4 x)) (\frac{1}{3} cos (3 x)))$

7.4: Trigonometric Substitution

The Idea

If you've been paying attention, then when you see integrals that involve $\sqrt{a^{2} - u^{2}}$ , $\sqrt{u^{2} - a^{2}}$ , or $\sqrt{a^{2} + u^{2}}$ , your first thought is an inverse trig function…alas, that won't always work. However, all is not lost! You may be able to replace the $u$ with sine, tangent or secant of an angle, which will allow the integral to simplify to the point that it can be worked out.

See page 506 of the text for the substitutions. When working definite integrals, be careful when switching the limits of integration…

Examples

[10.] Find $\int \frac{10}{x^{2} \sqrt{25 - x^{2}}} d x$ .

Well, this looks a bit like a candidate for arcsine (except for the $x^{2}$ out in front)…and the radicand is in the wrong order for this to be a candidate for arcsecant. Better try a trigonometric substitution…this fits the pattern for a sine: $x = 5 sin θ$ ; $x^{2} = 25 {sin}^{2} θ$ ; $d x = 5 cos θ d θ$ . That makes the integral $\int \frac{10}{(25 {sin}^{2} θ) \sqrt{25 - 25 {sin}^{2} θ}} 5 cos θ d θ = \int \frac{2 cos θ}{{sin}^{2} θ \sqrt{25 {cos}^{2} θ}} d θ$ . That cleans up nicely to $\frac{2}{5} \int \frac{1}{{sin}^{2} θ} d θ = \frac{2}{5} \int {csc}^{2} θ d θ$ . This is a direct integral—no need for any further substitutions (well, except for the negative sign): $\frac{2}{5} \int {csc}^{2} θ d θ = - \frac{2}{5} cot θ + C$ . Now, you might be tempted to exchange θ for an arsine…don't! Let's draw a picture of $x = 5 sin θ$ and see what we get:

An Image

From this, you can see that $- \frac{2}{5} cot θ + C = - \frac{2}{5} (\frac{\sqrt{25 - x^{2}}}{x}) + C$ .

[11.] Find $\int \frac{\sqrt{4 x^{2} + 9}}{x^{4}} d x$ .

The $4 x^{2} + 9$ makes me think of arctangents…but there's a radical around it; better use a trig substitution instead. How about $2 x = 3 tan θ$ ? That makes $4 x^{2} = 9 {tan}^{2} θ$ , $x^{4} = \frac{81}{16} {tan}^{4} θ$ and $d x = \frac{3}{2} {sec}^{2} θ d θ$ . Substitute: $\int \frac{\sqrt{4 x^{2} + 9}}{x^{4}} d x = \int \frac{\sqrt{9 {tan}^{2} θ + 9}}{\frac{81}{16} {tan}^{4} θ} \frac{3}{2} {sec}^{2} θ d θ$ . Simplify a bit: $\int \frac{\sqrt{4 x^{2} + 9}}{x^{4}} d x = \int \frac{\sqrt{9 {tan}^{2} θ + 9}}{\frac{81}{16} {tan}^{4} θ} \frac{3}{2} {sec}^{2} θ d θ = \frac{8}{27} \int \frac{\sqrt{9 {sec}^{2} θ}}{{tan}^{4} θ} {sec}^{2} θ d θ = \frac{8}{27} \int \frac{3 {sec}^{3} θ}{{tan}^{4} θ} d θ$ . This looks pretty nasty, but try switching everything to sines and cosines: $\frac{8}{27} \int \frac{3 {sec}^{3} θ}{{tan}^{4} θ} d θ = \frac{8}{9} \int \frac{1}{{cos}^{3} θ} \cdot \frac{{cos}^{4} θ}{{sin}^{4} θ} d θ = \frac{8}{9} \int \frac{cos θ}{{sin}^{4} θ} d θ$ . Easy! Let $u = sin θ$ so that $d u = cos θ d θ$ . The integral becomes $\frac{8}{9} \int \frac{1}{u^{4}} d u$ , which evaluates to $- \frac{8}{27} u^{- 3} + C = - \frac{8}{27} {(sin θ)}^{- 3} + C$ . Once again, be careful turning θ back into x:

An Image

So the answer is $- \frac{8}{27} {(\frac{2 x}{\sqrt{4 x^{2} + 9}})}^{- 3} + C$ .

[12.] Find $\int \frac{x^{3}}{\sqrt{x^{2} - 4}} d x$ .

Well, the denominator is almost in the pattern for arcsecant…but it's missing something. How about another substitution? Let $x = 2 sec θ$ , so that $x^{2} = 4 {sec}^{2} θ$ , $x^{3} = 8 {sec}^{3} θ$ , and $d x = 2 sec θ tan θ d θ$ . The integral becomes $\int \frac{8 {sec}^{3} θ}{\sqrt{4 {sec}^{2} θ - 4}} 2 sec θ tan θ d θ$ , which simplifies to $16 \int \frac{{sec}^{4} θ tan θ}{\sqrt{4 {tan}^{4} θ}} d θ = 16 \int \frac{{sec}^{4} θ tan θ}{2 tan θ} d θ = 8 \int {sec}^{4} θ d θ$ . Use the methods of the previous section on this: ${sec}^{4} θ = ({sec}^{2} θ) ({sec}^{2} θ) = ({tan}^{2} θ + 1) ({sec}^{2} θ) = {tan}^{2} θ {sec}^{2} θ + {sec}^{2} θ$ . The first piece of that is ripe for a u-substitution, and the second part is a direct integral! Let $u = tan θ$ , making $d u = {sec}^{2} θ$ . Now we have $8 \int {tan}^{2} θ {sec}^{2} θ d θ + 8 \int {sec}^{2} θ d θ = 8 \int u^{2} d u + 8 \int {sec}^{2} θ d θ$ . This becomes $\frac{8}{3} u^{3} + 8 tan θ + C = \frac{8}{3} {tan}^{3} θ + 8 tan θ + C$ . Once again, draw a triangle!

An Image

So the answer is $\frac{8}{3} (\frac{\sqrt{x^{2} - 4}}{2})^{3} θ + 8 \frac{\sqrt{x^{2} - 4}}{2} + C$ .

Page last validated:2005-12-27