MATH 2133 - SPRING 2001 - REVIEW FOR EXAM III

As usual this review is basically a checklist. You should refer to your book and class notes for more details. It is also a good idea to reread the parts of the review for Exam II on integration. In particular you should know all the standard forms for integrals that were listed there.

PRODUCTS OF POWERS OF SIN AND COS

In this section we consider ${\displaystyle\int \sin^mx\,\cos^nx\,dx}$, where $m\geq0$ and $n\geq0$.

1.

First recall that ${\displaystyle\int \cos x\,dx=\sin x+C}$ and ${\displaystyle\int \sin x\,dx=-\cos x+C}$.

2.

Next note that ${\displaystyle\int \sin^mx \cos x\,dx}$ can be done by setting $u=\sin x$ and $du=\cos x\,dx$, and that ${\displaystyle\int \cos^mx \sin x\,dx}$ can be done by setting $u=\cos x$ and $du=-\sin x\,dx$. These substitutions turn the integral into ${\displaystyle\pm\int u^m\,du}$.

3.

Now consider ${\displaystyle\int \sin^mx\,\cos^nx\,dx}$, where at least one of m or n is odd.

If m is odd, say m=2k+1, then $\sin^mx\cos^nx=\sin^{2k+1}x\cos^nx=(\sin^2x)^k\sin x\cos^nx$. Use the identity $\cos^2x+\sin^2x=1$ to set $\sin^2x=1-\cos^2x$. This gives $(1-\cos^2x)^k\cos^nx\sin x$. The substitution $u=\cos x$, $du=-\sin x\,dx$ then turns our integral into ${\displaystyle\int-(1-u^2)^ku^n\,du}$. Now we are integrating a polynomial; just multiply it out and then use the power rule.

If n is odd, say n=2k+1, then $\sin^mx\cos^nx=\sin^mx\cos^{2k+1}x= \sin^mx(\cos^2x)^k\cos x$. Use the identity $\cos^2x+\sin^2x=1$ to set $\cos^2x=1-\sin^2x$. This gives $\sin^mx(1-\sin^2x)^k\cos x$. The substitution $u=\sin x$, $du=\cos x\,dx$ then turns our integral into ${\displaystyle\int u^m(1-u^2)^k\,du}$. Again we are integrating a polynomial; as before just multiply it out and then use the power rule.

4.

Next consider ${\displaystyle\int \sin^mx\,\cos^nx\,dx}$, where m and n are both even, say m=2k and $n=2\ell$. Thus we have ${\displaystyle\int (\sin^2x)^k(\cos^2x)^{\ell}\,dx}$.

Recall the identity $\cos(2x)=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x$. We can use this to express $\cos^2 x$ and $\sin^2x$ in terms of $\cos(2x)$.

$$\cos^2x=\frac{1+\cos(2x)}{2},\,\,\,\,\,\,\sin^2x=\frac{1-\cos(2x)}{2}$$

We use these formulas to replace the $\cos^2 x$ and $\sin^2x$ in our integral. Then we multiply out the result. We will get a sum of integrals involving powers of $\cos(2x)$. The powers involved will be smaller, so the integrals will be easier to do. We do those integrals which do not involve even powers of $\cos(2x)$. For those which do have even powers we do the same trick; we replace $\cos^2(2x)$ by ${\displaystyle\frac{1+\cos(4x)}{2}}$. We continue until we can do all the integrals.

It is worth remembering the results for the integrals of $\cos^2 x$ and $\sin^2x$ themselves. We get

$$\int\cos^2x\,dx=\int\frac{1}{2}+\frac{\cos(2x)}{2}\,dx= \frac{x}{2}+\frac{\sin(2x)}{4}+C$$

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

PRODUCTS OF POWERS OF TAN AND SEC

In this section we consider ${\displaystyle\int \tan^mx\,\sec^nx\,dx}$, where $m\geq0$ and $n\geq0$.

1.

First recall that ${\displaystyle\int \sec^2 x\,dx=\tan x+C}$ and ${\displaystyle\int \sec x\tan x\,dx=\sec x+C}$.

2.

Next note that ${\displaystyle\int \tan^mx \sec^2 x\,dx}$ can be done by setting $u=\tan x$ and $du=\sec^2 x\,dx$, and that ${\displaystyle\int \sec^{m}x \sec x\tan x\,dx}$ can be done by setting $u=\sec x$ and $du=\sec x\tan x\,dx$. These substitutions turn the integral into ${\displaystyle\int u^m\,du}$.

3.

Now consider ${\displaystyle\int \tan^mx\,\sec^nx\,dx}$, where n is non-zero and even, say n=2k with k>0. We factor $\tan^mx\sec^{2k}x=\tan^mx(\sec^2x)^{k-1}\sec^2x$ and use the identity $1+\tan^2x=\sec^2x$ to set $(\sec^2x)^{k-1}=(1+\tan^2x)^{k-1}$ and thereby get $\tan^mx(1+\tan^2x)^{k-1}\sec^2x$. The substitution $u=\tan x$, $du=\sec^2 x\,dx$ then turns our integral into ${\displaystyle\int u^m(1+u^2)^{k-1}\,du}$. We are integrating a polynomial; just multiply it out and use the power rule.

4.

Consider ${\displaystyle\int\tan^mx\,dx}$. If m=1 we know that ${\displaystyle\int \tan x\,dx=-\ln\vert\cos x\vert+C}$. If m=2, then use the identity $1+\tan^2x=\sec^2x$ to write $\tan^2x=\sec^2x-1$, which is a function we know how to integrate. If m>2, then use the identity to write $\tan^mx=\tan^{m-2}x\tan^2x=\tan^{m-2}x(\sec^2x-1)=\tan^{m-2}x\sec^2x-\tan^{m-2}x$. The first function we integrate as above; the second function is a smaller power of $\tan x$, so we continue until we get something we can completely integrate.

5.

Now consider ${\displaystyle\int \tan^mx\sec^nx\,dx}$, where n is odd, say n=2q+1.

If m is also odd, say m=2k+1, then we have $\tan^mx\sec^nx= (\tan^2x)^k(\sec^2x)^q\sec x\tan x$. If k=0, then we have $\sec^{2q}x\sec x\tan x$, which we know how to integrate using $u=\sec x$, $du=\sec x\tan x\,dx$. If k>0, then we use the identity $1+\tan^2x=\sec^2x$ to get $(\sec^2x-1)^k(\sec^2x)^q\sec x\tan x$, which we can now integrate using the same substitution.

If m is even, then we have a problem. Neither of the substitutions $u=\tan x$ or $u=\sec x$ will work. The integral requires a different technique, namely (a tricky) integration by parts.

PRODUCTS OF POWERS OF COT AND CSC

This concerns ${\displaystyle\int \cot^mx\,\csc^nx\,dx}$, where $m\geq0$ and $n\geq0$. The pattern here is similar to that for $\tan^mx\sec^nx$. We use the identity $1+\cot^2x=\csc^2x$, the substitution $u=\cot x$, $du=-\csc^2x\,dx$, and the substitution $u=\csc x$, $du=-\csc x\cot x\,dx$. Note the presence of the minus signs in the differentials.

OTHER TRIG INTEGRALS

As a last resort, write everything in terms of sin and cos and hope you find something that works.

INTEGRALS INVOLVING THE INVERSE TRIG FUNCTIONS

1.

Suppose a>0. Then we have the following two standard forms:

$$\int\frac{du}{\sqrt{a^2-u^2}}=\mathrm{Arcsin}\left(\frac{u}{a... ...}{a^2+u^2}=\frac{1}{a}\mathrm{Arctan}\left(\frac{u}{a}\right)+C$$

Note that the Arctan integral has a factor of ${\displaystyle\frac{1}{a}}$ in front; the Arcsin integral does not.

2.

The key skill here is that of changing your given integral into a standard form. You identify the a² and u² and from there the a and the u. For example, given ${\displaystyle\int\frac{dx}{\sqrt{4-9x^2}}}$, you set a²=4 and u²=9x²; this gives a=2 and u=3x, thus $du=3\,dx$, and so we have ${\displaystyle\frac{1}{3}\int\frac{du}{\sqrt{2^2-u^2}}}$. If the coefficients are not perfect squares, then use the trick that for b>0 one has $b=(\sqrt{b})^2$. For example, given ${\displaystyle\int\frac{dx}{3+5x^2}}$, you set a²=3 and u²=5x²; this gives $a=\sqrt{3}$ and $u=\sqrt{5}x$, thus $du=\sqrt{5}\,dx$, and so we have ${\displaystyle\frac{1}{\sqrt{5}}\int\frac{du}{{(\sqrt{3})^2+u^2}}}$.

3.

Sometimes you need to complete the square to get the integral into the proper form. Recall that (x+b)²=x²+2bx+b².

Given a polynomial x²+2bx+c, we add and subtract b² to rewrite it as x²+2bx+b²-b²+c=(x+b)²+(c-b²); if c-b²>0, then we let u=x+b and $a=\sqrt{c-b^2}$ to get our u²+a². For example, given x²+10x+29, we recognize x²+10x as being the first two terms of (x+5)²=x²+10x+25 and write x²+10x+29=x²+10x+25-25+29=(x+5)²+4, so we have u²+a², with u=x+5 and a=2.

Given a polynomial p-2bx-x², we write it as p-(x²+2bx)=p-(x²+2bx+b²-b²)= p-((x+b)²-b²)=(p+b²)-(x+b)². If p+b²>0, then we let u=x+b and $a=\sqrt{p+b^2}$ to get a²-u². For example, given -21-10x-x², we rewrite it as -21-(x²+10x)=-21-(x²+10x+25-25)=-21-((x+5)²-25)= -21+25-(x+5)²=4-(x+5)², so we have a²-u², with u=x+5 and a=2.

4.

Don't be fooled by integrals that look like Arcsin or Arctan integrals but are not.

${\displaystyle\int\frac{u\,du}{\sqrt{a^2-u^2}}}$ is done by setting w=a²-u², so $dw=-2u\,du$, and thus we have

$$\int\frac{u\,du}{\sqrt{a^2-u^2}}=-\frac{1}{2}\int\frac{dw}{\s... ...1/2}\,dw=-\frac{1}{2}\frac{w^{1/2}}{(1/2)}+C= -\sqrt{a^2-u^2}+C$$

Thus this is a general power rule integral.

${\displaystyle\int\frac{u\,du}{{a^2+u^2}}}$ is done by setting w=a²+u², so $dw=2u\,du$, and thus we have

$$\int\frac{u\,du}{{a^2+u^2}}=\frac{1}{2}\int\frac{dw}{{w}}= \frac{1}{2}\ln\vert w\vert+C= \frac{1}{2}\ln\vert a^2+u^2\vert+C$$

Thus this is a logarithmic integral.

You should be able to identify an integral as an Arcsin, Arctan, general power, or logarithmic integral, as in problems 25-28 in section 8.6 of the book.

5.

Note that ${\displaystyle\int\frac{(bu+c)\,du}{\sqrt{a^2-u^2}}= b\int\frac{u\,du}{\sqrt{a^2-u^2}}+ c\int\frac{du}{\sqrt{a^2-u^2}}}$, and so can be done as a combination of a general power rule integral and an Arcsin integral.

Note also that ${\displaystyle\int\frac{(bu+c)\,du}{{a^2+u^2}}= b\int\frac{u\,du}{{a^2+u^2}}+ c\int\frac{du}{{a^2+u^2}}}$, and so can be done as a combination of a logarithmic integral and an Arctan integral.

INTEGRATION BY PARTS

1.

The integration by parts formula is

$$\int f(x)g^{\prime}(x)\,dx=f(x)g(x)-\int f^{\prime}(x)g(x)\,dx.$$

The general idea here is that f(x) should be some function with a simpler derivative $f^{\prime}(x)$, and $g^{\prime}(x)$ should be a function whose anti-derivative g(x) is fairly easy to find. The formula can also be expressed using differentials by setting u=f(x), $du=f^{\prime}(x)\,dx$, v=g(x), and $dv=g^{\prime}(x)\,dx$. We then have

$$\int u\,dv=uv-\int v\,du.$$

Sometimes you have to do integration by parts several times before finally getting everything integrated.

2.

One trick that is sometimes used on an integral of the form ${\displaystyle\int f(x)\,dx}$ is to set $g^{\prime}(x)=1$; then g(x)=x. For example,

$$\int \mathrm{Arctan}\, x \,dx= (\mathrm{Arctan}\, x)(x)-\int ... ...e} (x)\,dx=x \mathrm{Arctan}\, x-\int (\frac{1}{1+x^2})(x)\,dx=$$

$$x \mathrm{Arctan}\, x-\frac{1}{2}\ln\vert 1+x^2\vert+C.$$

(In the last integral we used the substitution w=1+x², $dw=2x\,dx$.)

3.

Another trick is to do integration by parts twice, writing the integral in terms of itself, solving for it, and then tacking on a +C. For example, given ${\displaystyle\int e^{2x}\sin(3x)\,dx}$, we choose u=e^2x and $dv=\sin(3x)\,dx$. Then $du=2e^{2x}\,dx$ and ${\displaystyle v=-\frac{1}{3}\cos(3x)}$, so we have

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

Next we do ${\displaystyle\int e^{2x}\cos(3x)\,dx}$ by setting u=e^2x and $dv=\cos(3x)\,dx$, so we have $du=2e^{2x}\,dx$ and ${\displaystyle v=\frac{1}{3}\sin(3x)}$. This gives

$$\int e^{2x}\cos(3x)\,dx=e^{2x}(\frac{1}{3}\sin(3x))-\int 2e^{... ...= \frac{1}{3}e^{2x}\sin(3x)-\frac{2}{3}\int e^{2x}\sin(3x)\,dx.$$

We now plug this into our previous formula to get ${\displaystyle\int e^{2x}\sin(3x)\,dx=-\frac{1}{3}e^{2x}\cos(3x)+\frac{2}{3}\int e^{2x}\cos(3x)\,dx}$

$$=-\frac{1}{3}e^{2x}\cos(3x)+\frac{2}{3}\left[\frac{1}{3}e^{2x}\sin(3x)-\frac{2}{3}\int e^{2x}\sin(3x)\,dx\right]$$

$$=-\frac{1}{3}e^{2x}\cos(3x)+\frac{2}{9}e^{2x}\sin(3x)-\frac{4}{9}\int e^{2x}\sin(3x)\,dx$$

We add ${\displaystyle\frac{4}{9}\int e^{2x}\sin(3x)\,dx}$ to both sides of the equation to get

$$\frac{13}{9}\int e^{2x}\sin(3x)\,dx= -\frac{1}{3}e^{2x}\cos(3... ...frac{2}{9}e^{2x}\sin(3x)= \frac{-3\cos(3x)+2\sin(3x)}{9}e^{2x}.$$

Multiplying both sides by ${\displaystyle\frac{9}{13}}$ and then tacking on a +C gives

$$\int e^{2x}\sin(3x)\,dx=e^{2x}\frac{-3\cos(3x)+2\sin(3x)}{13}+C.$$

Note that if we had made the other choice for u and dv in the second integral we would have just gotten that ${\displaystyle\int e^{2x}\sin(3x)\,dx=\int e^{2x}\sin(3x)\,dx}$! The secret is to not undo the differentiation and anti-differentiation that you have just done.

INTEGRATION BY LINEAR SUBSTITUTION

Given an integral involving (ax+b)^p/q, one thing to try is letting u= (ax+b)^1/q. Then instead of computing du directly, you first take the q^th power of both sides to get u^q=(ax+b), then take d of both sides to get du^q=d(ax+b), and thus $qu^{q-1}du=a\,dx$.

For example, given ${\displaystyle\int x\sqrt[3]{(x+1)^2}\,dx}$ we choose $u=\sqrt[3]{x+1}$. Then u³=x+1, $3u^2\,du=dx$, and (very important) x=u³-1. This turns our integral into

$${\displaystyle\int (u^3-1)u\,du=\int (u^4-u)\,du= \frac{u^5}{5}-\frac{u^2}{2}+C=\frac{(x+1)^{5/3}}{5}-\frac{(x+1)^{2/3}}{2}+C}.$$

TRIG SUBSTITUTIONS

These involve integrals containing $\sqrt{a^2-u^2}$, $\sqrt{a^2+u^2}$, or $\sqrt{u^2-a^2}$. In each case you set u equal to a times an appropriate trig function of $\theta$, deal with du and $d\theta$, do the $\theta$ integral, and finally express everything back in terms of u using inverse trig functions and triangles.

1.

If you have $\sqrt{a^2-u^2}$, set $u=a\sin\theta$, where ${\displaystyle-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}}$. Then $du=a\cos\theta\,d\theta$ and $\sqrt{a^2-u^2}=a\cos\theta$.

This will give you the integral of some function of $\theta$. Integrate it. This will give you another function of $\theta$, which you need to rewrite as a function of u. To handle trig functions of $\theta$ you use the fact that ${\displaystyle\sin \theta=\frac{u}{a}}$ and draw a right triangle which will give all the other trig functions of $\theta$.

For example, ${\displaystyle\tan\theta=\frac{u}{\sqrt{a^2-u^2}}}$.

(Technically, this triangle just covers the case ${\displaystyle0\leq \theta\leq\frac{\pi}{2}}$. Fortunately, the answers you get for the trig functions using this triangle will also work for ${\displaystyle-\frac{\pi}{2}\leq\theta\leq 0}$, so you don't need to consider that case separately.)

If you have any $\theta$ which does not appear inside a trig function, then set ${\displaystyle\theta=\mathrm{Arcsin}\left(\frac{u}{a}\right)}$.

2.

If you have $\sqrt{a^2+u^2}$, set $u=a\tan\theta$, where ${\displaystyle-\frac{\pi}{2}<\theta<\frac{\pi}{2}}$. Then $du=a\sec^2\theta\,d\theta$ and $\sqrt{a^2+u^2}=a\sec\theta$.

This will give you the integral of some function of $\theta$. Integrate it. This will give you another function of $\theta$, which you need to rewrite as a function of u. To handle trig functions of $\theta$ you use the fact that ${\displaystyle\tan \theta=\frac{u}{a}}$ and draw a right triangle which will give all the other trig functions of $\theta$.

For example, ${\displaystyle\sin\theta=\frac{u}{\sqrt{a^2+u^2}}}$.

(Technically, this triangle just covers the case ${\displaystyle0\leq \theta<\frac{\pi}{2}}$. Fortunately, the answers you get for the trig functions using this triangle will also work for ${\displaystyle-\frac{\pi}{2}<\theta\leq 0}$, so you don't need to consider that case separately.)

If you have any $\theta$ which does not appear inside a trig function, then set ${\displaystyle\theta=\mathrm{Arctan}\left(\frac{u}{a}\right)}$.

3.

Suppose you have $\sqrt{u^2-a^2}$. This case is a bit more complicated because you do different things depending on whether u is positive or negative. The best way to handle this is to first consider the case where u is positive. Then if you are in a situation where u is negative, just make the substitution w=-u, dw=-du; then w is positive, and you proceed as before.

So assume u is positive. Set $u=a\sec\theta$, where ${\displaystyle0\leq \theta<\frac{\pi}{2}}$. Then $du=a\sec\theta\tan\theta\,d\theta$ and $\sqrt{u^2-a^2}=a\tan\theta$.

This will give you the integral of some function of $\theta$. Integrate it. This will give you another function of $\theta$, which you need to rewrite as a function of u. To handle trig functions of $\theta$ you use the fact that ${\displaystyle\sec\theta=\frac{u}{a}}$ and draw a right triangle which will give all the other trig functions of $\theta$.

For example, ${\displaystyle\sin\theta=\frac{\sqrt{u^2-a^2}}{u}}$.

If you have any $\theta$ which does not appear inside a trig function, then set ${\displaystyle\theta=\mathrm{Arcsec}\left(\frac{u}{a}\right)}$.