W02 Notes

Trig power products

Videos

Review Videos

Videos, Math Dr. Bob:

• Trig power products: $\int {\cos }^{m}x{\sin }^{n}xdx$
• Trig differing frequencies: $\int \cos mx\sin nxdx$
• Trig tan and sec: $\int {\tan }^{m}x{\sec }^{n}xdx$
• Secant power: $\int {\sec }^{5}xdx$

Videos, Organic Chemistry Tutor:

• Trig power product techniques
• Trig substitution

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01 Theory

Theory 1

Review: trig identities

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

Trig power product: $\sin /\cos$

A $\sin /\cos$ power product has this form:

$$\int {\cos }^{m}x\cdot {\sin }^{n}xdx$$

for some integers $m$ and $n$ (even negative!).

To compute these integrals, use a sequence of these techniques:

• Swap an even bunch.
• $u$-sub for power-one.
• Power-to-frequency conversion.

Memorize these three techniques!

Examples of trig power products:

• ${\displaystyle \int \sin x\cdot {\cos }^{7}xdx}$
• ${\displaystyle \int {\sin }^{3}xdx}$
• ${\displaystyle \int {\sin }^{2}x\cdot {\cos }^{2}xdx}$

Swap an even bunch

If either ${\cos }^{m}x$ or ${\sin }^{n}x$ is an odd power, use

$$\begin{array}{cc}{\sin }^{2}x\gg \gg & 1-{\cos }^{2}x\\ & \\ \text{OR}{\cos }^{2}x\gg \gg & 1-{\sin }^{2}x\end{array}$$

(maybe repeatedly) to convert an even bunch to the opposite trig type.

An even bunch is all but one from the odd power.

For example:

$$\begin{array}{cc}{\sin }^{5}x\cdot {\cos }^{8}x& \gg \gg \sin x({\sin }^{2}x{)}^2\cdot {\cos }^{8}x\\ & \\ & \gg \gg \sin x(1-{\cos }^{2}x{)}^2\cdot {\cos }^{8}x\\ & \\ & \gg \gg \sin x(1-2{\cos }^{2}x+{\cos }^{4}x)\cdot {\cos }^{8}x\\ & \\ & \gg \gg \sin x({\cos }^{8}x-2{\cos }^{10}x+{\cos }^{12}x)\\ & \\ & \gg \gg \sin x{\cos }^{8}x-2\sin x{\cos }^{10}x+\sin x{\cos }^{12}x\end{array}$$

$u$-sub for power-one

If $m=1$ or $n=1$, perform $u$-substitution to do the integral.

The other trig power becomes a $u$ power; the power-one becomes $du$.

For example, using $u=\cos x$ and thus $du=-\sin xdx$ we can do:

$$\int \sin x{\cos }^{8}xdx\gg \gg \int -{\cos }^{8}x(-\sin xdx)\gg \gg -\int u^{8}du$$

By combining these tricks you can do any power product with at least one odd power! Make sure to leave a power-one from the odd power when swapping an even bunch.

Notice

Even powers: $1={\sin }^{0}x={\cos }^{0}x$. So the method works for $\int {\sin }^{3}xdx$ and similar.

Power-to-frequency conversion

Using these ‘power-to-frequency’ identities (maybe repeatedly):

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

change an even power (either type) into an odd power of cosine.

For example, consider the power product:

$${\sin }^{4}x\cdot {\cos }^{6}x$$

You can substitute appropriate powers of ${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$ and ${\cos }^{2}x=\frac{1}{2}(1+\cos 2x)$:

$$\begin{array}{cc}{\sin }^{4}x\cdot {\cos }^{6}x& \gg \gg ({\sin }^{2}x{)}^2\cdot ({\cos }^{2}x{)}^3\\ & \\ & \gg \gg (\frac{1}{2}(1-\cos 2x){)}^2\cdot (\frac{1}{2}(1+\cos 2x){)}^3\end{array}$$

By doing some annoying algebra, this expression can be expanded as a sum of smaller powers of $\cos 2x$:

$$\begin{array}{c}(\frac{1}{2}(1-\cos 2x){)}^2\cdot (\frac{1}{2}(1+\cos 2x){)}^3\\ \\ \gg \gg \frac{1}{32}(1+\cos (2x)-2{\cos }^2(2x)-2{\cos }^3(2x)+{\cos }^4(2x)+{\cos }^5(2x))\end{array}$$

Each of these terms can be integrated by repeating the same techniques.

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02 Illustration

Example - Power product - odd power

Power product - odd power

Compute the integral:

$$\int {\cos }^{2}x\cdot {\sin }^{5}xdx$$

Solution

(1) Swap over the even bunch:

Max even bunch leaving power-one is ${\sin }^{4}x$.

$${\sin }^{5}x\gg \gg \sin x({\sin }^{2}x{)}^2\gg \gg \sin x(1-{\cos }^{2}x{)}^2$$

Apply to ${\sin }^{5}x$ in the integrand:

$$\int {\cos }^{2}x\cdot {\sin }^{5}xdx\gg \gg \int {\cos }^{2}x\cdot \sin x(1-{\cos }^{2}x{)}^{2}dx$$

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(2) Perform $u$-substitution on the power-one integrand:

Set $u=\cos x$. Hence $du=\sin xdx$. Recognize this in the integrand and convert:

$$\begin{array}{cc}\int {\cos }^{2}x\cdot \sin x(1-{\cos }^{2}x{)}^{2}dx& \gg \gg \int {\cos }^{2}x\cdot (1{\cos }^{2}x{)}^2(\sin xdx)\\ & \\ & \gg \gg \int u^2\cdot (1-u^2{)}^{2}du\end{array}$$

---

(3) Integrate using power rule:

$$\int u^2\cdot {(1-u^2)}^{2}du=\frac{1}{3}{u}^3-\frac{2}{5}{u}^5+\frac{1}{7}{u}^7+C$$

Insert definition $u=\cos x$:

$$\begin{array}{c}\gg \gg \int {\cos }^{2}x\cdot {\sin }^{5}xdx\gg \gg \int u^2\cdot {(1-u^2)}^{2}du\\ \\ \gg \gg \frac{1}{3}{\cos }^{3}x-\frac{2}{5}{\cos }^{5}x+\frac{1}{7}{\cos }^{7}x+C\end{array}$$ Link to original

03 Theory

Theory 2

Trig power product: $\tan /\sec$ or $\cot /\csc$

A $\tan /\sec$ power product has this form:

$$\int {\tan }^{m}x\cdot {\sec }^{n}xdx$$

A $\cot /\csc$ power product has this form:

$$\int {\cot }^{m}x\cdot {\csc }^{n}xdx$$

To integrate these, swap an even bunch using:

• ${\tan }^{2}x+1={\sec }^{2}x$

OR:

• ${\cot }^{2}x+1={\csc }^{2}x$

Or do $u$-substitution using:

• $u=\tan x⇝du={\sec }^{2}xdx$
• $u=\sec x⇝du=\sec x\tan xdx$

OR:

• $u=\cot x⇝du=-{\csc }^{2}xdx$
• $u=\csc x⇝du=-\csc u\cot udx$

Note

There is no simple “power-to-frequency conversion” for tan / sec !

We can modify the power-one technique to solve some of these. We need to swap over an even bunch from the odd power so that exactly the $du$ factor is left behind.

Considering all the possibilities, one sees that this method works when:

• ${\tan }^{m}x$ is an odd power (with some secants present!)
• ${\sec }^{n}x$ is an even power

Quite a few cases escape this method:

• Any $\int {\tan }^{m}xdx$ with no power of $\sec x$
• Any $\int {\tan }^{m}x\cdot {\sec }^{n}xdx$ for $m$ even and $n$ odd

These tricks don’t work for $\int \tan xdx$ or $\int \sec xdx$ or $\int {\tan }^{4}x{\sec }^{5}xdx$, among others.

Special integrals: tan and sec

We have:

$$\begin{array}{cc}\int \tan xdx& =\ln |\sec x|+C\\ & \\ \int \sec xdx& =\ln |\sec x+\tan x|+C\end{array}$$

Note

These integrals should be memorized individually.

Extra - Deriving special integrals: tan and sec

The first formula can be found by $u$-substitution, considering that $\tan x=\frac{\sin x}{\cos x}$.

The second formula can be derived by multiplying $\sec x$ by a special “$1$”, computing instead $\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x}dx$ by expanding the numerator and doing $u$-sub on the denominator.

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04 Illustration

Example - Power product - tan and sec

Power product - tan and sec

Compute the integral:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx$$

Solution

(1) Try $du={\sec }^{2}xdx$:

Factor $du$ out of the integrand:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx\gg \gg \int {\tan }^{5}x\cdot \sec x({\sec }^{2}xdx)$$

We then must swap over remaining $\sec x$ into the $\tan x$ type.

Cannot do this because $\sec x$ has odd power. Need even to swap.

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(2) Try again: $du=\sec x\tan xdx$:

Factor $du$ out of the integrand:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx\gg \gg \int {\tan }^{4}x\cdot {\sec }^{2}x\left(\sec x\tan xdx\right)$$

Swap remaining $\tan x$ into $\sec x$ type:

$$\begin{array}{c}\gg \gg \int ({\tan }^{2}x{)}^2\cdot {\sec }^{2}x(\sec x\tan xdx)\\ \\ \gg \gg \int ({\sec }^{2}x-1{)}^2\cdot {\sec }^{2}x(\sec x\tan xdx)\end{array}$$

Substitute $u=\sec x$ and $du=\sec x\tan xdx$:

$$\gg \gg \int {(u^2-1)}^2\cdot u^{2}du$$

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(3) Integrate in $u$ and convert back to $x$:

$$\begin{array}{c}\gg \gg \int u^6-2u^4+u^{2}du\\ \\ \gg \gg \frac{u^7}{7}-2\frac{u^5}{5}+\frac{u^3}{3}+C\\ \\ \gg \gg \frac{{\sec }^{7}x}{7}-2\frac{{\sec }^{5}x}{5}+\frac{{\sec }^{3}x}{3}+C\end{array}$$ Link to original

Trig substitution

Videos

Review Videos

Videos, Math Dr. Bob:

• Trig sub 1: Basics and $\int \frac{1}{\sqrt{36-x^2}}dx$ and $\int \frac{x}{36+x^2}dx$ and $\int \frac{1}{\sqrt{x^2-36}}dx$
• Trig sub 2: $\int \frac{dx}{{(1+x^2)}^{5/2}}$
• Trig sub 3: $\int \frac{x^2}{\sqrt{1-4x^2}}dx$
• Trig sub 4: $\int \sqrt{e^{2x}-1}dx$
• Trig sub 5: $\int \frac{\sqrt{4-36x^2}}{x^2}dx$

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05 Theory

Theory 1

Certain algebraic expressions have a secret meaning that comes from the Pythagorean Theorem. This meaning has a very simple expression in terms of trig functions of a certain angle.

For example, consider the integral:

$$\int \frac{1}{x^2\sqrt{x^2-9}}dx$$

Now consider this triangle:

The triangle determines the relation $x=3\sec \theta$, and it implies $\sqrt{x^2-9}=3\tan \theta$.

Now plug these into the integrand above:

$$\frac{1}{x^2\sqrt{x^2-9^2}}\gg \gg \frac{1}{9{\sec }^2\theta \cdot 3\tan \theta }$$

Considering that $dx=3\sec \theta \tan \theta d\theta$, we obtain a very reasonable trig integral:

$$\begin{array}{cc}\int \frac{1}{x^2\sqrt{x^2-9^2}}dx\gg \gg & \int \frac{3\sec \theta \tan \theta }{27{\sec }^2\theta \tan \theta }d\theta \\ & \\ \gg \gg & \frac{1}{9}\int \cos \theta d\theta \gg \gg \frac{1}{9}\sin \theta +C\end{array}$$

We must rewrite this in terms of $x$ using $x=3\sec \theta$ to finish the problem. We need to find $\sin \theta$ assuming that $\sec \theta =\frac{x}{3}$. To do this, refer back to the triangle to see that $\sin \theta =\frac{\sqrt{x^2-9}}{x}$. Plug this in for our final value of the integral:

$$\frac{1}{9}\sin \theta +C\gg \gg \frac{\sqrt{x^2-9}}{9x}+C$$

Here is the moral of the story:

Pythagorean expressions

Re-express the Pythagorean expression using a triangle and a trig substitution.

In this way, we are able to eliminate square roots of quadratics.

There are always three steps for these trig sub problems:

• (1) Identify the trig sub: find the sides of a triangle and relevant angle $\theta$.
• (2) Solve a trig integral (often a power product).
• (3) Refer back to the triangle to convert the answer back to $x$.

To speed up your solution process for these problems, memorize these three transformations:

(1)

$$\sqrt{a^2-x^2}{\gg \gg }^{x=a\sin \theta }\sqrt{a^2-a^2{\sin }^2\theta }=a\cos \theta \text{from}1-{\sin }^2\theta ={\cos }^2\theta$$

(2)

$$\sqrt{a^2+x^2}{\gg \gg }^{x=a\tan \theta }\sqrt{a^2+a^2{\tan }^2\theta }=a\sec \theta \text{from}1+{\tan }^2\theta ={\sec }^2\theta$$

(3)

$$\sqrt{x^2-a^2}{\gg \gg }^{x=a\sec \theta }\sqrt{a^2{\sec }^2\theta -a^2}=a\tan \theta \text{from}{\sec }^2\theta -1={\tan }^2\theta$$

For a more complex quadratic with linear and constant terms, you will need to first complete the square for the quadratic and then do the trig substitution.

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06 Illustration

Example - Trig sub in quadratic: completing the square

Trig sub in quadratic - completing the square

Compute the integral:

$$\int \frac{dx}{\sqrt{x^2-6x+11}}$$

Solution

(1) Complete the square to obtain Pythagorean form:

$$x^2-6x+{\left(\frac{-6}{2}\right)}^2\gg \gg x^2-6x+9={(x-3)}^2$$

Add and subtract to get desired constant term:

$$\begin{array}{c}{x}^2-6x+11\gg \gg x^2-6x+9-9+11\\ \\ \gg \gg {(x-3)}^2+2\end{array}$$

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(2) Perform shift substitution:

Set $u=x-3$ as inside the square:

$${(x-3)}^2+2=u^2+2$$

Infer $du=dx$. Plug into integrand:

$$\int \frac{dx}{\sqrt{x^2-6x+11}}\gg \gg \int \frac{du}{\sqrt{u^2+2}}$$

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(3) Trig sub with $\tan \theta$:

Use substitution $u=\sqrt{2}\tan \theta$. (From triangle or memorized tip.)

Infer $du=\sqrt{2}{\sec }^2\theta d\theta$. Plug in data:

$$\int \frac{du}{\sqrt{u^2+2}}\gg \gg \int \frac{{\sec }^2\theta }{\sec \theta }d\theta =\int \sec \theta d\theta$$

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(4) Integrate using ad hoc memorized formula:

$$\int \sec \theta d\theta \gg \gg \ln |\tan \theta +\sec \theta |+C$$

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(5) Convert trig back to $x$:

First in terms of $u$, referring to the triangle:

$$\tan \theta =\frac{u}{\sqrt{2}},\sec \theta =\frac{\sqrt{u^2+2}}{\sqrt{2}}$$

Then in terms of $x$ using $u=x-3$. Plug everything in:

$$\ln |\tan \theta +\sec \theta |+C\gg \gg \ln |\frac{x-3}{\sqrt{2}}+\frac{\sqrt{{(x-3)}^2+2}}{\sqrt{2}}|+C$$

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(6) Simplify using log rules:

$$\ln \frac{f(x)}{a}\gg \gg \ln f(x)-\ln a$$

The common denominator $\frac{1}{\sqrt{2}}$ can be pulled outside as $-\ln \sqrt{2}$.

The new term $-\ln \sqrt{2}$ can be “absorbed into the constant” (redefine $C$).

So we write our final answer thus:

$$\ln |x-3+\sqrt{{(x-3)}^2+2}|+C$$Link to original