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.

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.