The below integration makes use of the following trig identities:
1. The Pythagorean identity: $\sin^2(x) + \cos^2(x) = 1$
2. The derivative of sine: $\frac{d}{dx}sin(x) = cos(x)$
3. The derivative of cosine: $\dfrac{d}{dx} \cos(x) = -\sin(x)$
4. Half angle cosine identity: $\cos^2(x) = \frac{1}{2}(1 + \cos(2x))$
5. Half angle sine identity: $\sin^2(x) = \frac{1}{2}(1 - \cos(2x))$
6. $tan^2(x) + 1 = sec^2(x)$
7. $\dfrac{d}{dx}(\tan(x)) = \sec^2(x) \Rightarrow \int \sec^2(x)dx = \tan(x) + C$ 
8. $\dfrac{d}{dx}(\sec x) = \sec(x)\tan(x) \Rightarrow \int\sec(x)\tan(x) dx = \sec(x) + C$
# Examples
> Evaluate the integral $\int\sin^5(x)dx$
1. With trig identities, it's common to work *backwards* with u-sub. In the above example, we can convert the equation into simpler cosine functions by setting $du$ to $-\sin(x)dx$. This means that $u$ is equal to $cos(x)$. 
$$ \int\sin^4(x)\sin(x)dx$$
2. Rewrite $sin^4(x)$ to be $(\sin^2(x))^2$ to take advantage of the trig identity $1 - \cos^2(x) = \sin^2(x)$ 
$$ \int(\sin^2x)^2 \sin(x)dx$$
3. Apply the above trig identity and substitute $u$:
$$ \int(1 - u^2)^2 (-du) $$
4. Foil out and move negative out of integral:
$$ -\int(1 - 2u^2 + u^4)du $$
5. Take advantage of the distributive property of integrals:
 $$ - (u - \frac{2}{3}u^3 + \frac{1}{5}u^5) + C $$
 6. Substituting $\cos(x)$ back in for $u$, we get the evaluated (but not entirely simplified) integral:
 $$-(\cos(x)- \frac{2}{3}\cos^3x + \frac{1}{5}\cos^5x) $$
# Trigonometric Substitutions
Trigonometric substitution is useful for equations containing  $\sqrt{a^2 + x^2}$ or $a^2 + x^2$, where $a$ is any constant. It removes any addition or subtraction.

The general process involves the use of a trig identity and multiplying everything in that identity by a constant.

Consider the identity:
$$ 1 + \tan^2(\theta) = \sec^2(\theta)$$
Multiplying both sides of the identity by $a^2$, we get:
$$a^2 + a^2\tan^2(\theta) = a^2\sec^2(\theta)$$
This enables us to make use of **substitution** to simplify many integrals.
- $x = a\tan \theta$
- $dx = a \sec^2\theta d\theta$
- for $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$
# Examples
> Evaluate the integral $\int\frac{3}{4+x^2}dx$ 
1. Move the constant coefficient out of the integral:
$$ \int \frac{3}{4+x^2}dx = 3\int \frac{1}{4 + x^2}dx$$
2. Let $x = 2tan\theta$ and $dx = (2sec^2\theta d\theta)$, substitute accordingly
$$ = 3\int\frac{1}{4 + 4\tan^2\theta}(2\sec^2\theta)d\theta$$
3. Factor $4$ out in the denominator
$$ = 3\int\frac{1}{4(1 + \tan^2\theta)}(2\sec^2\theta)d\theta$$
4. Considering the identity $1 + \tan^2 \theta = \sec^2 \theta$
$$ = 3\int\frac{1}{4(\sec^2\theta)}(2\sec^2\theta)d\theta$$
5. $\sec^2\theta$ is present in the numerator and the denominator, so we can cancel those out. This means that:
$$ 3\int\frac{2}{4}d\theta = \frac{3}{2} \theta + C$$
6. At this point, we want to determine what $\theta$ is equal to relative to $x$. 
	1. Look back to step 2 we defined $x = 2\tan\theta$
	2. Moving $2$ to the other side, we get $\frac{x}{2} = \tan\theta$ 
	3. Because we defined bounds for our definition of $\theta$, we can take advantage of $\arctan$, therefore:
$$ \theta = \arctan(\frac{x}{2}) $$
7. Rewriting the equation with $\theta$ in terms of x, we get:
$$ \frac{3}{2}\arctan(\frac{x}{2}) + C$$
This means that:
$$ \int\frac{3}{4+x^2}dx = \frac{3}{2}\arctan(\frac{x}{2}) + C $$