vault backup: 2025-09-05 13:04:16

education/calculus/precalculus/Composing Functions.md

@@ -1,3 +0,0 @@
-- To compose a function is to create a new function from multiple smaller functions.
-- They can be solved from the inside out
-- 

education/math/MATH1220 (calc II)/Integration with Trig Identities.md

@@ -49,5 +49,10 @@ $$ = 3\int\frac{1}{4(\sec^2\theta)}(2\sec^2\theta)d\theta$$
 $$ 3\int\frac{2}{4}d\theta = \frac{3}{2} \theta + C$$
 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 to,
-$$ $$
+	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:
+$$ \int\frac{3}{4+x^2}dx = \frac{3}{2}\arctan(\frac{x}{2}) + C $$