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

@@ -22,7 +22,7 @@ $$ -\int(1 - 2u^2 + u^4)du $$
  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.
+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.
 
@@ -41,4 +41,13 @@ This enables us to make use of **substitution** to simplify many integrals.
 $$ \int \frac{3}{4+x^2}dx = 3\int \frac{1}{4 + x^2}dx$$
 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 $4$ 
+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 to,
+$$ $$