Integrals Resulting in Inverse Trigonometric Functions - Revision history

Khanh: /* Example 2 */

2022-01-15T22:38:29Z

Example 2

Khanh at 22:37, 15 January 2022

2022-01-15T22:37:27Z

Lila: /* Resources */

2021-10-28T20:34:59Z

Resources

Lila: /* Example 2 */

2021-10-28T20:33:17Z

Example 2

Lila: /* Example 2 */

2021-10-28T20:30:12Z

Example 2

Lila: /* Example Problem */

2021-10-28T20:23:44Z

Example Problem

Lila at 20:21, 28 October 2021

2021-10-28T20:21:44Z

Lila at 20:20, 28 October 2021

2021-10-28T20:20:24Z

Lila at 20:19, 28 October 2021

2021-10-28T20:19:50Z

Lila at 20:19, 28 October 2021

2021-10-28T20:19:06Z

@@ Line 28: / Line 28: @@
 <p>This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral:</p>
-<p><math> \int \frac{4-x}{\sqrt{16-x^2}}\text{dx}  = \int \frac{4}{\sqrt{16-x^2}}\text{dx}  - \int \frac{x}{\sqrt{16-x^2}}\text{dx}  </math>/p>
+<p><math> \int \frac{4-x}{\sqrt{16-x^2}}\text{dx}  = \int \frac{4}{\sqrt{16-x^2}}\text{dx}  - \int \frac{x}{\sqrt{16-x^2}}\text{dx}  </math> </p>
 <p>The first integral is handled straightforward; the second integral is handled by substitution, with <math>u = 16-x^2</math>. We handle each separately. </p>

@@ Line 34: / Line 34: @@
 <p><math>\int \frac{4}{\sqrt{16-x^2}}\text{dx}  = 4\arcsin\frac{x}{4} + C.</math></p>
-<p><math>\int\frac{x}{\sqrt{16-x^2}}\text{dx} </math>: Set <math>u = 16-x^2</math>, so <math>\text{du}  = -2x\text{dx} <math> and <math>x\text{dx}  = -\text{du} /2</math>. We have</p>
+<p><math>\int\frac{x}{\sqrt{16-x^2}}\text{dx} </math>: Set <math>u = 16-x^2</math>, so <math>\text{du}  = -2x\text{dx} </math> and <math>x\text{dx}  = -\text{du} /2</math>. We have</p>
 <p><math>\begin{align} \int\frac{x}{\sqrt{16-x^2}}\text{dx}  =  \int\frac{-\text{du} /2}{\sqrt{u}}\\ = -\frac12\int \frac{1}{\sqrt{u}}\text{du}  \\ = - \sqrt{u} + C\\ = -\sqrt{16-x^2} + C.\end{align}</math></p>

@@ Line 43: / Line 43: @@
 ==Resources==
-[https://youtu.be/AE-0gXXx_j0 Integration into Inverse trigonometric functions using Substitution] by The Organic Chemistry Tutor
+*[https://youtu.be/AE-0gXXx_j0 Integration into Inverse trigonometric functions using Substitution] by The Organic Chemistry Tutor
-[https://youtu.be/MdsAvt9y5ds Integrating using Inverse Trigonometric Functions] by  patrickJMT
+*[https://youtu.be/MdsAvt9y5ds Integrating using Inverse Trigonometric Functions] by  patrickJMT

@@ Line 22: / Line 22: @@
 ===Example 2===
-<p>Evaluate <math> \int \frac{4-x}{\sqrt{16-x^2}}\dx </math>.</p>
+<p>Evaluate <math> \int \frac{4-x}{\sqrt{16-x^2}}\text{dx} </math>.</p>
 <p><strong>Solution</strong></p>
@@ Line 28: / Line 28: @@
 <p>This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral:</p>
-<p><math> \int \frac{4-x}{\sqrt{16-x^2}}\ dx = \int \frac{4}{\sqrt{16-x^2}}\ dx - \int \frac{x}{\sqrt{16-x^2}}\dx </math>/p>
+<p><math> \int \frac{4-x}{\sqrt{16-x^2}}\text{dx}  = \int \frac{4}{\sqrt{16-x^2}}\text{dx}  - \int \frac{x}{\sqrt{16-x^2}}\text{dx}  </math>/p>
 <p>The first integral is handled straightforward; the second integral is handled by substitution, with <math>u = 16-x^2</math>. We handle each separately. </p>
-<p><math>\int \frac{4}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac{x}{4} + C.</math></p>
+<p><math>\int \frac{4}{\sqrt{16-x^2}}\text{dx}  = 4\arcsin\frac{x}{4} + C.</math></p>
-<p><math>\int\frac{x}{\sqrt{16-x^2}}\ dx</math>: Set <math>u = 16-x^2</math>, so <math>du = -2xdx<math> and <math>xdx = -du/2</math>. We have</p>
+<p><math>\int\frac{x}{\sqrt{16-x^2}}\text{dx} </math>: Set <math>u = 16-x^2</math>, so <math>\text{du}  = -2x\text{dx} <math> and <math>x\text{dx}  = -\text{du} /2</math>. We have</p>
-<p><math>\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx &amp;= \int\frac{-du/2}{\sqrt{u}}\\ &amp;= -\frac12\int \frac{1}{\sqrt{u}}\ du \\ &amp;= - \sqrt{u} + C\\ &amp;= -\sqrt{16-x^2} + C.\end{align}</math></p>
+<p><math>\begin{align} \int\frac{x}{\sqrt{16-x^2}}\text{dx}  =  \int\frac{-\text{du} /2}{\sqrt{u}}\\ = -\frac12\int \frac{1}{\sqrt{u}}\text{du}  \\ = - \sqrt{u} + C\\ = -\sqrt{16-x^2} + C.\end{align}</math></p>
 <p>Combining these together, we have</p>
-<p><math> \int \frac{4-x}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.</math></p>
+<p><math> \int \frac{4-x}{\sqrt{16-x^2}}\text{dx}  = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.</math></p>
 ==Resources==

@@ Line 22: / Line 22: @@
 ===Example 2===
-<p>Evaluate \(\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\ dx\).</p>
+<p>Evaluate <math> \int \frac{4-x}{\sqrt{16-x^2}}\dx </math>.</p>
 <p><strong>Solution</strong></p>
@@ Line 28: / Line 28: @@
 <p>This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral:</p>
-<p>$$ \int \frac{4-x}{\sqrt{16-x^2}}\ dx = \int \frac{4}{\sqrt{16-x^2}}\ dx - \int \frac{x}{\sqrt{16-x^2}}\ dx.$$</p>
+<p><math> \int \frac{4-x}{\sqrt{16-x^2}}\ dx = \int \frac{4}{\sqrt{16-x^2}}\ dx - \int \frac{x}{\sqrt{16-x^2}}\dx </math>/p>
-<p>The first integral is handled using a straightforward application of Theorem \(\PageIndex{2}\); the second integral is handled by substitution, with \(u = 16-x^2\). We handle each separately.</p>
+<p>The first integral is handled straightforward; the second integral is handled by substitution, with <math>u = 16-x^2</math>. We handle each separately. </p>
-<p>\(\displaystyle \int \frac{4}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac{x}{4} + C.\)</p>
+<p><math>\int \frac{4}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac{x}{4} + C.</math></p>
-<p>\(\displaystyle \int\frac{x}{\sqrt{16-x^2}}\ dx\): Set \(u = 16-x^2\), so \(du = -2xdx\) and \(xdx = -du/2\). We have</p>
+<p><math>\int\frac{x}{\sqrt{16-x^2}}\ dx</math>: Set <math>u = 16-x^2</math>, so <math>du = -2xdx<math> and <math>xdx = -du/2</math>. We have</p>
-<p>\[\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx &amp;= \int\frac{-du/2}{\sqrt{u}}\\ &amp;= -\frac12\int \frac{1}{\sqrt{u}}\ du \\ &amp;= - \sqrt{u} + C\\ &amp;= -\sqrt{16-x^2} + C.\end{align}\]</p>
+<p><math>\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx &amp;= \int\frac{-du/2}{\sqrt{u}}\\ &amp;= -\frac12\int \frac{1}{\sqrt{u}}\ du \\ &amp;= - \sqrt{u} + C\\ &amp;= -\sqrt{16-x^2} + C.\end{align}</math></p>
 <p>Combining these together, we have</p>
-<p>$$ \int \frac{4-x}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.$$</p>
+<p><math> \int \frac{4-x}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.</math></p>
 ==Resources==

@@ Line 6: / Line 6: @@
-==Example Problem==
+===Example 1===
 <p>Evaluate the integral</p>
@@ Line 20: / Line 20: @@
 <p class="mt-indent-3" style="text-align:center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.</math></p>
 ==Resources==

← Older revision		Revision as of 20:21, 28 October 2021
Line 5:		Line 5:
	<math> \int\frac{du}{u\sqrt{u^2 - a^2}} =\frac{1}{a}\arcsec \left(\dfrac{\|u\|}{a}\right) + C </math>		<math> \int\frac{du}{u\sqrt{u^2 - a^2}} =\frac{1}{a}\arcsec \left(\dfrac{\|u\|}{a}\right) + C </math>

		+
		+	==Example Problem==
	<p>Evaluate the integral</p>		<p>Evaluate the integral</p>

@@ Line 17: / Line 17: @@
 <p>Applying the formula with <math> a=2, </math> we obtain</p>
-<p class="mt-indent-3" style="text-align:center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.\]</math></p>
+<p class="mt-indent-3" style="text-align:center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.</math></p>
 ==Resources==

@@ Line 15: / Line 15: @@
 <p style="text-align: center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}.</math></p>
-<p>Applying the formula with <math>\( a=2,\)</math> we obtain</p>
+<p>Applying the formula with <math> a=2, </math> we obtain</p>
-<p class="mt-indent-3" style="text-align:center;"><math>\[ \int;\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.\]</math></p>
+<p class="mt-indent-3" style="text-align:center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.\]</math></p>
 ==Resources==

@@ Line 11: / Line 11: @@
 <p><strong>Solution</strong></p>
-<p>Substitute <math>\( u=3x\)</math>. Then <math>\( du=3\,dx\)</math> and we have</p>
+<p>Substitute <math> u=3x</math>. Then <math> du=3dx</math> and we have</p>
-<p style="text-align: center;"><math>\[ \int;\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}.\nonumber\]</math></p>
+<p style="text-align: center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}.</math></p>
 <p>Applying the formula with <math>\( a=2,\)</math> we obtain</p>