Difference between revisions of "009B Sample Midterm 2, Problem 5"

Revision as of 14:26, 28 March 2016

Evaluate the integral:

\int \tan ^{4}x~dx

Foundations:
Recall:
1. Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sec ^{2}x=tan^{2}x+1}
2. $\int \sec ^{2}x~dx=\tan x+C$
How would you integrate Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \sec ^{2}(x)\tan(x)~dx} ?
You could use $u$ -substitution. Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\tan x} . Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=\sec ^{2}(x)dx} .
Thus, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \sec ^{2}(x)\tan(x)~dx=\int u~du={\frac {u^{2}}{2}}+C={\frac {\tan ^{2}x}{2}}+C} .

Solution:

Step 1:
First, we write Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \tan ^{4}(x)~dx=\int \tan ^{2}(x)\tan ^{2}(x)~dx} .
Using the trig identity $\sec ^{2}(x)=\tan ^{2}(x)+1$ , we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \tan ^{2}(x)=\sec ^{2}(x)-1} .
Plugging in the last identity into one of the Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \tan ^{2}(x)} , we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \tan ^{4}(x)~dx=\int \tan ^{2}(x)(\sec ^{2}(x)-1)~dx=\int \tan ^{2}(x)\sec ^{2}(x)~dx-\int \tan ^{2}(x)~dx=\int \tan ^{2}(x)\sec ^{2}(x)~dx-\int (\sec ^{2}x-1)~dx} ,
using the identity again on the last equality.

Step 2:
So, we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \tan ^{4}(x)~dx=\int \tan ^{2}(x)\sec ^{2}(x)~dx-\int (\sec ^{2}x-1)~dx} .
For the first integral, we need to use $u$ -substitution. Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\tan(x)} . Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=\sec ^{2}(x)dx} .
So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \tan ^{4}(x)~dx=\int u^{2}~du-\int (\sec ^{2}(x)-1)~dx} .

Step 3:
We integrate to get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \tan ^{4}(x)~dx={\frac {u^{3}}{3}}-(\tan(x)-x)+C={\frac {\tan ^{3}(x)}{3}}-\tan(x)+x+C} .

Final Answer:
${\frac {\tan ^{3}(x)}{3}}-\tan(x)+x+C$

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|Review <math>u</math>-substitution and
+|Recall:
 |-
-|trig identities
+|'''1.''' <math>\sec^2x=tan^2x+1</math>
+|-
+|'''2.''' <math>\int \sec^2 x~dx=\tan x+C</math>
+|-
+|How would you integrate <math>\int \sec^2(x)\tan(x)~dx</math>?
+|-
+|
+::You could use <math>u</math>-substitution. Let <math>u=\tan x</math>. Then, <math>du=\sec^2(x)dx</math>.
+|-
+|
+::Thus, <math>\int \sec^2(x)\tan(x)~dx=\int u~du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C</math>.
 |}