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2017-11-24T01:19:00Z Replaced content with "Evaluate the indefinite and definite integrals. (a) <math>\int x\ln x ~dx</math> (b) <math>\int_0..." Show changes

MathAdmin at 19:21, 9 April 2017

2017-04-09T19:21:08Z Show changes

MathAdmin: Created page with "Evaluate the indefinite and definite integrals. ::a) $\int \tan^3x ~dx$ ::b) $\int_0^\pi \sin^2x~dx<..."$

2016-04-19T00:59:03Z

Created page with "Evaluate the indefinite and definite integrals. ::a) <math>\int \tan^3x ~dx</math> ::b) <math>\int_0^\pi \sin^2x~dx<..."

New page

Evaluate the indefinite and definite integrals.

::a) <math>\int \tan^3x ~dx</math>
::b) <math>\int_0^\pi \sin^2x~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall the trig identities:
|-
|'''1.''' <math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
|-
|'''2.''' <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}</math>
|-
|How would you integrate <math style="vertical-align: -1px">\tan x~dx?</math>
|-
|
::You could use <math style="vertical-align: 0px">u</math>-substitution. First, write <math style="vertical-align: -14px">\int \tan x~dx=\int \frac{\sin x}{\cos x}~dx.</math>
|-
|
::Now, let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math> Thus,
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int \tan x~dx} & = & \displaystyle{\int \frac{-1}{u}~du}\\
&&\\
& = & \displaystyle{-\ln(u)+C}\\
&&\\
& = & \displaystyle{-\ln|\cos x|+C.}\\
\end{array}</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by writing
|-
|
::<math style="vertical-align: -14px">\int \tan^3x~dx=\int \tan^2x\tan x ~dx.</math>
|-
|Since <math style="vertical-align: -5px">\tan^2x=\sec^2x-1,</math> we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int (\sec^2x-1)\tan x ~dx}\\
&&\\
& = & \displaystyle{\int \sec^2x\tan x~dx-\int \tan x~dx.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we need to use <math>u</math>-substitution for the first integral.
|-
|
::Let <math style="vertical-align: -5px">u=\tan(x).</math> Then, <math style="vertical-align: -1px">du=\sec^2x~dx.</math> So, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int u~du-\int \tan x~dx}\\
&&\\
& = & \displaystyle{\frac{u^2}{2}-\int \tan x~dx}\\
&&\\
& = & \displaystyle{\frac{\tan^2x}{2}-\int \tan x~dx.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|For the remaining integral, we also need to use <math>u</math>-substitution.
|-
|First, we write
|-
|
::<math style="vertical-align: -14px">\int \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx.</math>
|-
|Now, we let <math style="vertical-align: 0px">u=\cos x.</math> Then, <math style="vertical-align: -1px">du=-\sin x~dx.</math> So, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\frac{\tan^2x}{2}+\int \frac{1}{u}~dx}\\
&&\\
& = & \displaystyle{\frac{\tan^2x}{2}+\ln |u|+C}\\
&&\\
& = & \displaystyle{\frac{\tan^2x}{2}+\ln |\cos x|+C.}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|One of the double angle formulas is <math style="vertical-align: -5px">\cos(2x)=1-2\sin^2(x).</math>
|-
|Solving for <math style="vertical-align: -5px">\sin^2(x),</math> we get <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}.</math>
|-
|Plugging this identity into our integral, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\int_0^\pi \frac{1-\cos(2x)}{2}~dx}\\
&&\\
& = & \displaystyle{\int_0^\pi \frac{1}{2}~dx-\int_0^\pi \frac{\cos(2x)}{2}~dx.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|If we integrate the first integral, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\left.\frac{x}{2}\right|_{0}^\pi-\int_0^\pi \frac{\cos(2x)}{2}~dx}\\
&&\\
& = & \displaystyle{\frac{\pi}{2}-\int_0^\pi \frac{\cos(2x)}{2}~dx.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|For the remaining integral, we need to use <math style="vertical-align: 0px">u</math>-substitution.
|-
|Let <math style="vertical-align: -1px">u=2x.</math> Then, <math style="vertical-align: -1px">du=2~dx</math> and <math style="vertical-align: -18px">\frac{du}{2}=dx.</math>
|-
|Also, since this is a definite integral and we are using <math style="vertical-align: 0px">u</math>-substitution, we need to change the bounds of integration.
|-
|We have <math style="vertical-align: -5px">u_1=2(0)=0</math> and <math style="vertical-align: -5px">u_2=2(\pi)=2\pi.</math>
|-
|So, the integral becomes
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du}\\
&&\\
& = & \displaystyle{\frac{\pi}{2}-\left.\frac{\sin(u)}{4}\right|_0^{2\pi}}\\
&&\\
& = & \displaystyle{\frac{\pi}{2}-\bigg(\frac{\sin(2\pi)}{4}-\frac{\sin(0)}{4}\bigg)}\\
&&\\
& = & \displaystyle{\frac{\pi}{2}.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)''' <math>\frac{\tan^2x}{2}+\ln |\cos x|+C</math>
|-
|   '''(b)''' <math>\frac{\pi}{2}</math>
|}

[[009B_Sample_Midterm_3|'''Return to Sample Exam''']]

009B Sample Midterm 3, Problem 5 - Revision history

MathAdmin: Replaced content with "Evaluate the indefinite and definite integrals. (a) \int x\ln x ~dx (b) \int_0..."

MathAdmin at 19:21, 9 April 2017

MathAdmin: Created page with "Evaluate the indefinite and definite integrals. ::a) \int \tan^3x ~dx ::b) \int_0^\pi \sin^2x~dx<..."

MathAdmin: Replaced content with "Evaluate the indefinite and definite integrals. (a) $\int x\ln x ~dx$ (b) $\int_0..."$

MathAdmin: Created page with "Evaluate the indefinite and definite integrals. ::a) $\int \tan^3x ~dx$ ::b) $\int_0^\pi \sin^2x~dx<..."$