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| | <span class="exam">(b) <math>\int \sin^3(x)\cos^2(x)~dx</math> | | <span class="exam">(b) <math>\int \sin^3(x)\cos^2(x)~dx</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Final 3, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | |'''1.''' Integration by parts tells us that | |
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| − | | <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
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| − | |'''2.''' Since <math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math> we have
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| − | | <math>\sin^2x=1-\cos^2x.</math>
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| − | '''Solution:''' | + | [[009B Sample Final 3, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |To calculate this integral, we use integration by parts.
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| − | |Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=\cos xdx.</math>
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| − | |Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: -1px">v=\sin x.</math>
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| − | |Therefore, we have
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| − | | <math>\int x\cos(x)~dx=x\sin x -\int \sin x~dx.</math>
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Then, we integrate to get
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| − | | <math> \int x\cos(x)~dx=x\sin x +\cos x+C.</math>
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| − | |}
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |First, we use the identity <math style="vertical-align: -1px">\sin^2 x=1-\cos^2 x</math> to get
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \sin^3(x)\cos^2(x)~dx} & = & \displaystyle{\int \sin^2 x (\cos^2 x) \sin x~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int (1-\cos^2 x)(\cos^2 x) \sin x~dx.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, we use <math style="vertical-align: 0px">u</math>-substitution.
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| − | |Let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx</math> and <math style="vertical-align: -5px">-du=\sin(x)dx.</math>
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| − | |Therefore, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \sin^3(x)\cos^2(x)~dx} & = & \displaystyle{\int (-1)(1-u^2)u^2~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\int u^4-u^2~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{u^5}{5}-\frac{u^3}{3}+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\cos^5 x}{5}-\frac{\cos^3 x}{3}+C.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(a)''' <math>x\sin x +\cos x+C</math>
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| − | | '''(b)''' <math>\frac{\cos^5 x}{5}-\frac{\cos^3 x}{3}+C</math>
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| − | |}
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| | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
