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| − | <span class="exam">Evaluate the integral: | + | <span class="exam"> Compute the following integrals: |
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| − | ::<math>\int \sin (\ln x)~dx.</math>
| + | <span class="exam">(a) <math>\int x^2\sin (x^3) ~dx</math> |
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| | + | <span class="exam">(b) <math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx</math> |
| | + | <hr> |
| | + | [[009B Sample Midterm 3, Problem 4 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |Integration by parts tells us that <math>\int u~dv=uv-\int vdu.</math>
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| − | |-
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| − | |How could we break up <math>\sin (\ln x)~dx</math> into <math>u</math> and <math>dv?</math>
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| − | |-
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| − | |
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| − | ::Notice that <math>\sin (\ln x)</math> is one term. So, we need to let <math>u=\sin (\ln x)</math> and <math>dv=dx.</math>
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| − | |}
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| − | '''Solution:'''
| + | [[009B Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We proceed using integration by parts. Let <math>u=\sin(\ln x)</math> and <math>dv=dx.</math> Then, <math>du=\cos(\ln x)\frac{1}{x}dx</math> and <math>v=x.</math>
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| − | |Therefore, we get
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| − | |-
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| − | |
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| − | ::<math>\int \sin (\ln x)~dx=x\sin(\ln x)-\int \cos(\ln x)~dx.</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 need to use integration by parts again. Let <math style="vertical-align: -6px">u=\cos(\ln x)</math> and <math style="vertical-align: -1px">dv=dx.</math> Then, <math style="vertical-align: -14px">du=-\sin(\ln x)\frac{1}{x}dx</math> and <math style="vertical-align: -1px">v=x.</math>
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| − | |-
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| − | |Therfore, we get
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int \sin (\ln x)~dx} & = & \displaystyle{x\sin(\ln x)-\bigg(x\cos(\ln x)+\int \sin(\ln x)~dx\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{x\sin(\ln x)-x\cos(\ln x)-\int \sin(\ln 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 3:
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| − | |-
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| − | |Notice that the integral on the right of the last equation is the same integral that we had at the beginning.
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| − | |So, if we add the integral on the right to the other side of the equation, we get
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| − | |-
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| − | |
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| − | ::<math>2\int \sin(\ln x)~dx=x\sin(\ln x)-x\cos(\ln x)</math>.
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| − | |-
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| − | |Now, we divide both sides by 2 to get
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| − | |-
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| − | ::<math>\int \sin(\ln x)~dx=\frac{x\sin(\ln x)}{2}-\frac{x\cos(\ln x)}{2}.</math>
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| − | |-
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| − | |Thus, the final answer is
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| − | |-
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| − | |
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| − | ::<math>\int \sin(\ln x)~dx=\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C.</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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| − | |<math>\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C</math>
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| − | |}
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| | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
