Science:Math Exam Resources/Courses/MATH103/April 2013/Question 02 (d)

MATH103 April 2013

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Question 02 (d)
Integration: Short Answer Problems - Evaluate the following integrals: state if a definite integral does not exist; use limits for improper integrals. (full marks for correct answer; work must be shown for partial marks) $\displaystyle I_{d}=\int \sin(\ln x)\,dx$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Start with the substitution u = ln x.

Hint 2
Next, do integration by parts twice to solve for the integral you wound up with after the substitution.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The integral will become less complicated if we make a substitution $u=\ln x$. First substitution If u = ln x then $du=(1/x)\ dx$ so $dx=x\ du=e^{u}\ du$. For the last equality we solved $u=\ln x$ for $x$. Therefore, $I=\int \sin(\ln x)\ dx=\int e^{u}\sin u\ du$ First integration by parts To find this integral, we will need to integrate by parts twice. Let $f=e^{u}$ with $df=e^{u}\ du$ and $dg=\sin u\ du$ with $g=-\cos u$. Then ${\begin{aligned}I&=fg-\int gdf=-e^{u}\cos u-\int e^{u}(-\cos u)\ du\\&=-e^{u}\cos u+\int e^{u}\cos u\ du\end{aligned}}$ Second integration by parts We repeat the process on the second integral, this time with $f=e^{u}$ and $df=e^{u}\ du$, and $dg=\cos u\ du$ with $g=\sin u$. Now we find: $I=-e^{u}\cos u+(e^{u}\sin u-\int e^{u}\sin u\ du)$ Inspect the result We recognize the integral above as I! We can now solve for I: $\displaystyle I=-e^{u}\cos u+e^{u}\sin u-I$ so $\displaystyle 2I=e^{u}\sin u-e^{u}\cos u$ Then $\displaystyle I={\frac {1}{2}}e^{u}(\sin u-\cos u)+C$ We added the arbitrary constant C because we are computing an indefinite integral. Bring x back Finally, the integral we desire can be found by replacing $u$ by $\ln x$ so that $e^{u}=x$: $\displaystyle I={\frac {1}{2}}x(\sin(\ln x)-\cos(\ln x))+C$

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