Science:Math Exam Resources/Courses/MATH105/April 2011/Question 01 (a)

MATH105 April 2011

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Question 01 (a)
The question in the exam asks you to determine the following indefinite integral: $\int \cos(\ln(x))\,{\text{d}}x$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
A possible solution involves applying once the substitution rule and twice integration by parts.

Checking a solution serves two purposes: helping you if, after having used the hint, 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 computation of the given integral involves several steps. Step 1. In the first step we apply the substitution rule with $u=\ln(x)$. This implies $du/dx=1/x$ and in particular $dx=xdu$. Hence, $\int \cos(\ln(x))\,dx=\int \cos(u)\cdot x\,du$ We have to rewrite $x$ in terms of $u$. This can be done by exponentiating the substitution equation: $e^{u}=x$. We arrive at $\int \cos(u)\cdot x\,du=\int \cos(u)e^{u}\,du$ Step 2. We apply integration by parts with $f(u)=\cos(u)$ and $g'(u)=e^{u}$: $\int \cos(u)e^{u}\,du=\cos(u)e^{u}+\int \sin(u)e^{u}\,du$ Integration by parts onto the integral in the right-hand side with $f(u)=\sin(u)$ and $g'(u)=e^{u}$ yields $\int \sin(u)e^{u}\,du=\sin(u)e^{u}-\int \cos(u)e^{u}\,du$ Step 3. Step 1 and Step 2 together results in the equation ${\begin{aligned}\int \cos(u)e^{u}\,du&=\cos(u)e^{u}+\int \sin(u)e^{u}\,du\\&=\cos(u)e^{u}+\sin(u)e^{u}-\int \cos(u)e^{u}\,du\end{aligned}}$ Step 4. We bring the integral on the right-hand side of the equation over to the left-hand side: $2\int \cos(u)e^{u}\,du=\cos(u)e^{u}+\sin(u)e^{u}=e^{u}(\sin(u)+\cos(u))$ Finally, we divide the equation by 2: $\int \cos(u)e^{u}\,du={\frac {e^{u}}{2}}(\sin(u)+\cos(u))+C$ where $C$ is an arbitrary constant Step 5. The last step is to rewrite our results in terms of $x$. Remember, that our very first step was the substitution $u=\ln(x)$. Therefore ${\begin{aligned}\int \cos(\ln(x))\,dx&=\int \cos(u)e^{u}\,du\\&={\frac {e^{u}}{2}}(\sin(u)+\cos(u))+C\\&={\frac {x}{2}}(\sin(\ln(x))+\cos(\ln(x)))+C\end{aligned}}$ where we used $e^{u}=x$.

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