Science:Math Exam Resources/Courses/MATH105/April 2011/Question 01 (a)/Solution 1

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 $e^{u}=x$.