Mathematics - Indefinite Integration Question with Solution | TestHub

Let $\mathrm{I}=\int \cos (\log \mathrm{x})\cdot 1\mathrm{d}\mathrm{x}\dots ..\text{ (i) }$ Use integral by parts, $\begin{array}{rl}& I=\cos (\log x)\cdot x-\int [-\sin (\log x)]\cdot \frac{1}{x}\cdot xdx\\ & =x\cdot \cos (\log x)+\int \sin (\log x)\cdot 1dx\\ & =x\cdot \cos (\log x)+\left[\sin (\log x)\cdot x-\int \cos (\log x)\cdot \frac{1}{x}\cdot xdx\right]+C\\ & =x\cdot (\log x)+\left[x\sin (\log x)-\int \cos (\log x)dx\right]+C\\ & =x\{\sin (\log x)+\cos (\log x)\}-I+C[from\text{ Eq. (i) }]\\ & \Rightarrow I=\frac{x}{2}\{\sin (\log x)+\cos (\log x)\}+C\end{array}$