Question

If $\int \frac{\cos \left(\ln x\right)}{x^3}dx=\frac{1}{a{x}^2}\left[b\sin \right(\ln x)-c\cos (\ln x\left)\right]+K,$ where $K$ is the constant of integration, then the value of $a+b+c$ is

Answer 1

Let $I=\int \frac{\cos \left(\ln x\right)}{x^3}dx=\int \frac{\cos \left(\ln x\right)}{x^2}\cdot \frac{dx}{x}$
Put $\ln x=t\Rightarrow \frac{1}{x}dx=dt$
$I=\int e^{-2t}\cos tdt=e^{-2t}\sin t-\int \sin t(-2e^{-2t})dt=e^{-2t}\sin t+2\int e^{-2t}\sin tdt=e^{-2t}\sin t+2I_1\dots \left(1\right)$

Now, we have
$I_1=\int e^{-2t}\sin tdt=e^{-2t}(-\cos t)-\int (-\cos t)(-2e^{-2t})dt=-e^{-2t}\cos t-2\int \cos t{e}^{-2t}dt=-e^{-2t}\cos t-2I$

From equation $\left(1\right),$
$I=e^{-2t}\sin t+2[-e^{-2t}\cos t-2I]\Rightarrow I=\frac{1}{5}{e}^{-2t}(\sin t-2\cos t)+K\Rightarrow I=\frac{1}{5x^2}\left[\sin \right(\ln x)-2\cos (\ln x\left)\right]+K$

On comparing, we get
$a=5,b=1,c=2$
$\therefore a+b+c=8$