Question

$\int \frac{\sin \left(\log x\right)}{x^3}dx=$

• A. $\frac{-1}{5x^2}\left(2\sin \right(\log x)+\cos (\log x\left)\right)+c$
• B. $\frac{1}{5x^2}\left(2\sin \right(\log x)+\cos (\log x\left)\right)+c$
• C. $\frac{1}{5x^2}\left(2\sin \right(\log x)-\cos (\log x\left)\right)+c$
• D. $\frac{1}{x^2}\left(2\sin \right(\log x)-\cos (\log x\left)\right)+c$

Answer 1

The correct option is A $\frac{-1}{5x^2}\left(2\sin \right(\log x)+\cos (\log x\left)\right)+c$

$I=\int \frac{\sin \left(\log x\right)}{x^3}dx=\sin \left(\log x\right)\int \frac{1}{x^3}dx-\int \cos \left(\log x\right)\frac{1}{x}(\int \frac{1}{x^3}dx)dx=-\sin \left(\log x\right)\frac{1}{2x^2}+\frac{1}{2}\int \frac{\cos \left(\log x\right)}{x^3}dx......................................1$
.

$I_1=\int \frac{\cos \left(\log x\right)}{x^3}dx=\cos \left(\log x\right)\int \frac{1}{x^3}dx+\int \sin \left(\log x\right)\frac{1}{x}(\int \frac{1}{x^3}dx)dx=-\cos \left(\log x\right)\frac{1}{2x^2}-\frac{1}{2}\int \frac{\sin \left(\log x\right)}{x^3}dx=-\cos \left(\log x\right)\frac{1}{2x^2}-\frac{I}{2}$

.By putting in $1$

$\therefore I=-\sin \left(\log x\right)\frac{1}{2x^2}+\frac{1}{2}(-\cos \left(\log x\right)\frac{1}{2x^2}-\frac{I}{2})+C\Rightarrow \frac{5I}{4}=-\frac{1}{2x^2}(\sin \left(\log x\right)+\frac{1}{2}\cos \left(\log x\right))+C\Rightarrow I=-\frac{1}{5x^2}(2\sin \left(\log x\right)+\cos \left(\log x\right))+c$