Let $\text{u}=\frac{\text{x}}{\sin\text{x}}$ and $\text{v}=\sin\text{x}$
$\therefore\ \frac{\text{du}}{\text{dx}}=\frac{\sin\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{x}-\text{x}\cdot\frac{\text{d}}{\text{dx}}\sin\text{x}}{(\sin\text{x})^2}$
$=\frac{\sin\text{x}-\text{x}\cos\text{x}}{\sin^2\text{x}}$
and $\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}$
$\therefore\ \frac{\text{du}}{\text{dv}}=\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{\sin\text{x}-\frac{\text{x}\cos\text{x}}{\sin^2\text{x}}}{\cos\text{x}}$
$=\frac{\sin\text{x}-\text{x}\cos\text{x}}{\sin^2\text{x}\cos\text{x}}$
$=\frac{\frac{\sin\text{x}-\text{x}\cos\text{x}}{\cos\text{x}}}{\frac{\sin^2\text{x}\cos\text{x}}{\cos\text{x}}}$
$=\frac{\tan\text{x}-\text{x}}{\sin^2\text{x}}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.