Solve sinθcosθ/2quadr^2lnr^2-r^2/2quaddr

$$\frac { \sin \theta \cos \theta } { 2 } \quad r ^ { 2 } \ln r ^ { 2 } - \frac { r ^ { 2 } } { 2 } \quad d r$$

Factor

$$\frac{r^{2}\left(\sin(2\theta )\ln(r^{2})-2dr\right)}{4}$$

Solution Steps

Factor out \frac{1}{2}.

$$\frac{r^{2}\left(\frac{1}{2}\ln(r^{2})\sin(2\theta )-dr\right)}{2}$$

Consider r^{2}\left(\frac{1}{2}\ln(r^{2})\sin(2\theta )-dr\right). Factor out r^{2}.

$$r^{2}\left(\frac{1}{2}\ln(r^{2})\sin(2\theta )-dr\right)$$

Rewrite the complete factored expression.

$$\frac{r^{2}\left(\frac{1}{2}\ln(r^{2})\sin(2\theta )-dr\right)}{2}$$

Evaluate

$$\frac{r^{2}\left(\sin(2\theta )\ln(r^{2})-2dr\right)}{4}$$