Question

$$\int\frac{1-\cos\ x}{(x-\sin^{2}x)^{2}}dx$$

Answer

$$(e^2*IM*n*t^2*g*r*a*d*x*(1-cos(x)))/(x-sin(x)^2)^2$$

Solution


Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{\imath ntegrate(1-\cos{x})dx}{{(x-\sin^{2}x)}^{2}}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{\imath n{t}^{2}{e}^{2}gra(1-\cos{x})dx}{{(x-\sin^{2}x)}^{2}}\]
Regroup terms.
\[\frac{{e}^{2}\imath n{t}^{2}gradx(1-\cos{x})}{{(x-\sin^{2}x)}^{2}}\]