Question

Evaluate: $\int x\frac{d}{dx}\left({\sin }^{2}x\right)dx$

• A. $\frac{1}{4}{\sin }^{2}x-\frac{1}{2}{\cos }^{2}x+c$
• B. $\frac{1}{2}\sin 2x-\frac{1}{4}\cos 2x+c$
• C. $\frac{1}{4}\sin 2x-\frac{1}{2}\cos 2x+c$
• D. $\frac{1}{2}{\sin }^{2}x-\frac{1}{4}{\cos }^{2}x+c$

Answer 1

The correct option is C $\frac{1}{4}\sin 2x-\frac{1}{2}\cos 2x+c$

$u\frac{dv}{dx}=\frac{duv}{dx}-\frac{vdu}{dx}u=xv={\sin }^{2}xx\frac{d}{dx}\left({\sin }^{2}x\right)=\frac{d}{dx}\left(x{\sin }^{2}x\right)-{\sin }^{2}x\frac{dx}{dx}I=\int x\frac{d}{dx}\left({\sin }^{2}x\right)dx=\int [\frac{d}{dx}\left(x{\sin }^{2}x\right)-{\sin }^{2}x\frac{dx}{dx}]dxI=\int \frac{d}{dx}\left(x{\sin }^{2}x\right)dx-\int {\sin }^{2}xdxI=\left(x{\sin }^{2}x\right)-\int \frac{1}{2}(1-\cos 2x)dx+KI=x\left[\frac{1}{2}(1-\cos 2x)\right]-\frac{1}{2}[x-\frac{\sin 2x}{2}]+CI=\frac{x}{2}-\frac{x\cos 2x}{2}-\frac{x}{2}+\frac{\sin 2x}{4}+C$

Hence the final and correct answer is;

$I=\frac{1}{4}\sin 2x-\frac{x}{2}\cos 2x+C$