# What is f(x) = int xsin2x-6cotx dx if f(pi/4)=1 ?

Feb 9, 2018

#### Answer:

$f \left(x\right) = - \frac{x}{2} \cdot \cos 2 x + \frac{1}{4} \cdot \sin 2 x - 6 \ln \left(\sin x\right) + 1 + \frac{\pi \sqrt{2}}{16} - 3 L n 2$

#### Explanation:

Due to $\int x \sin 2 x \cdot \mathrm{dx} = x \cdot \left(- \frac{1}{2} \cdot \cos 2 x\right) - \int \left(- \frac{1}{2}\right) \cdot \cos 2 x \cdot \mathrm{dx}$

=$- \frac{x}{2} \cdot \cos 2 x + \frac{1}{2} \int \cos 2 x \cdot \mathrm{dx}$

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

$f \left(x\right) = \int \left(x \sin 2 x - 6 \cot x\right) \cdot \mathrm{dx} = - \frac{x}{2} \cdot \cos 2 x + \frac{1}{4} \cdot \sin 2 x - 6 \ln \left(\sin x\right) + C$

After imposing $f \left(\frac{\pi}{4}\right) = 1$ condition,

$- \frac{\pi}{8} \cdot \sin \left(\frac{\pi}{4}\right) - 6 L n \left(\sin \left(\frac{\pi}{4}\right)\right) + C = 1$

$- \frac{\pi \sqrt{2}}{16} - 6 L n \left(\frac{\sqrt{2}}{2}\right) + C = 1$

$- \frac{\pi \sqrt{2}}{16} + 3 L n 2 + C = 1$

$C = 1 + \frac{\pi \sqrt{2}}{16} - 3 L n 2$

Thus,

$f \left(x\right) = - \frac{x}{2} \cdot \cos 2 x + \frac{1}{4} \cdot \sin 2 x - 6 \ln \left(\sin x\right) + 1 + \frac{\pi \sqrt{2}}{16} - 3 L n 2$