Use integration to find the exact value of $$\int_0^{\frac{\pi}{2}} x \sin 2x \, dx$$ - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 6

The integral of (\cos 2x) is: $\int \cos 2x \, dx = \frac{1}{2} \sin 2x + C$

Thus, we have: $\int x \sin 2x \, dx = -\frac{x}{2} \cos 2x + \frac{1}{4} \sin 2x$

Now, we will evaluate the integral from 0 to (\frac{\pi}{2}):

$\left[ -\frac{x}{2} \cos 2x + \frac{1}{4} \sin 2x \right]_0^{\frac{\pi}{2}}$

Evaluating at the bounds gives: At (x = \frac{\pi}{2}):\n$-\frac{\frac{\pi}{2}}{2} \cos \pi + \frac{1}{4} \sin \pi = \frac{\pi}{4}$\n At (x = 0):\n$0$

Thus, the value of the definite integral is: $\frac{\pi}{4} - 0 = \frac{\pi}{4}$