Let $\int_\pi ^a {\cos 2x{\text{d}}x}  = \frac{1}{2}{\text{, where }}\pi  < a < 2\pi$. Find the value of $a$.

correct integration (ignore absence of limits and “$+C$”)     (A1)

eg     $\frac{{\sin (2x)}}{2},{\text{ }}\int_\pi ^a {\cos 2x = \left[ {\frac{1}{2}\sin (2x)} \right]_\pi ^a}$

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg     $\frac{1}{2}\sin (2a) - \frac{1}{2}\sin (2\pi ),{\text{ }}\sin (2\pi ) - \sin (2a)$

$\sin (2\pi ) = 0$     (A1)

setting their result from an integrated function equal to $\frac{1}{2}$     M1

eg     $\frac{1}{2}\sin 2a = \frac{1}{2},{\text{ }}\sin (2a) = 1$

recognizing ${\sin ^{ - 1}}1 = \frac{\pi }{2}$     (A1)

eg     $2a = \frac{\pi }{2},{\text{ }}a = \frac{\pi }{4}$

correct value     (A1)

eg     $\frac{\pi }{2} + 2\pi ,{\text{ }}2a = \frac{{5\pi }}{2},{\text{ }}a = \frac{\pi }{4} + \pi$

$a = \frac{{5\pi }}{4}$     A1     N3

[7 marks]