Question

If $\underset{0}{\overset{\pi }{\int }}\sqrt{1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}}dx=4\sqrt{a}-2b+\frac{\pi }{a},$ then which of the following statements is/are true :
(where $a,b$ are prime numbers)

• A. $a+b=5$
• B. $a+b=8$
• C. $ab=6$
• D. $ab=12$

Answer 1

Answer: (A), (C)

$ab=6$

Let $I=\underset{0}{\overset{\pi }{\int }}\sqrt{1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}}dx$
$=\underset{0}{\overset{\pi }{\int }}|2\sin \frac{x}{2}-1|dx\left[\begin{array}{c}\sin \frac{x}{2}=\frac{1}{2}\\ \Rightarrow \frac{x}{2}=\frac{\pi }{6}\to x=\frac{\pi }{3}\end{array}\right]$
$=\underset{0}{\overset{\pi /3}{\int }}(1-2\sin \frac{x}{2})dx+\underset{\pi /3}{\overset{\pi }{\int }}(2\sin \frac{x}{2}-1)dx$
$={[x+4\cos \frac{x}{2}]}_0^{\pi /3}+{[-4\cos \frac{x}{2}-x]}_{\pi /3}^{\pi }$
$=\frac{\pi }{3}+4\frac{\sqrt{3}}{2}-4+(0-\pi +4\frac{\sqrt{3}}{2}+\frac{\pi }{3})$
$=4\sqrt{3}-4-\frac{\pi }{3}$
$\therefore a=3$ and $b=2$