Question

$\underset{0}{\overset{\pi }{\int }}|1+2\cos x|$ dx is equal to

• A. $\frac{2\pi }{3}$
• B. $\pi$
• C. 2
• D. $\frac{\pi }{3}+2\sqrt{3}$

Answer 1

Answer: (D)

$\frac{\pi }{3}+2\sqrt{3}$

${\int }_0^{\pi }|1+\cos x|.dx$

$={\int }_0^{2\pi /3}|1+\cos x|.dx+{\int }_{2\pi /3}^{\pi }|1+\cos x|.dx$

$={\int }_0^{2\pi /3}(1+\cos x).dx+{\int }_{2\pi /3}^{\pi }-(1+\cos x).dx$

$=[x+2\sin x{]}_0^{2\pi /3}-[x+2\sin x{]}_{\frac{2\pi }{3}}^{\pi }$

$=\frac{2\pi }{3}+2.\sin \frac{2\pi }{3}-\left[\right(\pi +0)-(\frac{2\pi }{3}+2\sin \frac{2\pi }{3})]$

$=(\frac{4\pi }{3}-\pi )+4.\sin (\pi -\frac{\pi }{3})$

$=\frac{\pi }{3}+4\times \frac{\sqrt{3}}{2}=\frac{\pi }{3}+2\sqrt{3}$