Question

The value of ${\int }_0^{\frac{\pi }{2}}\frac{\cos 3x+1}{2\cos x-1}dx$ is

• A. $2$
• B. $1$
• C. $\frac{1}{2}$
• D. $0$

Answer 1

The correct option is B $1$

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

$={\int }_0^{\pi /2}\frac{(4{\cos }^{3}x-3\cos \frac{\pi }{3})-(4{\cos }^3\frac{\pi }{3}-3\cos \frac{\pi }{3})}{2(\cos x-\cos \frac{\pi }{3})}dx$

$=2{\int }_0^{\pi /2}\frac{{\cos }^{3}x-{\cos }^3\frac{\pi }{3}}{(\cos x-\cos \frac{3\pi }{3})}dx-\frac{3}{2}{\int }_0^{\pi /2}\frac{\cos x-\cos \frac{\pi }{3}}{2(\cos x-\cos \frac{\pi }{3})}dx$

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

$={\int }_0^{\pi /2}(1+\cos x+\frac{1}{2}+\cos x)dx-\frac{3\pi }{4}=\frac{3\pi }{4}+1-\frac{3\pi }{4}=1$