Question

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

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

Answer 1

Answer: (B)

Integrand $=\frac{cos3x-cos\left(3\frac{\pi }{3}\right)}{2\left(cosx-cos\frac{\pi }{3}\right)}$
$=\frac{\left(4co{s}^{3}x-3cosx\right)-\left(4co{s}^3\alpha -3cos\alpha \right)}{2\left(cosx-cos\alpha \right)}$
(where $\alpha =\frac{\pi }{3})$
$=\frac{1}{2}\left[4\left(co{s}^{2}x+cosxcos\alpha +co{s}^2\alpha \right)-3\right]$
$\therefore$ given integral
$=\frac{4}{2}\underset{0}{\overset{\pi 2}{\int }}\left(co{s}^{2}x+cosxcos\alpha \right)dx-\frac{3}{2}\underset{0}{\overset{\pi /2}{\int }}dx$
$=2\left[\frac{1}{2}\cdot \frac{\pi }{2}+\frac{1}{2}\cdot 1+\frac{1}{4}\cdot \frac{\pi }{2}\right]-\frac{3}{2}\cdot \frac{\pi }{2}$
$=\frac{\pi }{2}+1+\frac{\pi }{4}-\frac{3\pi }{4}=1$