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 $= \frac{cos\,3x-cos\left(3\frac{\pi}{3}\right)}{2\left(cos\,x-cos \frac{\pi}{3}\right)}$
$=\frac{\left(4\,cos^{3}x-3 cos\,x\right)-\left(4 cos^{3} \alpha-3 cos\,\alpha\right)}{2\left(cos\,x-cos\,\alpha\right)}$
(where $\alpha=\frac{\pi}{3})$
$=\frac{1}{2}\left[4 \left(cos^{2}\,x+cos\,x cos\,\alpha+cos^{2}\,\alpha\right)-3\right]$
$\therefore $ given integral
$=\frac{4}{2} \int\limits_{0}^{\pi 2}\left(cos^{2}\,x+cos\,x\,cos\,\alpha\right)dx-\frac{3}{2} \int\limits_{0}^{\pi /2}dx$
$=2\left[\frac{1}{2}\cdot\frac{\pi}{2}+\frac{1}{2}\cdot1+\frac{1}{4}\cdot\frac{\pi}{2}\right]-\frac{3}{2}\cdot\frac{\pi}{2}$
$=\frac{\pi}{2}+1+\frac{\pi}{4}-\frac{3\pi}{4}=1$