Q86 / 150 · MHT CET 2007 · Mathematics

The value of $\int_{0}^{\pi / 2} \frac{\cos 3 x+1}{2 \cos x-1} d x$ is

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

Solution

Let $\begin{aligned} I=\int_{0}^{\pi / 2} \frac{\cos 3 x+1}{2 \cos x-1} d x \\ &=\int_{0}^{\pi / 2} \frac{\cos 3 x-\cos \frac{3 \pi}{3}}{2\left(\cos x-\cos \frac{\pi}{3}\right)} d x \\=\int_{0}^{\pi / 2} \frac{\left(4 \cos ^{3} x-3 \cos x\right)-\left(4 \cos ^{3} \frac{\pi}{3}-3 \cos \frac{\pi}{3}\right)}{2\left(\cos x-\cos \frac{\pi}{3}\right)} d x \\=& 2 \int_{0}^{\pi / 2}\left(\frac{\cos ^{3} x-\cos ^{3} \frac{\pi}{3}}{\cos x-\cos \frac{\pi}{3}}\right) d x \\=\frac{3}{2} \int_{0}^{\pi / 2}\left(\frac{\cos x-\cos \frac{\pi}{3}}{\cos x-\cos \frac{\pi}{3}}\right) d x \\=& \int_{0}^{\pi / 2}\left(1+\cos 2 x+\frac{1}{2}+\cos x\right) d x-\frac{3 \pi}{4} \\=& \frac{3 \pi}{4}+1-\frac{3 \pi}{4}=1 \end{aligned}$ $$

Hence, the correct answer is option: 'B'.

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