Q93 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics
The value of the definite integral $\int_0^{\pi}\dfrac{1}{5 + 4\cos x}\, dx$ is equal to ...
- A$\pi$
- B$\dfrac{\pi}{2}$
- Correct answerC$\dfrac{\pi}{3}$
- D$\dfrac{\pi}{4}$
Solution
Let $I = \int_0^{\pi}\dfrac{1}{5 + 4\cos x}\, dx$
Using the half-angle substitution $\cos x = \dfrac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$ and $dx = \dfrac{2\, dt}{1 + t^2}$ where $t = \tan(x/2)$
The limits change from $x = 0 \rightarrow \pi$ to $t = 0 \rightarrow \infty$
$I = \int_0^{\infty} \dfrac{1}{5 + 4\left(\dfrac{1 - t^2}{1 + t^2}\right)} \dfrac{2\, dt}{1 + t^2}$
$I = \int_0^{\infty} \dfrac{2\, dt}{5(1 + t^2) + 4(1 - t^2)}$
$I = \int_0^{\infty} \dfrac{2\, dt}{9 + t^2}$
$I = 2 \left[ \dfrac{1}{3} \tan^{-1}\left(\dfrac{t}{3}\right) \right]_0^{\infty}$
$I = \dfrac{2}{3} \left( \dfrac{\pi}{2} - 0 \right) = \dfrac{\pi}{3}$
Answer: $\dfrac{\pi}{3}$
Using the half-angle substitution $\cos x = \dfrac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$ and $dx = \dfrac{2\, dt}{1 + t^2}$ where $t = \tan(x/2)$
The limits change from $x = 0 \rightarrow \pi$ to $t = 0 \rightarrow \infty$
$I = \int_0^{\infty} \dfrac{1}{5 + 4\left(\dfrac{1 - t^2}{1 + t^2}\right)} \dfrac{2\, dt}{1 + t^2}$
$I = \int_0^{\infty} \dfrac{2\, dt}{5(1 + t^2) + 4(1 - t^2)}$
$I = \int_0^{\infty} \dfrac{2\, dt}{9 + t^2}$
$I = 2 \left[ \dfrac{1}{3} \tan^{-1}\left(\dfrac{t}{3}\right) \right]_0^{\infty}$
$I = \dfrac{2}{3} \left( \dfrac{\pi}{2} - 0 \right) = \dfrac{\pi}{3}$
Answer: $\dfrac{\pi}{3}$
Hence, the correct answer is option: 'C'.