Question

$\int_{0}^{\frac{\pi}{2}} \frac{dx}{5+4\cos x} =$

• A. $2 \tan^{-1} (\frac{1}{3})$
• B. $\tan^{-1} (\frac{1}{3})$
• C. $\frac{2}{3} \tan^{-1} (\frac{1}{3})$
• D. $\frac{1}{3} \tan^{-1} (\frac{1}{3})$

Answer 1

Answer: (C)

$\frac{2}{3} \tan^{-1}\left(\frac{1}{3}\right)$

Answer 2

We evaluate

$$I=\int_{0}^{\frac{\pi}{2}} \frac{dx}{5+4\cos x}.$$

Step 1: Use the Weierstrass substitution:

$$t=\tan\frac{x}{2}\quad \Rightarrow \quad dx=\frac{2\,dt}{1+t^2},\quad \cos x=\frac{1-t^2}{1+t^2}.$$

When $x=0$, $t=0$; when $x=\frac{\pi}{2}$, $t=\tan \frac{\pi}{4}=1$.

Step 2: Substitute in the integral:

$$5+4\cos x = 5+4\left(\frac{1-t^2}{1+t^2}\right)=\frac{5(1+t^2)+4(1-t^2)}{1+t^2}=\frac{9+t^2}{1+t^2}.$$

Thus,

$$I= \int_{0}^{1}\frac{2\,dt}{1+t^2}\cdot\frac{1+t^2}{9+t^2} =2\int_{0}^{1}\frac{dt}{9+t^2}.$$

Step 3: Integrate:

$$\int \frac{dt}{9+t^2} = \frac{1}{3}\tan^{-1}\left(\frac{t}{3}\right).$$

So,

$$I=2\left[\frac{1}{3}\tan^{-1}\left(\frac{t}{3}\right)\right]_{0}^{1}=\frac{2}{3}\tan^{-1}\left(\frac{1}{3}\right).$$