Evaluate the following integrals: _ 0 ^ 1 5+3 dx

Question

Evaluate the following integrals:
$\int\limits_{0}^{{\pi}}\frac{1}{5+3\cos\text{x}}\text{ dx}$

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Answer

We know that,
$\cos\text{x}=\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}$
$\Rightarrow\ \frac{1}{5+3\cos\text{x}}=\frac{1}{5+3\Bigg(\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)}\\=\frac{1+\tan^{2}\frac{\text{x}}{2}}{5\big(1+\tan^{2}\frac{\text{x}}{2}\big)+3\big(1-\tan^{2}\frac{\text{x}}{2}\big)}=\frac{\sec^2\frac{\text{x}}{2}\text{ dx}}{8+2\tan^{2}\frac{\text{x}}{2}}$
$\therefore\ \int_{0}^\limits{{\pi}}\frac{1}{5+3\cos\text{x}}\text{ dx}=\frac{1}{2}\int_{0}^\limits{{\pi}}\frac{\sec^2\frac{\text{x}}{2}}{2^2+2\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
Differentiating w.r.t. x, we get
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\pi\Rightarrow\text{t}=\infty$
$\therefore\ \frac{1}{2}\int^\limits\pi_0\bigg(\frac{\sec^2\frac{\text{x}}{2}\text{ dx}}{2^2+\tan^{2}\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int^\limits\infty_0\frac{\text{dt}}{2^2+\text{t}^2}$
$=\Big[\frac{1}{2}\tan^{-1}\Big(\frac{\text{t}}{2}\Big)\Big]^{\infty}_0$
$=\frac{1}{2}\Big[\tan^{-1}(\infty)-\tan^{-1}(0)\Big]$
$=\frac{1}{2}\Big[\tan^{-1}\Big(\tan\frac{\pi}{2}\Big)-\tan^{-1}\big(\tan0\big)\Big]$
$=\frac{1}{2}\Big[\frac{\pi}{2}-0\Big]$
$=\frac{\pi}{4}$

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