Question
Integrate the function in Exercise:
$\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}$

Answer

$\text{Ler I}=\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{ dx}\ \ \ \ ...\text{(i)}$
$\text{Putting }\tan\text{x}=\text{t}\ \ \Rightarrow\ \ \sec^2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \Rightarrow\ \ \sec^2\text{x}\text{ dx}=\text{dt}$
$\therefore\ \ \text{From eq. (i), }\text{ I}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+4}}=\int\frac{1}{\sqrt{\text{t}^2+(2)^2}}\text{ dt}$
$=\log\bigg|\text{t}+\sqrt{\text{t}^2+(2)^2}\bigg|+\text{c}$
$=\log\Big|\tan\text{x}+\sqrt{\tan^2\text{x}+4}\Big|+\text{c}$

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