Find: ^2x ^2x+4 dx. - Vidyadip

Question

Find: $\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{dx}.$

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Answer

$\text{I}=\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
So, $\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+4}}$
Or, $\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+2^2}}$
Since, we know
$\int\frac{\text{dx}}{\sqrt{\text{x}^2+\text{a}^2}}=\text{ln}\ \Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}$
$\text{I}=\text{ln}\ \Big|\text{t}+\sqrt{\text{t}^2+4}\Big|+\text{C}$
i.e., $\text{I}=\text{ln}|\tan\text{x}+\sqrt{{\tan}^2\text{x}+4}|+\text{C}$

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