Question
Evaluate the following integrals:$\int\frac{\cos\text{x}}{\sqrt{4-\sin^2\text{x}}}\text{ dx}$

Answer

$\int\frac{\cos\text{x}\text{ dx}}{\sqrt{4-\sin^2\text{x}}}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{\cos\text{x}\text{ dx}}{\sqrt{4-\sin^2\text{x}}}$
$=\int\frac{\text{dt}}{\sqrt{4-\text{t}^2}}$
$=\int\frac{\text{dt}}{\sqrt{2^2-\text{t}^2}}$
$=\sin^{-1}\Big(\frac{\text{t}}{2}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\sin\text{x}}{2}\Big)+\text{C}$

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