Question

Which expression is equal to \[\frac{1}{1+\sqrt{\tan x}}\] in the evaluation of \[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{\tan x}}\]?

Options

• \[\frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\]

• \[\frac{1}{\sqrt{\cos x}+\sqrt{\sin x}}\]

• \[\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\]

• \[\frac{\sqrt{\cos x}}{\sqrt{\cos x}-\sqrt{\sin x}}\]

Answer 1

Using \[\tan x=\frac{\sin x}{\cos x}\], the integrand is rewritten with \[\sqrt{\cos x}\] in the numerator. This produces the stated denominator \[\sqrt{\cos x}+\sqrt{\sin x}\].