# How do you evaluate the integral #int sec^2x/(sqrt(1-tanx))dx# from 0 to #pi/4#?

We have:

Thus we see:

Notice that we can take the negative sign to flip the order of the bounds, just to make things look nicer. Write the square root using fractional and negative exponents.

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To evaluate the integral (\int_0^{\pi/4} \frac{\sec^2 x}{\sqrt{1-\tan x}} , dx), we first observe that (\sec^2 x = \frac{1}{\cos^2 x}) and (\tan x = \frac{\sin x}{\cos x}).

Now, we perform the substitution (u = \sin x), so (du = \cos x , dx). When (x = 0), (u = \sin 0 = 0), and when (x = \frac{\pi}{4}), (u = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}).

So, the integral becomes:

[\int_0^{\pi/4} \frac{\sec^2 x}{\sqrt{1-\tan x}} , dx = \int_0^{\sqrt{2}/2} \frac{1}{\sqrt{1-u^2}} , du.]

This integral represents the area under the curve of the function (\frac{1}{\sqrt{1-u^2}}) from (0) to (\frac{\sqrt{2}}{2}), which is the arc length of the unit circle from (0) to (\frac{\pi}{4}).

This integral is a standard trigonometric integral. Using the trigonometric substitution (u = \sin \theta), we find that (du = \cos \theta , d\theta) and (\sqrt{1-u^2} = \cos \theta).

So, the integral becomes:

[\int \frac{1}{\sqrt{1-u^2}} , du = \int \frac{1}{\cos \theta} \cdot \cos \theta , d\theta = \int d\theta = \theta + C.]

Thus, evaluating from (0) to (\frac{\sqrt{2}}{2}) gives:

[\left[\theta\right]_0^{\sqrt{2}/2} = \frac{\pi}{4}.]

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