Calculate the integral: integral from pi/4 to (pi/3)

Calculate the integral: $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{dx}{\cos^2 x(1+\tan x)^2}$.

We will use substitution to simplify this integral.

Step 1: Choose a substitution

Let $u = \tan x$. Then $du = \sec^2 x \, dx = \frac{1}{\cos^2 x} \, dx$, so $\frac{dx}{\cos^2 x} = du$.

Step 2: Change the limits of integration

When $x = \frac{\pi}{4}$: $u = 1$. When $x = \frac{\pi}{3}$: $u = \sqrt{3}$.

Step 3: Rewrite the integral

\[\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{dx}{\cos^2 x(1+\tan x)^2} = \int_1^{\sqrt{3}} \frac{1}{(1+u)^2} \, du\]

Step 4: Integrate

\[\int \frac{1}{(1+u)^2} \, du = -\frac{1}{1+u} + C\]

Step 5: Apply the Newton-Leibniz formula

\[\int_1^{\sqrt{3}} \frac{1}{(1+u)^2} \, du = \left[-\frac{1}{1+u}\right]_1^{\sqrt{3}}\]

Step 6: Evaluate at the limits

At $u=\sqrt3$: $-\frac{1}{1+\sqrt3}$. At $u=1$: $-\frac12$.

Step 7: Calculate the final result

\[\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{dx}{\cos^2 x(1+\tan x)^2} = -\frac{1}{1+\sqrt{3}} + \frac{1}{2} = \frac{1}{2} - \frac{1}{1+\sqrt{3}}\]

Step 8: Rationalize the denominator

\[\frac{1}{1+\sqrt{3}} = \frac{1-\sqrt{3}}{(1+\sqrt{3})(1-\sqrt{3})} = \frac{1-\sqrt{3}}{-2} = \frac{\sqrt{3}-1}{2}\]

Step 9: Substitute back

\[\frac{1}{2} - \frac{\sqrt{3}-1}{2} = \frac{1 - \sqrt{3} + 1}{2} = \frac{2 - \sqrt{3}}{2}\]

Verification using the antiderivative:

The antiderivative of $\frac{1}{\cos^2 x(1+\tan x)^2}$ is $-\frac{1}{1+\tan x}$. Let $F(x)=-\frac1{1+\tan x}$; then $F'(x)=\frac{\sec^2x}{(1+\tan x)^2}=\frac1{\cos^2x(1+\tan x)^2}$. Using this:

\[\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{dx}{\cos^2 x(1+\tan x)^2} = \left[-\frac{1}{1+\tan x}\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = -\frac{1}{1+\sqrt{3}} + \frac{1}{2} = \frac{2-\sqrt{3}}{2}\]

Numerical value:

Using $\sqrt{3} \approx 1.732$: $\frac{2-\sqrt{3}}{2} \approx 0.134$.

Final answer:

\[\boxed{\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{dx}{\cos^2 x(1+\tan x)^2} = \frac{2-\sqrt{3}}{2}}\]

Or equivalently:

\[\boxed{\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{dx}{\cos^2 x(1+\tan x)^2} = 1 - \frac{\sqrt{3}}{2}}\]