Key Concepts and Formulas

• Trigonometric Identities: The half-angle identity $1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$ is crucial for simplifying the integrand. The reciprocal identity $\frac{1}{\cos^2 \theta} = \sec^2 \theta$ is also essential.
• Standard Integration Formula: The integral of $\sec^2 u$ is $\tan u$. Specifically, $\int \sec^2(ax) \, dx = \frac{1}{a}\tan(ax) + C$.
• Fundamental Theorem of Calculus: For a definite integral $\int_a^b f(x) \, dx = F(b) - F(a)$, where $F(x)$ is an antiderivative of $f(x)$.
• Tangent Half-Angle Values: Knowledge of $\tan(\pi/8) = \sqrt{2} - 1$ and $\tan(3\pi/8) = \sqrt{2} + 1$ is required.

Step-by-Step Solution

We need to evaluate the definite integral: $I = \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}}$

Step 1: Simplify the Integrand The denominator $1 + \cos x$ can be simplified using the half-angle identity $1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$. This identity is useful because it transforms the sum into a single squared trigonometric term, which is easier to integrate. Substituting this into the integral, we get: $I = \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}\left(\frac{x}{2}\right)}}}$ Using the reciprocal identity $\frac{1}{\cos^2 \theta} = \sec^2 \theta$, we can rewrite the integrand: $I = {1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}\left(\frac{x}{2}\right)\,dx}$

Step 2: Perform the Indefinite Integration Now we integrate $\sec^2\left(\frac{x}{2}\right)$. The standard integral for $\sec^2 u$ is $\tan u$. For $\sec^2(ax)$, the integral is $\frac{1}{a}\tan(ax)$. In our case, $a = \frac{1}{2}$. Therefore, the antiderivative of $\sec^2\left(\frac{x}{2}\right)$ is $\frac{1}{1/2}\tan\left(\frac{x}{2}\right) = 2\tan\left(\frac{x}{2}\right)$. Applying this to the definite integral: $I = {1 \over 2}\left[ 2\tan\left(\frac{x}{2}\right) \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$ Simplifying the constant factor, we get: $I = \left[ \tan\left(\frac{x}{2}\right) \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$

Step 3: Apply the Limits of Integration Using the Fundamental Theorem of Calculus, we substitute the upper and lower limits of integration into the antiderivative and subtract. $I = \tan\left(\frac{1}{2} \cdot \frac{3\pi}{4}\right) - \tan\left(\frac{1}{2} \cdot \frac{\pi}{4}\right)$ $I = \tan\left(\frac{3\pi}{8}\right) - \tan\left(\frac{\pi}{8}\right)$

Step 4: Evaluate the Trigonometric Values We need to find the values of $\tan\left(\frac{3\pi}{8}\right)$ and $\tan\left(\frac{\pi}{8}\right)$. To find $\tan\left(\frac{\pi}{8}\right)$, we use the double angle formula for tangent, $\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$. Let $\theta = \frac{\pi}{8}$, so $2\theta = \frac{\pi}{4}$. $\tan\left(\frac{\pi}{4}\right) = 1 = \frac{2\tan\left(\frac{\pi}{8}\right)}{1-\tan^2\left(\frac{\pi}{8}\right)}$ Let $t = \tan\left(\frac{\pi}{8}\right)$. Then $1 = \frac{2t}{1-t^2}$, which gives $1-t^2 = 2t$, or $t^2 + 2t - 1 = 0$. Solving this quadratic equation for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $t = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$ Since $\frac{\pi}{8}$ is in the first quadrant, $\tan\left(\frac{\pi}{8}\right)$ must be positive. Thus, $\tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1$.

To find $\tan\left(\frac{3\pi}{8}\right)$, we use the complementary angle identity $\tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta$. $\tan\left(\frac{3\pi}{8}\right) = \tan\left(\frac{\pi}{2} - \frac{\pi}{8}\right) = \cot\left(\frac{\pi}{8}\right)$ Since $\cot\theta = \frac{1}{\tan\theta}$, we have: $\tan\left(\frac{3\pi}{8}\right) = \frac{1}{\tan\left(\frac{\pi}{8}\right)} = \frac{1}{\sqrt{2}-1}$ Rationalizing the denominator: $\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1$ So, $\tan\left(\frac{3\pi}{8}\right) = \sqrt{2} + 1$.

Step 5: Calculate the Final Value Substitute the values of $\tan\left(\frac{3\pi}{8}\right)$ and $\tan\left(\frac{\pi}{8}\right)$ back into the expression for $I$: $I = (\sqrt{2} + 1) - (\sqrt{2} - 1)$ $I = \sqrt{2} + 1 - \sqrt{2} + 1$ $I = 2$

Common Mistakes & Tips

• Trigonometric Identity Errors: Ensure correct application of half-angle and reciprocal identities. A common mistake is to forget the factor of 2 in $1 + \cos x = 2\cos^2(x/2)$.
• Integration of $\sec^2(ax)$: Remember the chain rule when integrating, leading to the $\frac{1}{a}$ factor. Forgetting this factor is a frequent error.
• Evaluating Tangent Values: If you don't recall $\tan(\pi/8)$ and $\tan(3\pi/8)$, be careful when deriving them using the double angle formula or complementary angle identities to avoid algebraic errors.

Summary

The integral was evaluated by first simplifying the integrand using the trigonometric identity $1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$. This transformed the integral into a form involving $\sec^2\left(\frac{x}{2}\right)$, which is a standard integral. After performing the indefinite integration to get $\tan\left(\frac{x}{2}\right)$, the Fundamental Theorem of Calculus was applied using the given limits of integration. Finally, the specific values of $\tan\left(\frac{3\pi}{8}\right)$ and $\tan\left(\frac{\pi}{8}\right)$ were calculated and substituted to find the definite integral's value.

The final answer is \boxed{2}.