To evaluate the integral

$I = \int_{0}^{\pi/4} \sec^3 \theta \, d\theta,$

we use the standard formula for the integral of $\sec^3 \theta$:

$\int \sec^3 \theta \, d\theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta| + C.$

Step-by-step solution:

1. Apply the formula for $\sec^3 \theta$: $\int \sec^3 \theta \, d\theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta|.$

2. Evaluate at the bounds $\theta = 0$ and $\theta = \frac{\pi}{4}$.

At $\theta = \frac{\pi}{4}$:

• $\sec \frac{\pi}{4} = \sqrt{2}$,
• $\tan \frac{\pi}{4} = 1$.

Thus:

= \frac{1}{2} (\sqrt{2} \cdot 1) + \frac{1}{2} \ln |\sqrt{2} + 1|.$$ This simplifies to: $$\frac{\sqrt{2}}{2} + \frac{1}{2} \ln (\sqrt{2} + 1).$$ #### At $$\theta = 0$$: - $$\sec 0 = 1$$, - $$\tan 0 = 0$$. Thus: $$\frac{1}{2} \sec 0 \tan 0 + \frac{1}{2} \ln |\sec 0 + \tan 0| = 0 + \frac{1}{2} \ln 1 = 0.$$ 3. Subtract the results at the bounds: $$I = \left[\frac{\sqrt{2}}{2} + \frac{1}{2} \ln (\sqrt{2} + 1)\right] - 0.$$ ### Final Answer: $$I = \frac{\sqrt{2}}{2} + \frac{1}{2} \ln (\sqrt{2} + 1).$$ Let me know if you'd like further clarification or details! --- ### Related Questions: 1. What is the indefinite integral of $$\sec^3 \theta$$? 2. How do you compute the integral of $$\sec^2 \theta$$? 3. Can you derive the formula for $$\int \sec^3 \theta \, d\theta$$ from scratch? 4. How does the logarithmic term arise in the integral of $$\sec^3 \theta$$? 5. What are other common integrals involving secant functions? ### Tip: When working with integrals of secant powers, always check if reducing the power simplifies the computation.