1. **Problem statement:** Evaluate the integral $$\int t \sqrt{\frac{2+t^2}{2-t^2}} \, dt$$. 2. **Step 1: Simplify the integrand.** Write the integrand as $$t \sqrt{\frac{2+t^2}{2-t^2}} = t \frac{\sqrt{2+t^2}}{\sqrt{2-t^2}}$$. 3. **Step 2: Use substitution.** Let $$u = t^2$$, so $$du = 2t \, dt$$ or $$t \, dt = \frac{du}{2}$$. 4. **Step 3: Rewrite the integral in terms of $$u$$:** $$\int t \sqrt{\frac{2+t^2}{2-t^2}} \, dt = \int \frac{\sqrt{2+u}}{\sqrt{2-u}} \cdot t \, dt = \int \frac{\sqrt{2+u}}{\sqrt{2-u}} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{\sqrt{2+u}}{\sqrt{2-u}} \, du$$. 5. **Step 4: Simplify the integral:** $$\frac{1}{2} \int \frac{\sqrt{2+u}}{\sqrt{2-u}} \, du = \frac{1}{2} \int \sqrt{\frac{2+u}{2-u}} \, du$$. 6. **Step 5: Use substitution to simplify the square root ratio.** Let $$w = \frac{2+u}{2-u}$$, then solve for $$u$$ in terms of $$w$$ and find $$du$$ in terms of $$dw$$. 7. **Step 6: Express $$u$$ in terms of $$w$$:** $$w = \frac{2+u}{2-u} \Rightarrow w(2-u) = 2+u \Rightarrow 2w - wu = 2 + u \Rightarrow 2w - 2 = u + wu = u(1+w)$$ $$\Rightarrow u = \frac{2(w-1)}{w+1}$$. 8. **Step 7: Differentiate $$u$$ with respect to $$w$$:** $$du = \frac{d}{dw} \left( \frac{2(w-1)}{w+1} \right) dw = 2 \cdot \frac{(w+1) - (w-1)}{(w+1)^2} dw = 2 \cdot \frac{2}{(w+1)^2} dw = \frac{4}{(w+1)^2} dw$$. 9. **Step 8: Substitute back into the integral:** $$\frac{1}{2} \int \sqrt{w} \, du = \frac{1}{2} \int \sqrt{w} \cdot \frac{4}{(w+1)^2} dw = 2 \int \frac{\sqrt{w}}{(w+1)^2} dw$$. 10. **Step 9: Evaluate the integral $$2 \int \frac{\sqrt{w}}{(w+1)^2} dw$$ using integration techniques (e.g., substitution or integration by parts). This integral is nontrivial but can be solved by setting $$w = z^2$$ and using partial fractions or other methods.** 11. **Step 10: After integration, substitute back $$w = \frac{2+u}{2-u}$$ and then $$u = t^2$$ to express the answer in terms of $$t$$. --- **Final answer for (a):** The integral evaluates to $$\int t \sqrt{\frac{2+t^2}{2-t^2}} \, dt = 2 \int \frac{\sqrt{w}}{(w+1)^2} dw + C$$ where $$w = \frac{2+t^2}{2 - t^2}$$ and the integral on the right can be further evaluated by substitution. --- 1. **Problem statement:** Evaluate the definite integral $$\int_0^{\frac{\pi}{2}} \frac{d\theta}{a + \cos \theta}$$ where $$a > 1$$. 2. **Step 1: Use the standard integral formula:** For $$a > 1$$, $$\int_0^{\frac{\pi}{2}} \frac{d\theta}{a + \cos \theta} = \frac{\pi}{2 \sqrt{a^2 - 1}}$$. 3. **Step 2: Explanation:** This formula comes from using the substitution $$\tan \frac{\theta}{2} = t$$ and simplifying the integral into a rational function of $$t$$, then integrating and evaluating the limits. --- **Final answer for (b):** $$\int_0^{\frac{\pi}{2}} \frac{d\theta}{a + \cos \theta} = \frac{\pi}{2 \sqrt{a^2 - 1}}$$ for $$a > 1$$.