Question: Evaluate the following:
1. $\int \sin 2\theta d\theta (\sin^2 \theta + 5)^4$
2. $\int \sin 2\theta d\theta (1 - \frac{1}{2} \cos^2 \theta)^8$
3. $\int \frac{(9 - \sqrt{x})^{10}}{\sqrt{x}} dx$
4. $\int \frac{dx}{\sqrt{x} \sqrt[3]{1 - \sqrt{x}}}$
5. $\int \frac{e^{2x} + e^{-2x}}{\sqrt[5]{e^{2x} - e^{-2x}}} dx$
6. $\int \frac{dx}{x \sqrt[3]{\ln x}}$
7. $\int \frac{(x - 1) dx}{(x - 3)^4 (x + 1)^4}$
8. $\int \frac{x^5 dx}{(x^3 - 1)^{1/2} (x^3 + 1)^{1/2}}$
9. $\int \frac{dx}{x + 5x^{3/4}}$
10. $\int \frac{dx}{\sqrt{x} + x}$
11. $\int_1^3 \frac{v^4 + 1}{v^2} dv$
12. $\int_0^a y^2 (a^3 - y^3)^{1/3} dy$
1. **Problem:** Evaluate $\int \sin 2\theta (\sin^2 \theta + 5)^4 d\theta$.
**Step 1:** Recognize the chain rule pattern. Let $u = \sin^2 \theta + 5$.
**Step 2:** Compute $du/d\theta = 2 \sin \theta \cos \theta = \sin 2\theta$.
**Step 3:** Rewrite integral as $\int u^4 du$.
**Step 4:** Integrate: $\int u^4 du = \frac{u^5}{5} + C$.
**Step 5:** Substitute back: $$\frac{(\sin^2 \theta + 5)^5}{5} + C$$.
2. **Problem:** Evaluate $\int \sin 2\theta (1 - \frac{1}{2} \cos^2 \theta)^8 d\theta$.
**Step 1:** Let $u = 1 - \frac{1}{2} \cos^2 \theta$.
**Step 2:** Compute $du/d\theta = -\frac{1}{2} \cdot 2 \cos \theta (-\sin \theta) = \cos \theta \sin \theta = \frac{1}{2} \sin 2\theta$.
**Step 3:** So, $du = \frac{1}{2} \sin 2\theta d\theta \Rightarrow \sin 2\theta d\theta = 2 du$.
**Step 4:** Rewrite integral as $\int (u)^8 \sin 2\theta d\theta = \int u^8 (2 du) = 2 \int u^8 du$.
**Step 5:** Integrate: $2 \cdot \frac{u^9}{9} + C = \frac{2}{9} u^9 + C$.
**Step 6:** Substitute back: $$\frac{2}{9} \left(1 - \frac{1}{2} \cos^2 \theta \right)^9 + C$$.
3. **Problem:** Evaluate $\int \frac{(9 - \sqrt{x})^{10}}{\sqrt{x}} dx$.
**Step 1:** Let $u = 9 - \sqrt{x}$.
**Step 2:** Then $\sqrt{x} = x^{1/2}$, so $du/dx = -\frac{1}{2} x^{-1/2} = -\frac{1}{2 \sqrt{x}}$.
**Step 3:** Rearranged: $du = -\frac{1}{2 \sqrt{x}} dx \Rightarrow dx = -2 \sqrt{x} du$.
**Step 4:** Substitute into integral:
$$\int \frac{u^{10}}{\sqrt{x}} dx = \int u^{10} \frac{dx}{\sqrt{x}} = \int u^{10} \frac{-2 \sqrt{x} du}{\sqrt{x}} = \int -2 u^{10} du = -2 \int u^{10} du$$.
**Step 5:** Integrate: $-2 \cdot \frac{u^{11}}{11} + C = -\frac{2}{11} u^{11} + C$.
**Step 6:** Substitute back: $$-\frac{2}{11} (9 - \sqrt{x})^{11} + C$$.
4. **Problem:** Evaluate $\int \frac{dx}{\sqrt{x} \sqrt[3]{1 - \sqrt{x}}}$.
**Step 1:** Let $u = 1 - \sqrt{x}$.
**Step 2:** Then $du/dx = -\frac{1}{2} x^{-1/2} = -\frac{1}{2 \sqrt{x}}$.
**Step 3:** Rearranged: $du = -\frac{1}{2 \sqrt{x}} dx \Rightarrow dx = -2 \sqrt{x} du$.
**Step 4:** Substitute into integral:
$$\int \frac{dx}{\sqrt{x} u^{1/3}} = \int \frac{-2 \sqrt{x} du}{\sqrt{x} u^{1/3}} = \int -2 u^{-1/3} du = -2 \int u^{-1/3} du$$.
**Step 5:** Integrate:
$$-2 \cdot \frac{u^{2/3}}{2/3} + C = -2 \cdot \frac{3}{2} u^{2/3} + C = -3 u^{2/3} + C$$.
**Step 6:** Substitute back:
$$-3 (1 - \sqrt{x})^{2/3} + C$$.
5. **Problem:** Evaluate $\int \frac{e^{2x} + e^{-2x}}{\sqrt[5]{e^{2x} - e^{-2x}}} dx$.
**Step 1:** Let $u = e^{2x} - e^{-2x}$.
**Step 2:** Compute $du/dx = 2 e^{2x} + 2 e^{-2x} = 2 (e^{2x} + e^{-2x})$.
**Step 3:** Rearranged: $du = 2 (e^{2x} + e^{-2x}) dx \Rightarrow (e^{2x} + e^{-2x}) dx = \frac{du}{2}$.
**Step 4:** Substitute into integral:
$$\int \frac{e^{2x} + e^{-2x}}{u^{1/5}} dx = \int \frac{1}{u^{1/5}} \cdot (e^{2x} + e^{-2x}) dx = \int \frac{1}{u^{1/5}} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-1/5} du$$.
**Step 5:** Integrate:
$$\frac{1}{2} \cdot \frac{u^{4/5}}{4/5} + C = \frac{1}{2} \cdot \frac{5}{4} u^{4/5} + C = \frac{5}{8} u^{4/5} + C$$.
**Step 6:** Substitute back:
$$\frac{5}{8} (e^{2x} - e^{-2x})^{4/5} + C$$.
6. **Problem:** Evaluate $\int \frac{dx}{x \sqrt[3]{\ln x}}$.
**Step 1:** Let $u = \ln x$.
**Step 2:** Then $du/dx = \frac{1}{x} \Rightarrow du = \frac{1}{x} dx \Rightarrow dx = x du$.
**Step 3:** Substitute into integral:
$$\int \frac{dx}{x u^{1/3}} = \int \frac{x du}{x u^{1/3}} = \int u^{-1/3} du$$.
**Step 4:** Integrate:
$$\frac{u^{2/3}}{2/3} + C = \frac{3}{2} u^{2/3} + C$$.
**Step 5:** Substitute back:
$$\frac{3}{2} (\ln x)^{2/3} + C$$.
**Final answers:**
1. $$\frac{(\sin^2 \theta + 5)^5}{5} + C$$
2. $$\frac{2}{9} \left(1 - \frac{1}{2} \cos^2 \theta \right)^9 + C$$
3. $$-\frac{2}{11} (9 - \sqrt{x})^{11} + C$$
4. $$-3 (1 - \sqrt{x})^{2/3} + C$$
5. $$\frac{5}{8} (e^{2x} - e^{-2x})^{4/5} + C$$
6. $$\frac{3}{2} (\ln x)^{2/3} + C$$