1. Problem: Evaluate the integral $$\int \frac{\cos 2x}{\sqrt{\sin 2x + 2}}\,dx$$ Step 1: Use substitution. Let $$u = \sin 2x + 2$$. Then, $$\frac{du}{dx} = 2\cos 2x \Rightarrow du = 2\cos 2x\,dx$$. Step 2: Rewrite integral in terms of $$u$$: $$\int \frac{\cos 2x}{\sqrt{\sin 2x + 2}} dx = \int \frac{\cos 2x}{\sqrt{u}} dx = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-1/2} du$$ Step 3: Integrate: $$\frac{1}{2} \int u^{-1/2} du = \frac{1}{2} \cdot 2 u^{1/2} + C = \sqrt{u} + C$$ Step 4: Substitute back: $$\sqrt{\sin 2x + 2} + C$$ 2. Problem: Evaluate $$\int \frac{-\sin x}{\sqrt{\cos x + 1}} dx$$ Step 1: Let $$u = \cos x + 1$$, then $$du = -\sin x dx$$. Step 2: Substitute: $$\int \frac{-\sin x}{\sqrt{\cos x + 1}} dx = \int \frac{1}{\sqrt{u}} du = \int u^{-1/2} du$$ Step 3: Integrate: $$2u^{1/2} + C = 2 \sqrt{\cos x + 1} + C$$ 3. Problem: Evaluate $$\int \frac{\cos \sqrt{x}}{\sqrt{x}} dx$$ Step 1: Set $$t = \sqrt{x}$$, then $$x = t^2$$ and $$dx = 2t dt$$. Step 2: Substitute: $$\int \frac{\cos t}{t} dx = \int \frac{\cos t}{t} (2t dt) = 2 \int \cos t dt$$ Step 3: Integrate: $$2 \sin t + C = 2 \sin \sqrt{x} + C$$ 4. Problem: Evaluate $$\int (x + 1) \sin(x^2 + 2x) dx$$ Step 1: Let $$u = x^2 + 2x$$, then $$du = (2x + 2) dx = 2(x + 1) dx$$. Step 2: Thus, $$ (x + 1) dx = \frac{du}{2}$$. Step 3: Rewrite integral: $$\int (x + 1) \sin(u) dx = \int \sin u \cdot \frac{du}{2} = \frac{1}{2} \int \sin u du$$ Step 4: Integrate: $$-\frac{1}{2} \cos u + C = -\frac{1}{2} \cos(x^2 + 2x) + C$$ 5. Problem: Evaluate $$\int \frac{\tan x}{\cos^2 x \sqrt{\tan^2 x + 1}} dx$$ Step 1: Note that $$\tan^2 x + 1 = \sec^2 x$$, so $$\sqrt{\tan^2 x + 1} = |\sec x|$$. Assume positive in domain so $$\sqrt{\tan^2 x + 1} = \sec x$$. Step 2: Re-express integral: $$\int \frac{\tan x}{\cos^2 x \sec x} dx = \int \frac{\tan x}{\cos^2 x \cdot \frac{1}{\cos x}} dx = \int \frac{\tan x \cos x}{\cos^2 x} dx = \int \frac{\tan x \cos x}{\cos^2 x} dx = \int \frac{\tan x}{\cos x} dx$$ Step 3: Since $$\tan x = \frac{\sin x}{\cos x}$$, $$\frac{\tan x}{\cos x} = \frac{\sin x}{\cos^2 x}$$. Step 4: Integrate: $$\int \frac{\sin x}{\cos^2 x} dx$$. Letting $$t = \cos x$$, $$dt = -\sin x dx$$, so $$-dt = \sin x dx$$ Thus, $$\int \frac{\sin x}{\cos^2 x} dx = \int \frac{-dt}{t^2} = -\int t^{-2} dt = -(-t^{-1}) + C = \frac{1}{\cos x} + C = \sec x + C$$ 6. Problem: Evaluate $$\int \frac{1}{\cos^2 x \sqrt{\tan x + 1}} dx$$ Step 1: Let $$u = \tan x + 1$$, then $$du = \sec^2 x dx = \frac{1}{\cos^2 x} dx$$. Step 2: Substitute: $$\int \frac{1}{\cos^2 x \sqrt{\tan x + 1}} dx = \int u^{-1/2} du = 2 u^{1/2} + C = 2 \sqrt{\tan x + 1} + C$$ 7. Problem: Evaluate $$\int \frac{x^2}{\sqrt{x^3 + 2}} dx$$ Step 1: Let $$u = x^3 + 2$$, then $$du = 3x^2 dx$$. Step 2: Thus, $$x^2 dx = \frac{du}{3}$$. Step 3: Substitute: $$\int \frac{x^2}{\sqrt{u}} dx = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{3} = \frac{1}{3} \int u^{-1/2} du$$ Step 4: Integrate: $$\frac{1}{3} \cdot 2 u^{1/2} + C = \frac{2}{3} \sqrt{x^3 + 2} + C$$ 8. Problem: Evaluate $$\int \frac{x}{\sqrt{3 - x^2}} dx$$ Step 1: Let $$u = 3 - x^2$$, then $$du = -2x dx$$ Step 2: So, $$x dx = -\frac{du}{2}$$ Step 3: Substitute: $$\int \frac{x}{\sqrt{u}} dx = \int \frac{1}{\sqrt{u}} \cdot x dx = \int \frac{1}{\sqrt{u}} \cdot \left(-\frac{du}{2}\right) = -\frac{1}{2} \int u^{-1/2} du$$ Step 4: Integrate: $$-\frac{1}{2} \cdot 2 u^{1/2} + C = -\sqrt{3 - x^2} + C$$ 9. Problem: Evaluate $$\int \frac{4x^3 + 3x^2}{\sqrt{x^4 + x^3 + 1}} dx$$ Step 1: Let $$u = x^4 + x^3 + 1$$, then $$du = (4x^3 + 3x^2) dx$$ Step 2: Substitute: $$\int \frac{4x^3 + 3x^2}{\sqrt{u}} dx = \int \frac{du}{\sqrt{u}} = \int u^{-1/2} du$$ Step 3: Integrate: $$2 u^{1/2} + C = 2 \sqrt{x^4 + x^3 + 1} + C$$ 10. Problem: Evaluate $$\int \frac{6x^2 + 4x}{\sqrt{x^3 + x^2 -1}} dx$$ Step 1: Let $$u = x^3 + x^2 -1$$, then $$du = (3x^2 + 2x) dx$$. Step 2: Notice the numerator is $$6x^2 + 4x = 2 (3x^2 + 2x)$$, so $$6x^2 + 4x dx = 2 du$$ Step 3: Rewrite: $$\int \frac{6x^2 + 4x}{\sqrt{u}} dx = \int \frac{2 du}{\sqrt{u}} = 2 \int u^{-1/2} du$$ Step 4: Integrate: $$2 \cdot 2 u^{1/2} + C = 4 \sqrt{x^3 + x^2 -1} + C$$ --- Final answers: 1. $$\sqrt{\sin 2x + 2} + C$$ 2. $$2 \sqrt{\cos x + 1} + C$$ 3. $$2 \sin \sqrt{x} + C$$ 4. $$-\frac{1}{2} \cos(x^2 + 2x) + C$$ 5. $$\sec x + C$$ 6. $$2 \sqrt{\tan x + 1} + C$$ 7. $$\frac{2}{3} \sqrt{x^3 + 2} + C$$ 8. $$-\sqrt{3 - x^2} + C$$ 9. $$2 \sqrt{x^4 + x^3 + 1} + C$$ 10. $$4 \sqrt{x^3 + x^2 - 1} + C$$