1. Evaluate $$\int \cos x \ e^{\sin x} \, dx$$ Use substitution: let $$u = \sin x$$, then $$du = \cos x \, dx$$. The integral becomes $$\int e^u \, du = e^u + C = e^{\sin x} + C$$. 2. Evaluate $$\int \sin 2\theta \ e^{\sin^2 \theta} \, d\theta$$ Use substitution: let $$u = \sin^2 \theta$$, then $$du = 2 \sin \theta \cos \theta \, d\theta = \sin 2\theta \, d\theta$$. The integral becomes $$\int e^u \frac{du}{2} = \frac{1}{2} \int e^u \, du = \frac{1}{2} e^u + C = \frac{1}{2} e^{\sin^2 \theta} + C$$. 3. Evaluate $$\int \frac{1}{\sqrt{x e^{-\sqrt{x}}}} \sec^2(e^{\sqrt{x}} + 1) \, dx$$ Rewrite denominator: $$\sqrt{x e^{-\sqrt{x}}} = \sqrt{x} e^{-\frac{\sqrt{x}}{2}}$$. Let $$t = e^{\sqrt{x}} + 1$$, then $$dt = e^{\sqrt{x}} \frac{1}{2 \sqrt{x}} dx$$. Rewrite $$dx$$: $$dx = 2 \sqrt{x} e^{-\sqrt{x}} dt$$. Substitute into integral: $$\int \frac{1}{\sqrt{x} e^{-\frac{\sqrt{x}}{2}}} \sec^2(t) \cdot 2 \sqrt{x} e^{-\sqrt{x}} dt = \int 2 e^{-\frac{\sqrt{x}}{2}} e^{-\sqrt{x}} \sec^2(t) dt = \int 2 e^{-\frac{3\sqrt{x}}{2}} \sec^2(t) dt$$. This is complicated; instead, use substitution $$u = e^{\sqrt{x}}$$, then $$du = e^{\sqrt{x}} \frac{1}{2 \sqrt{x}} dx$$. Rewrite integral in terms of $$u$$: $$\int \frac{1}{\sqrt{x} e^{-\sqrt{x}/2}} \sec^2(u + 1) dx = \int \frac{1}{\sqrt{x} e^{-\sqrt{x}/2}} \sec^2(u + 1) dx$$. Using $$dx = 2 \sqrt{x} e^{-\sqrt{x}} du$$, the integral simplifies to $$\int 2 e^{-\frac{\sqrt{x}}{2}} \sec^2(u + 1) du$$. Since $$e^{-\frac{\sqrt{x}}{2}} = u^{-1/2}$$, the integral becomes $$\int 2 u^{-1/2} \sec^2(u + 1) du$$. This integral is non-elementary; leave as is or use numerical methods. 4. Evaluate $$\int \frac{1}{x^2} e^{1/x} \sec(1 + e^{1/x}) \tan(1 + e^{1/x}) \, dx$$ Let $$t = 1 + e^{1/x}$$, then $$dt = -\frac{1}{x^2} e^{1/x} dx$$. Rewrite integral: $$\int \frac{1}{x^2} e^{1/x} \sec(t) \tan(t) dx = -\int \sec(t) \tan(t) dt = -\sec(t) + C = -\sec(1 + e^{1/x}) + C$$. 5. Evaluate $$\int \frac{dx}{x \ln x}$$ Let $$u = \ln x$$, then $$du = \frac{1}{x} dx$$. Integral becomes $$\int \frac{1}{u} du = \ln |u| + C = \ln |\ln x| + C$$. 6. Evaluate $$\int \frac{\ln \sqrt{t}}{t} dt$$ Rewrite $$\ln \sqrt{t} = \frac{1}{2} \ln t$$. Integral becomes $$\frac{1}{2} \int \frac{\ln t}{t} dt$$. Use substitution: let $$u = \ln t$$, then $$du = \frac{1}{t} dt$$. Integral becomes $$\frac{1}{2} \int u \, du = \frac{1}{2} \frac{u^2}{2} + C = \frac{(\ln t)^2}{4} + C$$. 7. Evaluate $$\int \frac{dz}{1 + e^z}$$ Rewrite denominator: $$1 + e^z$$. Let $$u = 1 + e^z$$, then $$du = e^z dz$$. Rewrite integral: $$\int \frac{dz}{u}$$ but no direct substitution; instead, write as $$\int \frac{1}{1 + e^z} dz = \int \frac{e^{-z}}{1 + e^{-z}} dz$$. Let $$w = e^{-z}$$, then $$dw = -e^{-z} dz = -w dz$$, so $$dz = -\frac{dw}{w}$$. Integral becomes $$\int \frac{w}{1 + w} \left(-\frac{dw}{w}\right) = -\int \frac{1}{1 + w} dw = -\ln|1 + w| + C = -\ln(1 + e^{-z}) + C$$. 8. Evaluate $$\int \frac{dx}{x \sqrt{x^4 - 1}}$$ Let $$t = x^2$$, then $$dt = 2x dx$$, so $$dx = \frac{dt}{2x}$$. Rewrite integral: $$\int \frac{1}{x \sqrt{t^2 - 1}} \cdot \frac{dt}{2x} = \int \frac{1}{2 x^2 \sqrt{t^2 - 1}} dt = \int \frac{1}{2 t \sqrt{t^2 - 1}} dt$$. Use substitution $$u = \frac{1}{t}$$ or trigonometric substitution; result is $$\frac{1}{\sqrt{2}} \sec^{-1}(x^2) + C$$. 9. Evaluate $$\int \frac{5}{9 + 4r^2} dr$$ Rewrite denominator as $$3^2 + (2r)^2$$. Use formula $$\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C$$. Integral becomes $$5 \int \frac{dr}{9 + 4r^2} = 5 \cdot \frac{1}{3} \tan^{-1} \frac{2r}{3} + C = \frac{5}{3} \tan^{-1} \frac{2r}{3} + C$$. 10. Evaluate $$\int \frac{1}{\sqrt{e^{2\theta} - 1}} d\theta$$ Let $$u = e^\theta$$, then $$du = e^\theta d\theta = u d\theta$$, so $$d\theta = \frac{du}{u}$$. Rewrite integral: $$\int \frac{1}{\sqrt{u^2 - 1}} \cdot \frac{du}{u}$$. Use substitution $$u = \sec x$$, then $$du = \sec x \tan x dx$$. Integral becomes $$\int \frac{1}{\sqrt{\sec^2 x - 1}} \cdot \frac{\sec x \tan x dx}{\sec x} = \int \frac{1}{\tan x} \tan x dx = \int dx = x + C$$. Back substitute $$x = \sec^{-1} u = \sec^{-1} e^\theta$$. Final answer: $$\sec^{-1} e^\theta + C$$.