1. Evaluate the integral (a) $\int x^2 \cos 2x \, dx$. We use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = x^2$, so $du = 2x \, dx$. Let $dv = \cos 2x \, dx$, so $v = \frac{\sin 2x}{2}$. Then, $$\int x^2 \cos 2x \, dx = x^2 \cdot \frac{\sin 2x}{2} - \int \frac{\sin 2x}{2} \cdot 2x \, dx = \frac{x^2 \sin 2x}{2} - \int x \sin 2x \, dx.$$ Now evaluate $\int x \sin 2x \, dx$ by parts again: Let $u = x$, $du = dx$. Let $dv = \sin 2x \, dx$, $v = -\frac{\cos 2x}{2}$. So, $$\int x \sin 2x \, dx = -\frac{x \cos 2x}{2} + \int \frac{\cos 2x}{2} \, dx = -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} + C.$$ Substitute back: $$\int x^2 \cos 2x \, dx = \frac{x^2 \sin 2x}{2} - \left(-\frac{x \cos 2x}{2} + \frac{\sin 2x}{4}\right) + C = \frac{x^2 \sin 2x}{2} + \frac{x \cos 2x}{2} - \frac{\sin 2x}{4} + C.$$ 2. Evaluate (b) $\int x^4 e^{3x} \, dx$. Use integration by parts repeatedly or tabular method. Let $u = x^4$, $dv = e^{3x} dx$, so $du = 4x^3 dx$, $v = \frac{e^{3x}}{3}$. Then, $$\int x^4 e^{3x} dx = \frac{x^4 e^{3x}}{3} - \int \frac{4x^3 e^{3x}}{3} dx = \frac{x^4 e^{3x}}{3} - \frac{4}{3} \int x^3 e^{3x} dx.$$ Repeat integration by parts on $\int x^3 e^{3x} dx$ similarly until the power of $x$ reduces to zero. 3. Evaluate (c) $\int x^4 \log(x + 4) \, dx$. Use integration by parts with $u = \log(x+4)$, $dv = x^4 dx$, so $du = \frac{1}{x+4} dx$, $v = \frac{x^5}{5}$. Then, $$\int x^4 \log(x+4) dx = \frac{x^5}{5} \log(x+4) - \int \frac{x^5}{5(x+4)} dx.$$ Simplify the integral and evaluate. 4. Evaluate (d) $\int \frac{\sin^{-1} x}{\sqrt{1 - x^2}} dx$. Let $t = \sin^{-1} x$, so $x = \sin t$, $dx = \cos t dt$, and $\sqrt{1 - x^2} = \cos t$. Rewrite integral: $$\int \frac{t}{\cos t} \cdot \cos t dt = \int t dt = \frac{t^2}{2} + C = \frac{(\sin^{-1} x)^2}{2} + C.$$ 5. Evaluate (e) $\int e^{2x} \cos 4x \, dx$. Use formula for $\int e^{ax} \cos bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) + C$. Here, $a=2$, $b=4$, so $$\int e^{2x} \cos 4x \, dx = \frac{e^{2x}}{2^2 + 4^2} (2 \cos 4x + 4 \sin 4x) + C = \frac{e^{2x}}{20} (2 \cos 4x + 4 \sin 4x) + C.$$ 6. Evaluate (f) $\int \sinh 7x \, dx$. Recall $\int \sinh kx \, dx = \frac{\cosh kx}{k} + C$. So, $$\int \sinh 7x \, dx = \frac{\cosh 7x}{7} + C.$$ 7. Evaluate (g) $\int \sqrt{7x - 1} \, dx$. Let $t = 7x - 1$, so $dt = 7 dx$, $dx = \frac{dt}{7}$. Rewrite integral: $$\int \sqrt{t} \cdot \frac{dt}{7} = \frac{1}{7} \int t^{1/2} dt = \frac{1}{7} \cdot \frac{2}{3} t^{3/2} + C = \frac{2}{21} (7x - 1)^{3/2} + C.$$ 8. Evaluate (h) $\int x (5x + 6)^6 \, dx$. Let $t = 5x + 6$, so $dt = 5 dx$, $dx = \frac{dt}{5}$. Express $x$ in terms of $t$: $x = \frac{t - 6}{5}$. Rewrite integral: $$\int \frac{t - 6}{5} t^6 \cdot \frac{dt}{5} = \frac{1}{25} \int (t - 6) t^6 dt = \frac{1}{25} \int (t^7 - 6 t^6) dt = \frac{1}{25} \left( \frac{t^8}{8} - 6 \frac{t^7}{7} \right) + C = \frac{(5x + 6)^8}{200} - \frac{6 (5x + 6)^7}{175} + C.$$ 9. Evaluate (i) $\int x^2 \sqrt{x + 1} \, dx$. Let $t = x + 1$, so $x = t - 1$, $dx = dt$. Rewrite integral: $$\int (t - 1)^2 \sqrt{t} dt = \int (t^2 - 2t + 1) t^{1/2} dt = \int (t^{5/2} - 2 t^{3/2} + t^{1/2}) dt.$$ Integrate termwise: $$\frac{2}{7} t^{7/2} - \frac{4}{5} t^{5/2} + \frac{2}{3} t^{3/2} + C = \frac{2}{7} (x+1)^{7/2} - \frac{4}{5} (x+1)^{5/2} + \frac{2}{3} (x+1)^{3/2} + C.$$ 10. Evaluate (j) $\int x (5 + x^2)^8 \, dx$. Let $t = 5 + x^2$, so $dt = 2x dx$, $x dx = \frac{dt}{2}$. Rewrite integral: $$\int x (5 + x^2)^8 dx = \int t^8 \cdot \frac{dt}{2} = \frac{1}{2} \int t^8 dt = \frac{1}{2} \cdot \frac{t^9}{9} + C = \frac{(5 + x^2)^9}{18} + C.$$ 11. Evaluate (k) $\int x^2 \sqrt{7 - 4x^3} \, dx$. Let $t = 7 - 4x^3$, so $dt = -12 x^2 dx$, $x^2 dx = -\frac{dt}{12}$. Rewrite integral: $$\int x^2 \sqrt{t} dx = \int \sqrt{t} \cdot x^2 dx = \int \sqrt{t} \cdot \left(-\frac{dt}{12}\right) = -\frac{1}{12} \int t^{1/2} dt = -\frac{1}{12} \cdot \frac{2}{3} t^{3/2} + C = -\frac{(7 - 4x^3)^{3/2}}{18} + C.$$ 12. Evaluate (l) $\int \cot x \, dx$. Recall $\int \cot x \, dx = \log |\sin x| + C$. Final answers are as above for each integral.