1. Problem: Evaluate $$\int_0^{\frac{\pi}{2}} \sin^n x \, dx$$ and find specific values for $$n=6$$ and $$n=5$$. Step 1: Use the reduction formula for $$I_n = \int_0^{\frac{\pi}{2}} \sin^n x \, dx$$: $$I_n = \frac{n-1}{n} I_{n-2}$$ with $$I_0 = \frac{\pi}{2}$$ and $$I_1 = 1$$. Step 2: For $$n=6$$: $$I_6 = \frac{5}{6} I_4$$, $$I_4 = \frac{3}{4} I_2$$, $$I_2 = \frac{1}{2} I_0 = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}$$. Step 3: Calculate upward: $$I_4 = \frac{3}{4} \times \frac{\pi}{4} = \frac{3 \pi}{16}$$, $$I_6 = \frac{5}{6} \times \frac{3 \pi}{16} = \frac{15 \pi}{96} = \frac{5 \pi}{32}$$. Step 4: For $$n=5$$ use the formula: $$I_5 = \frac{4}{5} I_3$$, $$I_3 = \frac{2}{3} I_1 = \frac{2}{3} \times 1 = \frac{2}{3}$$, Step 5: Thus: $$I_5 = \frac{4}{5} \times \frac{2}{3} = \frac{8}{15}$$. --- 2. Problem: Evaluate $$\int_0^{\frac{\pi}{2}} \cos^n x \, dx$$ and find for $$n=8$$ and $$n=7$$. Step 1: Since $$\cos x$$ and $$\sin x$$ are symmetric in these bounds, $$\int_0^{\frac{\pi}{2}} \cos^n x \, dx = \int_0^{\frac{\pi}{2}} \sin^n x \, dx = I_n$$ as defined previously. Step 2: Use reduction formula same as for sine: $$I_n = \frac{n-1}{n} I_{n-2}$$. Step 3: For $$n=8$$: $$I_8 = \frac{7}{8} I_6 = \frac{7}{8} \times \frac{5\pi}{32} = \frac{35 \pi}{256}$$. Step 4: For $$n=7$$: $$I_7 = \frac{6}{7} I_5 = \frac{6}{7} \times \frac{8}{15} = \frac{48}{105} = \frac{16}{35}$$. --- 3. Problem: Evaluate $$\int_0^{\frac{\pi}{2}} \sin^5 \theta \cos^6 \theta \, d\theta$$. Step 1: Rewrite as $$\int_0^{\frac{\pi}{2}} (\sin^5 \theta)(\cos^6 \theta) \, d\theta$$. Step 2: Use Beta function identity: $$\int_0^{\frac{\pi}{2}} \sin^m \theta \cos^n \theta d\theta = \frac{1}{2} B\left(\frac{m+1}{2}, \frac{n+1}{2}\right)$$. Step 3: Here, $$m=5$$, $$n=6$$, so: $$= \frac{1}{2} B\left(3, \frac{7}{2}\right) = \frac{1}{2} \times \frac{\Gamma(3) \Gamma(\frac{7}{2})}{\Gamma(\frac{13}{2})}$$. Step 4: Compute Gamma values: $$\Gamma(3) = 2! = 2$$, $$\Gamma\left(\frac{7}{2}\right) = \frac{15 \sqrt{\pi}}{8}$$, $$\Gamma\left(\frac{13}{2}\right) = \frac{10395 \sqrt{\pi}}{64}$$. Step 5: Substitute and simplify: $$= \frac{1}{2} \times \frac{2 \times \frac{15 \sqrt{\pi}}{8}}{\frac{10395 \sqrt{\pi}}{64}} = \frac{1}{2} \times \frac{30/8}{10395/64} = \frac{1}{2} \times \frac{30 \times 64}{8 \times 10395} = \frac{1}{2} \times \frac{1920}{83160} = \frac{1920}{166320} = \frac{8}{693}$$. --- 4. Problem: Evaluate $$\int_0^{\frac{\pi}{2}} \sin^4 \theta \cos^2 \theta \, d\theta$$. Step 1: Using Beta function again, $$m=4, n=2$$, $$= \frac{1}{2} B\left(\frac{5}{2}, \frac{3}{2}\right) = \frac{1}{2} \frac{\Gamma(\frac{5}{2}) \Gamma(\frac{3}{2})}{\Gamma(4)}$$. Step 2: Calculate Gamma values: $$\Gamma\left(\frac{5}{2}\right) = \frac{3 \sqrt{\pi}}{4}$$, $$\Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}$$, $$\Gamma(4) = 3! = 6$$. Step 3: Substitute: $$= \frac{1}{2} \times \frac{\frac{3 \sqrt{\pi}}{4} \times \frac{\sqrt{\pi}}{2}}{6} = \frac{1}{2} \times \frac{3 \pi /8}{6} = \frac{3 \pi}{96} = \frac{\pi}{32}$$. --- 5. Problem: Evaluate $$\int_0^a \frac{x^7}{\sqrt{a^2 - x^2}} dx$$. Step 1: Substitute $$x = a \sin \theta$$ so $$dx = a \cos \theta d\theta$$. Step 2: Rewrite integral limits: when $$x=0, \theta=0$$ and when $$x=a, \theta=\frac{\pi}{2}$$. Step 3: Substitute in integral: $$ \int_0^{\frac{\pi}{2}} \frac{(a \sin \theta)^7}{\sqrt{a^2 - a^2 \sin^2 \theta}} a \cos \theta d\theta = a^8 \int_0^{\frac{\pi}{2}} \frac{\sin^7 \theta \cos \theta}{a \cos \theta} d\theta = a^7 \int_0^{\frac{\pi}{2}} \sin^7 \theta d\theta$$. Step 4: Use reduction formula for $$\int_0^{\frac{\pi}{2}} \sin^n x dx$$: For $$n=7$$, $$I_7 = \frac{6}{7} I_5$$, from step 1: $$I_5 = \frac{8}{15}$$, thus: $$I_7 = \frac{6}{7} \times \frac{8}{15} = \frac{48}{105} = \frac{16}{35}$$. Step 5: Final answer: $$= a^7 \times \frac{16}{35} = \frac{16 a^7}{35}$$. --- 6. Problem: Evaluate $$\int_0^\infty \frac{x^2}{(1 + x^2)^{7/2}} dx$$. Step 1: Substitute $$x = \tan \theta$$, $$dx = \sec^2 \theta d\theta$$, Limits from $$0$$ to $$\pi/2$$. Step 2: Transform integral: $$\int_0^{\pi/2} \frac{\tan^2 \theta \sec^2 \theta}{(1 + \tan^2 \theta)^{7/2}} d\theta = \int_0^{\pi/2} \frac{\tan^2 \theta \sec^2 \theta}{(\sec^2 \theta)^{7/2}} d\theta = \int_0^{\pi/2} \tan^2 \theta \sec^2 \theta \sec^{-7} \theta d\theta = \int_0^{\pi/2} \tan^2 \theta \sec^{-5} \theta d\theta$$. Step 3: Using $$\sec = \frac{1}{\cos}$$ and $$\tan = \frac{\sin}{\cos}$$, rewrite: $$\int_0^{\pi/2} \left(\frac{\sin^2 \theta}{\cos^2 \theta}\right) \cos^5 \theta d\theta = \int_0^{\pi/2} \sin^2 \theta \cos^3 \theta d\theta$$. Step 4: Use Beta function identity: $$\int_0^{\pi/2} \sin^m \theta \cos^n \theta d\theta = \frac{1}{2} B\left( \frac{m+1}{2}, \frac{n+1}{2} \right)$$. Here, $$m=2, n=3$$, $$= \frac{1}{2} B\left(\frac{3}{2}, 2\right) = \frac{1}{2} \frac{\Gamma(\frac{3}{2}) \Gamma(2)}{\Gamma(\frac{7}{2})}$$. Step 5: Values: $$\Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}, \quad \Gamma(2) = 1!, \quad \Gamma\left(\frac{7}{2}\right) = \frac{15 \sqrt{\pi}}{8}$$. Step 6: Substitute: $$= \frac{1}{2} \times \frac{\frac{\sqrt{\pi}}{2} \times 1}{\frac{15 \sqrt{\pi}}{8}} = \frac{1}{2} \times \frac{\frac{\sqrt{\pi}}{2}}{\frac{15 \sqrt{\pi}}{8}} = \frac{1}{2} \times \frac{8}{30} = \frac{4}{30} = \frac{2}{15}$$. --- 7. Problem: Evaluate $$\int_0^{2a} x^{5/2} \sqrt{2a - x} \, dx$$. Step 1: Use substitution $$x = 2a t$$, so $$dx = 2a dt$$. Step 2: Rewrite integral limits: $$x=0 \Rightarrow t=0$$, $$x=2a \Rightarrow t=1$$. Step 3: Integral becomes: $$ \int_0^1 (2a t)^{5/2} \sqrt{2a - 2a t} 2a dt = 2a \times (2a)^{5/2} \int_0^1 t^{5/2} \sqrt{2a(1 - t)} dt$$. Step 4: Take powers of $$2a$$ outside: $$= 2a \times (2a)^{5/2} \times (2a)^{1/2} \int_0^1 t^{5/2} (1-t)^{1/2} dt = 2a \times (2a)^3 \int_0^1 t^{5/2} (1-t)^{1/2} dt = 2a \times 8a^3 \int_0^1 t^{5/2} (1-t)^{1/2} dt = 16 a^4 \int_0^1 t^{5/2} (1-t)^{1/2} dt$$. Step 5: Recognize integral as Beta function: $$\int_0^1 t^p (1-t)^q dt = B(p+1, q+1) = \frac{\Gamma(p+1) \Gamma(q+1)}{\Gamma(p+q+2)}$$. Here, $$p=\frac{5}{2}, q=\frac{1}{2}$$. Step 6: Compute Beta function: $$= \frac{\Gamma(\frac{7}{2}) \Gamma(\frac{3}{2})}{\Gamma(5)} = \frac{\frac{15 \sqrt{\pi}}{8} \times \frac{\sqrt{\pi}}{2}}{24} = \frac{15 \pi / 16}{24} = \frac{15 \pi}{384}$$. Step 7: Multiply back: $$= 16 a^4 \times \frac{15 \pi}{384} = a^4 \times \frac{240 \pi}{384} = a^4 \times \frac{5 \pi}{8}$$. --- 8. Problem: Evaluate $$\int_0^\infty x^{9/2} e^{-x} dx$$. Step 1: This is a Gamma integral: $$\int_0^\infty x^{m} e^{-x} dx = \Gamma(m+1)$$. Step 2: Here $$m = \frac{9}{2} = 4.5$$, so $$= \Gamma\left(\frac{11}{2}\right)$$. Step 3: Use formula: $$\Gamma\left(n + \frac{1}{2}\right) = \frac{(2n)!}{4^n n!} \sqrt{\pi}$$ for integer $$n$$: Take $$n=5$$: $$\Gamma\left(5 + \frac{1}{2}\right) = \frac{10!}{4^5 5!} \sqrt{\pi}$$. Step 4: Calculate: $$10! = 3628800, \quad 4^5 = 1024, \quad 5! = 120$$, $$\Gamma\left(\frac{11}{2}\right) = \frac{3628800}{1024 \times 120} \sqrt{\pi} = \frac{3628800}{122880} \sqrt{\pi} = 29.53125 \sqrt{\pi}$$. --- 9. Problem: Evaluate $$\int_0^1 \frac{x}{\sqrt{\log_1 x}} dx$$. Note: The expression $$\log_1 x$$ is undefined (log base 1 is not valid). Therefore, the problem likely intends natural log or log base e. Assuming the problem means $$\int_0^1 \frac{x}{\sqrt{-\log x}} dx$$ where $$\log$$ is natural log and the negative is to ensure positivity under root. Step 1: Use substitution $$t = -\log x$$, then $$x = e^{-t}$$, $$dx = - e^{-t} dt$$. Step 2: Limits change from $$x=0 \Rightarrow t= \infty$$, to $$x=1 \Rightarrow t=0$$. Step 3: Substituting: $$ \int_\infty^0 \frac{e^{-t}}{\sqrt{t}} (-e^{-t}) dt = \int_0^\infty \frac{e^{-2t}}{\sqrt{t}} dt $$. Step 4: Recognize integral: $$\int_0^\infty t^{-\frac{1}{2}} e^{-2t} dt = \Gamma\left(\frac{1}{2}\right)/2^{1/2} = \frac{\sqrt{\pi}}{\sqrt{2}}$$. Step 5: Final answer: $$= \frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}$$. --- 10. Problem: Evaluate $$\int_0^\infty \frac{x^{9}}{9^x} dx$$. Step 1: Rewrite $$9^x = e^{x \log 9}$$. Step 2: Integral becomes: $$\int_0^\infty x^{9} e^{-x \log 9} dx = \frac{\Gamma(10)}{(\log 9)^{10}}$$. Step 3: Since $$\Gamma(10) = 9! = 362880$$. Step 4: Final answer: $$= \frac{362880}{(\log 9)^{10}}$$. --- 11. Problem: Evaluate $$\int_0^\infty e^{-\sqrt{x}} x^{1/4} dx$$. Step 1: Substitute $$t = \sqrt{x}$$, so $$x = t^2$$, $$dx = 2t dt$$. Step 2: Rewrite integral: $$\int_0^\infty e^{-t} (t^2)^{1/4} 2 t dt = 2 \int_0^\infty e^{-t} t^{\frac{1}{2}} t dt = 2 \int_0^\infty e^{-t} t^{\frac{3}{2}} dt$$. Step 3: Simplify: $$2 \int_0^\infty e^{-t} t^{\frac{3}{2}} dt = 2 \Gamma\left(\frac{5}{2}\right)$$. Step 4: Recall: $$\Gamma\left(\frac{5}{2}\right) = \frac{3 \sqrt{\pi}}{4}$$. Step 5: Final answer: $$= 2 \times \frac{3 \sqrt{\pi}}{4} = \frac{3 \sqrt{\pi}}{2}$$. --- 12. Problem: Evaluate $$\int_0^{\pi/2} \sqrt{\cot \theta} d\theta$$. Step 1: Write $$\sqrt{\cot \theta} = \frac{\sqrt{\cos \theta}}{\sqrt{\sin \theta}}$$. Step 2: Substitute $$t = \sin^2 \theta$$, $$dt = 2 \sin \theta \cos \theta d\theta$$. Step 3: Express in terms of $$t$$: After substitution, this integral reduces to Beta function: $$\int_0^{\pi/2} (\cot \theta)^{1/2} d\theta = \int_0^{\pi/2} \sin^{-1/2} \theta \cos^{1/2} \theta d\theta = \frac{1}{2} B\left(\frac{1}{4}, \frac{3}{4}\right)$$. Step 4: Use Beta function formula: $$B(p,q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}$$ and $$\Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{3}{4}\right) = \pi \sqrt{2}$$. Step 5: So, $$= \frac{1}{2} \times \frac{\pi \sqrt{2}}{\Gamma(1)} = \frac{\pi \sqrt{2}}{2}$$. --- 13. Problem: Evaluate $$\int_0^2 x (8 - x^3)^{1/3} dx$$. Step 1: Substitute $$t = 8 - x^3$$, then $$dt = -3 x^2 dx$$. Step 2: Express $$x dx$$ in terms of $$dt$$: Rewrite integral as: $$\int_0^2 x (8 - x^3)^{1/3} dx = \int_8^0 (t)^{1/3} \frac{dt}{-3 x}$$. This is complicated; try another substitution: Step 3: Set $$u = x^3$$, so $$x = u^{1/3}$$, $$dx = \frac{1}{3} u^{-2/3} du$$. Step 4: Then $$x (8 - x^3)^{1/3} dx = u^{1/3} (8 - u)^{1/3} \frac{1}{3} u^{-2/3} du = \frac{1}{3} u^{-1/3} (8 - u)^{1/3} du$$. Step 5: Limits change: $$u(0) = 0$$, $$u(2) = 8$$. Step 6: Integral becomes: $$\frac{1}{3} \int_0^8 u^{-1/3} (8 - u)^{1/3} du$$. Step 7: Use Beta function form: $$\int_0^a x^{m} (a - x)^n dx = a^{m+n+1} B(m+1, n+1)$$. Here, $$m = -\frac{1}{3}, n= \frac{1}{3}, a=8$$. Step 8: $$= \frac{1}{3} \times 8^{\frac{-1}{3} + \frac{1}{3} + 1} B\left(\frac{2}{3}, \frac{4}{3}\right) = \frac{1}{3} \times 8^{1} B\left(\frac{2}{3}, \frac{4}{3}\right) = \frac{8}{3} B\left(\frac{2}{3}, \frac{4}{3}\right)$$. Step 9: Beta function: $$= \frac{8}{3} \times \frac{\Gamma(\frac{2}{3}) \Gamma(\frac{4}{3})}{\Gamma(2)} = \frac{8}{3} \times \Gamma\left(\frac{2}{3}\right) \Gamma\left(\frac{4}{3}\right)$$. Step 10: Use $$\Gamma(z+1) = z \Gamma(z)$$, \( \Gamma\left(\frac{4}{3}\right) = \frac{1}{3} \Gamma\left(\frac{1}{3}\right) \) . Hence, $$= \frac{8}{3} \times \Gamma\left(\frac{2}{3}\right) \times \frac{1}{3} \Gamma\left(\frac{1}{3}\right) = \frac{8}{9} \Gamma\left(\frac{1}{3}\right) \Gamma\left(\frac{2}{3}\right)$$. Step 11: Recall identity: $$\Gamma\left(\frac{1}{3}\right) \Gamma\left(\frac{2}{3}\right) = \frac{\pi}{\sin \left( \frac{\pi}{3} \right)} = \frac{\pi}{\frac{\sqrt{3}}{2}} = \frac{2 \pi}{\sqrt{3}}$$. Step 12: Final answer: $$= \frac{8}{9} \times \frac{2 \pi}{\sqrt{3}} = \frac{16 \pi}{9 \sqrt{3}}$$. --- 14. Problem: Evaluate $$\int_0^1 x^3 (1 - \sqrt{x})^5 dx$$. Step 1: Substitute $$t = \sqrt{x}$$, then $$x = t^2$$, $$dx = 2 t dt$$. Step 2: Transform integral: $$\int_0^1 (t^2)^3 (1 - t)^5 2 t dt = 2 \int_0^1 t^{6} (1-t)^5 t dt = 2 \int_0^1 t^{7} (1-t)^5 dt$$. Step 3: Beta function: $$2 B(8,6) = 2 \frac{\Gamma(8) \Gamma(6)}{\Gamma(14)}$$. Step 4: Use factorial values: $$\Gamma(n) = (n-1)!$$, so, $$\Gamma(8) = 7! = 5040, \quad \Gamma(6) = 5! = 120, \quad \Gamma(14) = 13! = 6227020800$$. Step 5: Calculate: $$= 2 \times \frac{5040 \times 120}{6227020800} = 2 \times \frac{604800}{6227020800} = 2 \times \frac{1}{10.29...} = \frac{1}{5.15...} \approx 0.194$$. Exact fraction: $$= \frac{2 \times 604800}{6227020800} = \frac{1209600}{6227020800} = \frac{1}{5148}$$ (approximation). To simplify: $$\frac{1209600}{6227020800} = \frac{1}{5148}$$ approximately. --- 15. Problem: Evaluate $$\int_0^\infty \frac{1}{1 + x^4} dx$$. Step 1: Known integral: $$\int_0^\infty \frac{dx}{1 + x^4} = \frac{\pi}{2 \sqrt{2}}$$. --- 16. Problem: Evaluate $$\int_0^1 \frac{dx}{\sqrt{1 - x^3}}$$. Step 1: Substitute $$t = x^3$$, $$x = t^{1/3}$$, $$dx = \frac{1}{3} t^{-2/3} dt$$. Step 2: The integral becomes: $$\int_0^1 \frac{\frac{1}{3} t^{-2/3} dt}{\sqrt{1 - t}} = \frac{1}{3} \int_0^1 t^{-2/3} (1 - t)^{-1/2} dt$$. Step 3: Beta function form: $$= \frac{1}{3} B\left(1 - \frac{2}{3}, \frac{1}{2}\right) = \frac{1}{3} B\left(\frac{1}{3}, \frac{1}{2}\right) = \frac{1}{3} \frac{\Gamma(\frac{1}{3}) \Gamma(\frac{1}{2})}{\Gamma(\frac{5}{6})}$$. Step 4: Values: $$\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$$. Exact closed form is complex but this is the solution format. --- 17. Problem: Find area between $$y = x^2$$ and $$y = -x^2 + 1$$ in $$[-1,1]$$. Step 1: Find intersection points: Set $$x^2 = -x^2 + 1 \Rightarrow 2 x^2 = 1 \Rightarrow x^2 = \frac{1}{2}$$. Since $$-1 \leq x \leq 1$$, they intersect inside the interval. Step 2: Area between curves: $$\int_{-1}^1 \bigl((-x^2 + 1) - x^2\bigr) dx = \int_{-1}^1 (1 - 2 x^2) dx$$. Step 3: Calculate integral: $$= \left[ x - \frac{2 x^3}{3} \right]_{-1}^1 = (1 - \frac{2}{3}) - (-1 + \frac{2}{3}) = \left(1 - \frac{2}{3}\right) + \left(1 - \frac{2}{3}\right) = 2 \times \frac{1}{3} = \frac{2}{3}$$. --- Final Answers Summary: 1. $$\int_0^{\frac{\pi}{2}} \sin^6 x dx = \frac{5 \pi}{32}$$, $$\int_0^{\frac{\pi}{2}} \sin^5 x dx = \frac{8}{15}$$. 2. $$\int_0^{\frac{\pi}{2}} \cos^8 x dx = \frac{35 \pi}{256}$$, $$\int_0^{\frac{\pi}{2}} \cos^7 x dx = \frac{16}{35}$$. 3. $$\int_0^{\frac{\pi}{2}} \sin^5 \theta \cos^6 \theta d\theta = \frac{8}{693}$$. 4. $$\int_0^{\frac{\pi}{2}} \sin^4 \theta \cos^2 \theta d\theta = \frac{\pi}{32}$$. 5. $$\int_0^a \frac{x^7}{\sqrt{a^2 - x^2}} dx = \frac{16 a^7}{35}$$. 6. $$\int_0^\infty \frac{x^2}{(1+x^2)^{7/2}} dx = \frac{2}{15}$$. 7. $$\int_0^{2a} x^{5/2} \sqrt{2a-x} dx = \frac{5 \pi a^4}{8}$$. 8. $$\int_0^\infty x^{9/2} e^{-x} dx = \Gamma\left(\frac{11}{2}\right) = 29.53125 \sqrt{\pi}$$ (exact value using Gamma). 9. Assuming $$\log_1 x$$ means natural log, $$\int_0^1 \frac{x}{\sqrt{-\log x}} dx = \sqrt{\frac{\pi}{2}}$$. 10. $$\int_0^\infty \frac{x^9}{9^x} dx = \frac{362880}{(\log 9)^{10}}$$. 11. $$\int_0^\infty e^{-\sqrt{x}} x^{1/4} dx = \frac{3 \sqrt{\pi}}{2}$$. 12. $$\int_0^{\pi/2} \sqrt{\cot \theta} d\theta = \frac{\pi \sqrt{2}}{2}$$. 13. $$\int_0^2 x (8 - x^3)^{1/3} dx = \frac{16 \pi}{9 \sqrt{3}}$$. 14. $$\int_0^1 x^3 (1 - \sqrt{x})^5 dx = \frac{1209600}{6227020800} = \frac{1}{5148}$$ approx. 15. $$\int_0^\infty \frac{1}{1+x^4} dx = \frac{\pi}{2 \sqrt{2}}$$. 16. $$\int_0^1 \frac{dx}{\sqrt{1 - x^3}} = \frac{1}{3} \frac{\Gamma\left(\frac{1}{3}\right) \sqrt{\pi}}{\Gamma\left(\frac{5}{6}\right)}$$. 17. Area between $$y=x^2$$ and $$y=-x^2 + 1$$ on $$[-1,1]$$ is $$\frac{2}{3}$$.