(9) $\\int\\cos 5xdx$ (10) $\\int\\sqrt{3x + 5}dx$

Explanation:

Step1: Solve $\int\cos5xdx$ using substitution

Let $u = 5x$, then $du=5dx$, and $dx=\frac{1}{5}du$. $$\int\cos5xdx=\frac{1}{5}\int\cos udu$$

Step2: Integrate $\cos u$

Since $\int\cos udu=\sin u + C$ (where $C$ is the constant of integration), substituting back $u = 5x$ gives: $$\frac{1}{5}\int\cos udu=\frac{1}{5}\sin u + C=\frac{1}{5}\sin5x + C$$

Step3: Solve $\int\sqrt{3x + 5}dx$ using substitution

Let $t=3x + 5$, then $dt = 3dx$, and $dx=\frac{1}{3}dt$. $$\int\sqrt{3x + 5}dx=\frac{1}{3}\int\sqrt{t}dt$$

Step4: Integrate $\sqrt{t}$

Since $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n=\frac{1}{2}$, we have $\int\sqrt{t}dt=\int t^{\frac{1}{2}}dt=\frac{2}{3}t^{\frac{3}{2}}+C$. Substituting back $t = 3x + 5$ gives: $$\frac{1}{3}\int\sqrt{t}dt=\frac{1}{3}\times\frac{2}{3}t^{\frac{3}{2}}+C=\frac{2}{9}(3x + 5)^{\frac{3}{2}}+C$$

Answer:

(9) $\frac{1}{5}\sin5x + C$; (10) $\frac{2}{9}(3x + 5)^{\frac{3}{2}}+C$