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(a) Although a quadratic under a square root often suggests trigonometric substitution, in this case we have an easier substitution. Specifically, let $y=9+x^2$. Then $\mathrm{d}y=2x\mathrm{d}x,\ x\mathrm{d}x=\frac{\mathrm{d}y}{2},\ y(0)=9,$ and $y(4)=25$.

$$\begin{aligned} \int_0^4\frac{x}{\sqrt{9+x^2}}\,\mathrm{d}x &=\int_9^{25}\frac{1}{\sqrt{y}}\,\frac{\mathrm{d}y}{2} =\frac{1}{2}\cdot\frac{\sqrt{y}}{1/2}\Big|_9^{25} =5-3= 2 \end{aligned}$$

(b) The power of cosine is odd, so we can reserve one cosine for the differential and change the rest to sines. Substituting $y=\sin x,$ $\mathrm{d}y=\cos x$, $\mathrm{d}x,$ $y(0)=0,$ $y(\pi/2)=1,$ $\cos^2x=1-y^2$:

$$\begin{aligned} \int_0^{\pi/2}\cos^3x\ \sin^2x\,\mathrm{d}x &=\int_0^{\pi/2}\cos^2x\ \sin^2x\ \cos x\,\mathrm{d}x =\int_0^1\!(1-y^2)y^2\,\mathrm{d}y =\int_0^1(y^2-y^4)\,\mathrm{d}y \\ &=\bigg[\frac{y^3}{3}-\frac{y^5}{5}\bigg]_0^1 =\frac{1}{3}-\frac{1}{5} \\ &= \frac{2}{15} \end{aligned}$$

(c) The integrand is the product of two different kinds of functions, with no obvious substitution or simplification. If we differentiate $\log x$, it will match better with the polynomial nature of the rest of the integrand. So, integrate by parts with $u(x)=\log x$ and $\mathrm{d}v=x^3\,\mathrm{d}x$, then $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$ and $v=x^4/4$.

$$\begin{aligned} \int_1^{e}x^3\log x\,\mathrm{d}x &=\frac{x^4}{4}\log x\bigg|_1^e -\int_1^{e}\frac{x^4}{4}\cdot\frac{1}{x}\,\mathrm{d}x =\frac{e^4}{4}-\int_1^{e}\frac{x^3}{4}\,\mathrm{d}x =\frac{e^4}{4}-\frac{x^4}{16}\bigg|_1^e \\ &=\frac{3e^4}{16}+\frac{1}{16} \end{aligned}$$