Hello Anh Nguyen,
Before we work the given problems, let's develop two formulas we can use.
First, let's consider:
$$I=\int e^{x}\sin(x)\,dx$$
Let's use integration by parts, where:
$$u=\sin(x)\,\therefore\,du=\cos(x)\,dx$$
$$dv=e^{x}\,dx\,\therefore\,v=e^x$$
Hence:
$$I=\sin(x)e^{x}-\int e^{x}\cos(x)\,dx$$
Let's use integration by parts again, where:
$$u=\cos(x)\,\therefore\,du=-\sin(x)\,dx$$
$$dv=e^{x}\,dx\,\therefore\,v=e^x$$
Hence:
$$I=\sin(x)e^{x}-\left(\cos(x)e^{x}+\int e^{x}\sin(x)\,dx \right)$$
$$I=e^{x}\left(\sin(x)-\cos(x) \right)-I$$
Add $I$ to both sides:
$$2I=e^{x}\left(\sin(x)-\cos(x) \right)$$
Divide through by $2$ and append the constant of integration:
$$I=\frac{e^{x}}{2}\left(\sin(x)-\cos(x) \right)+C$$
Thus, we have:
(1) $$\int e^{x}\sin(x)\,dx=\frac{e^{x}}{2}\left(\sin(x)-\cos(x) \right)+C$$
Next, let's consider:
$$I=\int e^{x}\cos(x)\,dx$$
Above, we found:
$$I=\cos(x)e^{x}+\int e^{x}\sin(x)\,dx$$
Using (1), we may state:
$$I=\cos(x)e^{x}+\frac{e^{x}}{2}\left(\sin(x)-\cos(x) \right)+C$$
$$I=\frac{e^{x}}{2}\left(\sin(x)+\cos(x) \right)+C$$
Hence, we may state:
(2) $$\int e^{x}\cos(x)\,dx=\frac{e^{x}}{2}\left(\sin(x)+\cos(x) \right)+C$$
Now, armed with these two formulas, let's work the given problems.
1.) We are given to evaluate:
$$I=\int -6e^{2t}\sin(t)\cos(t)\,dt$$
Using the double-angle identity for sine, we may write:
$$I=-3\int e^{2t}\sin(2t)\,dt$$
Using the substitution $$w=2t\,\therefore\,dw-2\,dt$$ we have:
$$I=-\frac{3}{2}\int e^{w}\sin(w)\,dw$$
Using (1), we obtain:
$$I=-\frac{3}{2}\left(\frac{e^{w}}{2}\left(\sin(w)-\cos(w) \right) \right)+C$$
$$I=\frac{3}{4}e^{w}\left(\cos(w)-\sin(w) \right)+C$$
Back-substituting for $w$ and $I$, we have:
$$\int -6e^{2t}\sin(t)\cos(t)\,dt=\frac{3}{4}e^{2t}\left(\cos(2t)-\sin(2t) \right)+C$$
2.) We are given to evaluate:
$$I=\int 5e^{2t}\sin^2(t)\,dt$$
Using a power-reduction identity for sine, we may write:
$$I=\frac{5}{2}\int e^{2t}\left(1-\cos(2t) \right)\,dt$$
Using the substitution $$w=2t\,\therefore\,dw-2\,dt$$ we have:
$$I=\frac{5}{4}\int e^{w}\left(1-\cos(w) \right)\,dw$$
$$I=\frac{5}{4}\left(\int e^{w}\,dw-\int e^{w}\cos(w)\,dw \right)$$
Using (2), we may state:
$$I=\frac{5}{4}\left(e^{w}-\frac{e^{w}}{2}\left(\sin(w)+\cos(w) \right) \right)+C$$
$$I=\frac{5}{8}e^{w}\left(2-\sin(w)-\cos(w) \right)+C$$
Back-substituting for $w$ and $I$, we have:
$$\int 5e^{2t}\sin^2(t)\,dt=\frac{5}{8}e^{2t}\left(2-\sin(2t)-\cos(2t) \right)+C$$
3.) We are given to evaluate:
$$I=\int -e^{2t}\cos^2(t)\,dt=\int e^{2t}\left(\sin^2(t)-1 \right)\,dt=\int e^{2t}\sin^2(t)\,dt-\int e^{2t}\,dt$$
Using the result of problem 2.) we may write:
$$I=\frac{1}{8}e^{2t}\left(2-\sin(2t)-\cos(2t) \right)-\frac{1}{2}e^{2t}+C$$
And so we conclude:
$$\int -e^{2t}\cos^2(t)\,dt=-\frac{1}{8}e^{2t}\left(2+\sin(2t)+\cos(2t) \right)+C$$