Anh Nguyen's questions regarding indefinite integrals (integration by parts)

[MarkFL]

Here are the questions:

I need help to solve following integrals problems?


How to solve these problems:

1) integral of -6e^2tsintcost dt

2) integral of 5e^2tsin^2t dt

3) integral of -e^2tcos^2t dt

I have posted a link there to this thread so the OP can see my work.

 

[MarkFL]

Hello Anh Nguyen,

Before we work the given problems, let's develop two formulas we can use.

First, let's consider:

\(\displaystyle I=\int e^{x}\sin(x)\,dx\)

Let's use integration by parts, where:

\(\displaystyle u=\sin(x)\,\therefore\,du=\cos(x)\,dx\)

\(\displaystyle dv=e^{x}\,dx\,\therefore\,v=e^x\)

Hence:

\(\displaystyle I=\sin(x)e^{x}-\int e^{x}\cos(x)\,dx\)

Let's use integration by parts again, where:

\(\displaystyle u=\cos(x)\,\therefore\,du=-\sin(x)\,dx\)

\(\displaystyle dv=e^{x}\,dx\,\therefore\,v=e^x\)

Hence:

\(\displaystyle I=\sin(x)e^{x}-\left(\cos(x)e^{x}+\int e^{x}\sin(x)\,dx \right)\)

\(\displaystyle I=e^{x}\left(\sin(x)-\cos(x) \right)-I\)

Add $I$ to both sides:

\(\displaystyle 2I=e^{x}\left(\sin(x)-\cos(x) \right)\)

Divide through by $2$ and append the constant of integration:

\(\displaystyle I=\frac{e^{x}}{2}\left(\sin(x)-\cos(x) \right)+C\)

Thus, we have:

(1) \(\displaystyle \int e^{x}\sin(x)\,dx=\frac{e^{x}}{2}\left(\sin(x)-\cos(x) \right)+C\)

Next, let's consider:

\(\displaystyle I=\int e^{x}\cos(x)\,dx\)

Above, we found:

\(\displaystyle I=\cos(x)e^{x}+\int e^{x}\sin(x)\,dx\)

Using (1), we may state:

\(\displaystyle I=\cos(x)e^{x}+\frac{e^{x}}{2}\left(\sin(x)-\cos(x) \right)+C\)

\(\displaystyle I=\frac{e^{x}}{2}\left(\sin(x)+\cos(x) \right)+C\)

Hence, we may state:

(2) \(\displaystyle \int e^{x}\cos(x)\,dx=\frac{e^{x}}{2}\left(\sin(x)+\cos(x) \right)+C\)

Now, armed with these two formulas, let's work the given problems.

1.) We are given to evaluate:

\(\displaystyle I=\int -6e^{2t}\sin(t)\cos(t)\,dt\)

Using the double-angle identity for sine, we may write:

\(\displaystyle I=-3\int e^{2t}\sin(2t)\,dt\)

Using the substitution \(\displaystyle w=2t\,\therefore\,dw-2\,dt\) we have:

\(\displaystyle I=-\frac{3}{2}\int e^{w}\sin(w)\,dw\)

Using (1), we obtain:

\(\displaystyle I=-\frac{3}{2}\left(\frac{e^{w}}{2}\left(\sin(w)-\cos(w) \right) \right)+C\)

\(\displaystyle I=\frac{3}{4}e^{w}\left(\cos(w)-\sin(w) \right)+C\)

Back-substituting for $w$ and $I$, we have:

\(\displaystyle \int -6e^{2t}\sin(t)\cos(t)\,dt=\frac{3}{4}e^{2t}\left(\cos(2t)-\sin(2t) \right)+C\)

2.) We are given to evaluate:

\(\displaystyle I=\int 5e^{2t}\sin^2(t)\,dt\)

Using a power-reduction identity for sine, we may write:

\(\displaystyle I=\frac{5}{2}\int e^{2t}\left(1-\cos(2t) \right)\,dt\)

Using the substitution \(\displaystyle w=2t\,\therefore\,dw-2\,dt\) we have:

\(\displaystyle I=\frac{5}{4}\int e^{w}\left(1-\cos(w) \right)\,dw\)

\(\displaystyle I=\frac{5}{4}\left(\int e^{w}\,dw-\int e^{w}\cos(w)\,dw \right)\)

Using (2), we may state:

\(\displaystyle I=\frac{5}{4}\left(e^{w}-\frac{e^{w}}{2}\left(\sin(w)+\cos(w) \right) \right)+C\)

\(\displaystyle I=\frac{5}{8}e^{w}\left(2-\sin(w)-\cos(w) \right)+C\)

Back-substituting for $w$ and $I$, we have:

\(\displaystyle \int 5e^{2t}\sin^2(t)\,dt=\frac{5}{8}e^{2t}\left(2-\sin(2t)-\cos(2t) \right)+C\)

3.) We are given to evaluate:

\(\displaystyle I=\int -e^{2t}\cos^2(t)\,dt=\int e^{2t}\left(\sin^2(t)-1 \right)\,dt=\int e^{2t}\sin^2(t)\,dt-\int e^{2t}\,dt\)

Using the result of problem 2.) we may write:

\(\displaystyle I=\frac{1}{8}e^{2t}\left(2-\sin(2t)-\cos(2t) \right)-\frac{1}{2}e^{2t}+C\)

And so we conclude:

\(\displaystyle \int -e^{2t}\cos^2(t)\,dt=-\frac{1}{8}e^{2t}\left(2+\sin(2t)+\cos(2t) \right)+C\)