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

MarkFL

Staff member
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

Staff member
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:

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