[SPLIT] int [e^(ax)sin(bx)] dx, etc.

[warwick]

21. integral of e^(ax)sin(bx)dx

= [(-e^(ax)cos(bx))/b] + [(ae^(ax)sin(bx))/b^2] - [a^2/b^2] integral of e^(ax)sin(bx)dx

I'm getting stuck here. I'm also getting suck on these after the first Parts substitution.

35. integral of [sec (theta^(1/2)] ^(-1) d theta from 2 to 4

39. integral of x^(1/2) [tan (x^(1/2)] ^(-1) dx from 1 to 2

41. Integrate by u-substitution and then by parts.

a) integral of e^(x^1/2) dx b) integral of cos (x^1/2) dx

 

[galactus]

Since you've appear to be struggling with this one, I'll go ahead and help with #1.

Let \(\displaystyle \L\\u=e^{ax}, \;\ dv=sin(bx)dx, \;\ du=ae^{ax}dx, \;\ v=\frac{-1}{b}cos(bx)\)

\(\displaystyle \L\\\int{e^{ax}sin(bx)}dx\)

\(\displaystyle \L\\\frac{-1}{b}e^{ax}cos(bx)+\frac{a}{b}\int{e^{ax}cos(bx)}dx\)

Now, let \(\displaystyle \L\\u=e^{ax}, \;\ dv=cos(bx)dx\) and get:

\(\displaystyle \L\\\int{e^{ax}cos(bx)}dx=\frac{1}{b}e^{ax}sin(bx)-\frac{a}{b}\int{e^{ax}sin(bx)}dx\)

So,

\(\displaystyle \L\\\int{e^{ax}sin(bx)}dx=\frac{-1}{b}e^{ax}cos(bx)+\frac{a}{b^{2}}e^{ax}sin(bx)-\frac{a^{2}}{b^{2}}\int{e^{ax}sin(bx)}dx\)

\(\displaystyle \L\\\int{e^{ax}sin(bx)}dx=\fbox{\frac{e^{ax}}{a^{2}+b^{2}}(a\cdot{sin(bx)}-b\cdot{cos(bx)})+C}\)


For 41a:

\(\displaystyle \L\\\int{e^{\sqrt{x}}}dx\)

Try this:

Let \(\displaystyle \L\\z=\sqrt{x}, \;\ x=z^{2}, \;\ dx=2zdz\)

\(\displaystyle \L\\2\int{ze^{z}}dz\)

Now, use parts:

Let \(\displaystyle \L\\u=z, \;\ dv=e^{z}dz, \;\ du=dz, \;\ v=e^{z}\)

Now, finish up?.

 

[warwick]

Since you've appear to be struggling with this one, I'll go ahead and help with #1.

Let \(\displaystyle \L\\u=e^{ax}, \;\ dv=sin(bx)dx, \;\ du=ae^{ax}dx, \;\ v=\frac{-1}{b}cos(bx)\)

\(\displaystyle \L\\\int{e^{ax}sin(bx)}dx\)

\(\displaystyle \L\\\frac{-1}{b}e^{ax}cos(bx)+\frac{a}{b}\int{e^{ax}cos(bx)}dx\)

Now, let \(\displaystyle \L\\u=e^{ax}, \;\ dv=cos(bx)dx\) and get:

\(\displaystyle \L\\\int{e^{ax}cos(bx)}dx=\frac{1}{b}e^{ax}sin(bx)-\frac{a}{b}\int{e^{ax}sin(bx)}dx\)

So,

\(\displaystyle \L\\\int{e^{ax}sin(bx)}dx=\frac{-1}{b}e^{ax}cos(bx)+\frac{a}{b^{2}}e^{ax}sin(bx)-\frac{a^{2}}{b^{2}}\int{e^{ax}sin(bx)}dx\)

\(\displaystyle \L\\\int{e^{ax}sin(bx)}dx=\fbox{\frac{e^{ax}}{a^{2}+b^{2}}(a\cdot{sin(bx)}-b\cdot{cos(bx)})+C}\)

Thanks. I see what I did differently. Here, I left the a and b in the integrand, so that made my u's and v's not very pretty. I'll rework it myself now.

 

[tkhunny]

This falls under my rule #2.

 

[warwick]

This falls under my rule #2.

For which one?

Ok, guys, I worked out finished up #21 and #41. This is one of my longest problem sets for the class. Looking at the syllabus, I have a few more sets as long as this one. Now, I just have 35 and 39 and the last two of the set. I'll probably ask the professor to do the last two in class tomorrow.