Integration by parts

[Johnson]

Hello. I need help with a few questions.

1) Integral [x^3 * e^(x^2) dx]

I know that the rule for integration by parts is uv-integral[u'v], but how do I know which to choose for u and which to choose for v? I tried both but I can't seem to get the answer.

2) Solve the definite integral of arctanx dx, using 0 and 1 as the limits of integration. I have no idea what the integral of arctan is, and it seems like mathematica couldn't give me anything either.

3) Integral [ (sinx)^2 dx] using integration by parts and compare it to the form you get from integral [(1-cos2x)/2 dx]. I did the second part and got x/2 +
sin(2x)/4 + c, but I can't do the first part.[/tex]

 

[galactus]

Do you have to use parts for #1. It's better to use a u-substitution

\(\displaystyle \L\\\int{x^{3}e^{x^{2}}}dx\)

Let \(\displaystyle \L\\u=e^{x^{2}};\;\ x^{2}=ln(u);\;\ du=2xe^{x^{2}}dx;\;\ \frac{du}{2}=xe^{x^{2}}dx\)

Make the substitutions:

\(\displaystyle \L\\\frac{1}{2}\int{ln(u)}du\)

\(\displaystyle \L\\\frac{1}{2}[uln(u)-u]\)

Resub:

\(\displaystyle \L\\\frac{1}{2}[e^{x^{2}}ln(e^{x^{2}})-e^{x^{2}}]\)

\(\displaystyle \L\\\frac{e^{x^{2}}}{2}[x^{2}-1]\)


For #2:

Use parts:

Let \(\displaystyle \L\\u=tan^{-1}(x);\;\ dv=dx;\;\ du=\frac{1}{x^{2}+1}dx;\;\ v=x\)

\(\displaystyle \L\\xtan^{-1}(x)-\int\frac{x}{x^{2}+1}dx\)

Now, make the substitution \(\displaystyle \L\\u=x^{2}+1;\;\ du=2xdx;\;\ \frac{du}{2}=xdx\)

 

[soroban]

Hello, Johnson!

#3 is interesting . . .

\(\displaystyle 3)\;\L\int \sin^2x\,dx\;\) using integration by parts
and compare it to the form you get from: \(\displaystyle \,\L\int\frac{1\,-\,\cos2x}{2}\,dx\)

Let \(\displaystyle \L\,I\;=\;\int\sin^2x\,dx\)

Let: \(\displaystyle \;u \,=\,\sin x\;\;\;\;\;dv\,=\,\sin x\,dx\)

Then: \(\displaystyle \,du\,=\,\cos x\,dx\;\;\;v\,=\,-\cos x\)


We have: \(\displaystyle \L\,I\;=\;-\sin x\cos x \,-\,\int(-\cos x)(\cos x\,dx)\)

. . . . . . . \(\displaystyle \L I\;=\;-\sin x\cos x \,+\,\int\cos^2x\,dx\)

. . . . . . . \(\displaystyle \L I\;=\;-\sin x\cos x \,+\,\int\left(1 - \sin^2x\right)\,dx\)

. . . . . . . \(\displaystyle \L I \;= \;-\sin x\cos x \,+\,\int dx \,-\,\underbrace{\int\sin^2x\,dx}\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . This is \(\displaystyle I\)

So we have: \(\displaystyle \L\,I \;= \;-\sin x\cos x\,+\,x\,-\,I\,+\,C\)

Hence: \(\displaystyle \L\,2I\;=\;-\sin x\cos x\,+\,x\,+\,C\)

And we have: \(\displaystyle \L\,I\;=\;\frac{1}{2}\left(x\,-\,\sin x\cos x\right)\,+\,C\)


Therefore: \(\displaystyle \L\,\int\sin^2x\,dx\;=\;\frac{1}{2}(x\,-\,\sin x\cos x)\,+\,C\)

 

[soroban]

Johnson!

With #1, the choice should be obvious (well, almost) . . .

\(\displaystyle \L 1)\:\int x^3e^{x^2}\,dx\)

If we let \(\displaystyle u \,= \,e^{x^2}\), then \(\displaystyle du\,=\,2x\,e^{x^2}\,dx\)

The derivative is "worse"!
\(\displaystyle \;\;\)We want to pick \(\displaystyle u\) so the derivative is "simpler".

If we let: \(\displaystyle \,u\,=\,x^3\) and \(\displaystyle dv = e^{x^2}\,dx\)
then: \(\displaystyle \,du\,=\,3x^2\,dx\;\;\;v\,=\,\int e^{x^2}dx\,=\,?\)
\(\displaystyle \;\;\)and we find that we can't integrate \(\displaystyle e^{x^2}\,dx\)

But we can integrate: \(\displaystyle \,x\cdot e^{x^2}\,dx\)

So there are the assignments:
\(\displaystyle \;\;\)Let: \(\displaystyle \,u\,=\,x^2\;\;\;\;dv\,=\,x\cdot e^{x^2}\,dx\)
\(\displaystyle \;\)Then: \(\displaystyle du = 2x\,dx\;\;\;v\,=\,\frac{1}{2}e^{x^2}\)

And we have: \(\displaystyle \L\,\frac{1}{2}x^2e^{x^2} \,-\,\int\left(\frac{1}{2}e^{x^2}\right)\left(2x\,dx\right) \;= \;\frac{1}{2}x^2e^{x^2} \,-\,\int x\cdot e^{x^2}\,dx\)

And you can finish this now . . . right?

 

[skeeter]

\(\displaystyle \L \int x^3e^{x^2} dx\)

\(\displaystyle \L t = x^2\)

\(\displaystyle \L dt = 2x dx\)

\(\displaystyle \L \frac{1}{2} \int (2x)x^2 e^{x^2} dx\)

\(\displaystyle \L \frac{1}{2} \int t e^t dt\)

use tabular integration ...

+ t ...... e^t
- 1 ...... e^t
0 ....... e^t

\(\displaystyle \L \frac{1}{2} \int t e^t dt = \frac{1}{2}[t e^t - e^t] + C = \frac{e^t}{2}[t - 1] + C\)

back substitute ...

\(\displaystyle \L \frac{e^{x^2}}{2}[x^2 - 1] + C\)

 

[Johnson]

Thanks very much, everyone.