Integration by Parts substitution

  • Thread starter revolve
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  • #1
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Homework Statement



[tex]
\int\arctan(4t)dt
[/tex]

Homework Equations





The Attempt at a Solution



[tex]
\int\arctan(4t)dt = t\arctan(4t) -4 \int \frac{t}{1+16t^2}dt
[/tex]

I'm stuck at this point. I think I need to make a substitution for the denominator, but I'm not sure how to go about doing so.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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31
[tex]\int \frac{t}{1+16t^2}dt[/tex]


f we let u=1+16t2 what is du equal to?
 
  • #3
26
0

Homework Statement



[tex]
\int\arctan(4t)dt
[/tex]

Homework Equations





The Attempt at a Solution



[tex]
\int\arctan(4t)dt = t\arctan(4t) -4 \int \frac{t}{1+16t^2}dt
[/tex]

I'm stuck at this point. I think I need to make a substitution for the denominator, but I'm not sure how to go about doing so.

It's good so far.

You can substitute for [tex] V = {1+16t^2} [/tex], so [tex] dV = 32tdt[/tex].

Then, you'll have [tex] \frac{1}{32} \int \frac{1}{V}dV[/tex].

That equates to [tex] \frac {ln(|V|)}{32}[/tex].

Multiply by -4, to obtain [tex] \frac {-ln(|V|)}{8}[/tex].

Replace V and you're done.

Good luck,

Marc.
 
  • #4
19
0
Got it. Thank you both! Very helpful.
 

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