Integration by Parts substitution

In summary, the conversation discusses solving the integral \int\arctan(4t)dt by using a substitution for the denominator. The substitution V = {1+16t^2} and its corresponding derivative are mentioned, and ultimately, the solution is found to be \frac {-ln(|V|)}{8}.
  • #1
revolve
19
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.
 
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  • #2
[tex]\int \frac{t}{1+16t^2}dt[/tex]


f we let u=1+16t2 what is du equal to?
 
  • #3
revolve said:

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
Got it. Thank you both! Very helpful.
 

Related to Integration by Parts substitution

What is Integration by Parts substitution?

Integration by Parts substitution is a technique used in calculus to evaluate integrals that involve products of functions. It involves choosing one function to differentiate and another function to integrate.

When is Integration by Parts substitution used?

Integration by Parts substitution is used when the integrand (the function being integrated) is a product of two functions and cannot be easily evaluated using other techniques like substitution or the power rule.

How does Integration by Parts substitution work?

This technique uses the formula ∫u*dv = uv - ∫v*du, where u and v are the chosen functions. The goal is to simplify the integral on the right side until it can be evaluated using basic integration rules.

What are some tips for choosing u and dv in Integration by Parts substitution?

When choosing u and dv, it is helpful to choose u as the function that becomes simpler when differentiated and dv as the function that becomes simpler when integrated. Common choices for u include logarithmic, inverse trigonometric, and algebraic functions.

Can Integration by Parts substitution be used for definite integrals?

Yes, Integration by Parts substitution can be used for definite integrals by applying the formula for the indefinite integral and then evaluating the limits of integration at the end.

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