Integration by parts computation

In summary, the conversation discusses the computation of an integral and its asymptotic expansion as x approaches infinity. The process involves using integration by parts and substituting variables to find the solution.
  • #1
the_kid
116
0

Homework Statement


Consider the following integral:

I=[itex]\int^{\pi/4}_{0}[/itex]cos(xt[itex]^{2}[/itex])tan[itex]^{2}[/itex](t)dt

I'm trying to compute as many terms as possible of its asymptotic expansion as x[itex]\rightarrow\infty[/itex].


Homework Equations



x

The Attempt at a Solution


Let u=cos(xt[itex]^{2}[/itex]). And dv=tan[itex]^{2}[/itex](t)dt.
Then du=-2xtcos(xt[itex]^{2}[/itex])dt and v=[itex]\int[/itex]tan[itex]^{2}[/itex](t)dt=tan(t)-t+C.

Integration by parts yields:

I=cos(xt[itex]^{2}[/itex])tan(t)-tcos(xt[itex]^{2}[/itex])+[itex]\int[/itex][2xtcos(xt[itex]^{2}[/itex])tan(t)-2xt[itex]^{2}[/itex]cos(xt[itex]^{2}[/itex])]dt,
where all terms are evaluated from 0 to pi/4, obviously.

This feels wrong to me. Can anyone give me some help?
 
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  • #2
hi the_kid! :smile:

(try using the X2 button just above the Reply box :wink:)
the_kid said:
Let u=cos(xt[itex]^{2}[/itex]). And dv=tan[itex]^{2}[/itex](t)dt.
Then du=-2xtcos(xt[itex]^{2}[/itex])dt

erm :redface:

du=-2xtsin(xt2)dt :wink:
 

Related to Integration by parts computation

1. What is integration by parts computation?

Integration by parts computation is a method used in calculus to evaluate the integral of a product of two functions. It involves applying the product rule of differentiation to rewrite the integral in a different form, making it easier to solve.

2. When is integration by parts computation used?

Integration by parts computation is used when the integral of a product of two functions cannot be easily evaluated using other methods, such as substitution or trigonometric identities. It is also useful for solving integrals involving logarithmic and exponential functions.

3. How do you use integration by parts computation?

To use integration by parts computation, you need to identify the two functions in the integrand and choose one as the "u" function and the other as the "dv" function. Then, apply the formula ∫u dv = uv - ∫v du to rewrite the integral in a different form. Finally, solve the new integral and substitute back in the original variables.

4. What are the key steps in integration by parts computation?

The key steps in integration by parts computation are: 1) identifying the two functions in the integrand, 2) choosing one as the "u" function and the other as the "dv" function, 3) applying the formula ∫u dv = uv - ∫v du, 4) solving the new integral, and 5) substituting back in the original variables.

5. Are there any tips for solving integrals using integration by parts computation?

Yes, there are several tips for solving integrals using integration by parts computation. These include choosing the "u" function to be the most complicated or difficult to integrate, choosing the "dv" function to be the simplest or easiest to integrate, and repeating the process if the integral becomes more complicated after the first application of the formula. It is also important to carefully choose the "du" and "v" terms when using the formula.

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