Integration by Parts: $\int u\cos(u)\,\mathrm{d}u$

Click For Summary

Discussion Overview

The discussion revolves around the integral $\int e^{2x} \cos(e^x) \,\mathrm{d}x$, focusing on the method of integration by parts and various approaches to solve it. Participants explore different substitutions and integration techniques, including potential issues with u-substitution.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes using integration by parts with the substitution $u = e^x$, leading to the integral $\int u \cos(u) \,\mathrm{d}u$.
  • Another participant challenges the initial substitution, noting that the differential $du = 2e^{2x} \,\mathrm{d}x$ was not properly accounted for.
  • There is a suggestion to use a different substitution, $u = \cos(2x)$ and $dv = e^{2x}$, and to apply integration by parts in that context.
  • Participants express uncertainty about the effectiveness of their approaches and seek clarification on the use of LaTeX for formatting equations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to solve the integral, with multiple competing approaches and some disagreement on the validity of the initial substitution.

Contextual Notes

There are unresolved issues regarding the correct application of u-substitution and integration by parts, as well as the proper handling of differentials in the context of the integral.

Who May Find This Useful

This discussion may be useful for students and practitioners interested in integration techniques, particularly those involving integration by parts and substitution methods in calculus.

brunette15
Messages
58
Reaction score
0
I have the following integral \int e(2x) cos(ex).

Let u = ex

Do integration by parts:
\int u2cos(u) du = u2sin(u) - \int (2usin(u) du

Do integration by parts again for \int (2usin(u) du:

\int (2usin(u) du = -2ucos(u) - \int -2cos(u) du

Putting it all together:

\int e(2x) cos(ex) = e(2x)sin(ex) - 2excos(ex) + 2sin(ex)
 
Physics news on Phys.org
Hi brunette15,

I don't think you u-substitution quite works, because $du=2e^{2x}$ that you haven't accounted for. Let's try another way.

Let $u=\cos(2x)$ and $dv = e^{2x}$. Then $$\int udv=uv-\int vdu$$.

What do you get when you try this?

Also, you might consider using $\LaTeX$. It takes a little time to learn but it makes your equations look very nice and clean. We can help you out if you are interested. :)
 
Jameson said:
Hi brunette15,

I don't think you u-substitution quite works, because $du=2e^{2x}$ that you haven't accounted for. Let's try another way.

Let $u=\cos(2x)$ and $dv = e^{2x}$. Then $$\int udv=uv-\int vdu$$.

What do you get when you try this?

Also, you might consider using $\LaTeX$. It takes a little time to learn but it makes your equations look very nice and clean. We can help you out if you are interested. :)

I see now! Thankyou! I have tried using latex however I am not too sure how to use it :/
 
brunette15 said:
I have the following integral \int e(2x) cos(ex).

Let u = ex

Do integration by parts:
\int u2cos(u) du = u2sin(u) - \int (2usin(u) du

Do integration by parts again for \int (2usin(u) du:

\int (2usin(u) du = -2ucos(u) - \int -2cos(u) du

Putting it all together:

\int e(2x) cos(ex) = e(2x)sin(ex) - 2excos(ex) + 2sin(ex)

If your integral is $\displaystyle \begin{align*} \int{ \mathrm{e}^{2x}\cos{ \left( \mathrm{e}^x \right) } \,\mathrm{d}x} \end{align*}$, write it as $\displaystyle \begin{align*} \int{ \mathrm{e}^x\cos{ \left( \mathrm{e}^x \right) } \,\mathrm{e}^x\,\mathrm{d}x} \end{align*}$ and then let $\displaystyle \begin{align*} u = \mathrm{e}^x \implies \mathrm{d}u = \mathrm{e}^x\,\mathrm{d}x \end{align*}$, giving $\displaystyle \begin{align*} \int{ u\cos{ (u)}\,\mathrm{d}u} \end{align*}$. Now you can use Integration by Parts.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K