Integrating Double Integral of Cos(x+y) from 0 to pi: Step-by-Step Guide

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In summary, the conversation discusses how to integrate the double integral cos(x+y) dy dx from 0 to pi and from 0 to pi again, with the given answer being -4. The solution involves using the substitution u=x+y and the trigonometric addition formula sin(a+b) = sin(a)cos(b) + cos(a)sin(b). The final step is to integrate again, similar to the first integral.
  • #1
Math10
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Homework Statement


How to integrate the double integral cos(x+y) dy dx from 0 to pi and from 0 to pi again.

Homework Equations


The answer is -4.

The Attempt at a Solution


Here's the work:
u=x+y
du=dy
cos(u)du=sin(x+y)
integral of [sin(x+y)] evaluate from 0 to pi dx from 0 to pi
integral of (sin(x+pi)-sin(x))dx from 0 to pi
And what's next?
 
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  • #2
Math10 said:

Homework Statement


How to integrate the double integral cos(x+y) dy dx from 0 to pi and from 0 to pi again.

Homework Equations


The answer is -4.

The Attempt at a Solution


Here's the work:
u=x+y
du=dy
cos(u)du=sin(x+y)
integral of [sin(x+y)] evaluate from 0 to pi dx from 0 to pi
integral of (sin(x+pi)-sin(x))dx from 0 to pi
And what's next?

Integrate again, similar to what you did for the first integral.
 
  • #3
Er...doesn't the final step that you list give you the answer already?
 
  • #4
But I can't. Because sin(x+pi)-sin(x)=sin(x)+sin(pi)-sin(x)=sin(pi)=0. The integral of 0 is?
 
  • #5
Math10 said:
But I can't. Because sin(x+pi)-sin(x)=sin(x)+sin(pi)-sin(x)=sin(pi)=0. The integral of 0 is?
##\sin(a+b)\ne \sin a +\sin b##
 
  • #6
Math10 said:
Because sin(x+pi)=sin(x)+sin(pi)
That is most certainly not correct. Can you check your compound angle formula?
 
  • #7
Then what's sin(a+b)?
 
  • #9
@Fightfish , thank you so much for the info! I got it!
 

Related to Integrating Double Integral of Cos(x+y) from 0 to pi: Step-by-Step Guide

1. What is integration?

Integration refers to the process of finding the area under a curve on a graph. It is an important concept in calculus and is used to solve various mathematical problems.

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