How do I Evaluate the Integral using Substitution and Simplification?

  • Thread starter Nyasha
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In summary: This will help me a lot!In summary, the textbook states that y=2sin\theta should be used for substitution in solving homework equations, but the equation becomes very difficult to solve when this substitution is made. Changing the limits of integration to remember that the y values are at the 0% and 100% limits of the integral, or sketching a graph to see where the two sin^2's and cos^2's intersect, can help solve the equation.
  • #1
Nyasha
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Homework Statement



[tex]
18\int_0^4 \sqrt{4- (y-2)^2}dy
[/tex]

Homework Equations



According to the textbook l am supposed to use [tex]y=2sin\theta [/tex] for substitution

The Attempt at a Solution



[tex]y=2sin\theta [/tex]

[tex]dy=2cos\theta d\theta [/tex]

[tex]
18\int_0^4 \sqrt{4- (2sin\theta-2)^2}2cos\theta d\theta
[/tex]
( I am stuck at converting the limits of my integral using [tex]y=2sin\theta [/tex] )


[tex]
18\int_0^4 \sqrt{4- (4sin^2\theta-4sin\theta-4sin\theta+4)}2cos\theta d\theta
[/tex]

[tex]
18\int_0^4 \sqrt{4- (4sin^2\theta-8sin\theta+4)}2cos\theta d\theta
[/tex]


( How do l simplify this integral )
 
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  • #2
Nyasha said:

Homework Statement



[tex]
18\int_0^4 \sqrt{4- (y-2)^2}dy
[/tex]

Homework Equations



According to the textbook l am supposed to use [tex]y=2sin\theta [/tex] for substitution

Ermmmm... how can [itex]y[/itex] ever equal 4 with this substitution?
 
  • #3
gabbagabbahey said:
Ermmmm... how can [itex]y[/itex] ever equal 4 with this substitution?
That is where l am stuck and also on how to simply this integral
 
  • #4
Try a different substitution!:smile:

...maybe [itex]y=4\sin^2\theta[/itex] :wink:
 
  • #5
I would make two substitutions: u = y - 2, so du = dy, and then the trig substitution.

As far as your limits of integration, you can carry them along all the way through until you finally undo both substitutions, when your antiderivative will be in terms of y.

You can remind yourself that the limits of integration are y values by doing adding "y = " on the lower limit of integration, like so:
[tex] 18\int_{y = 0}^4 \sqrt{4- (2sin\theta-2)^2}2cos\theta d\theta [/tex]

That should help you remember that these are y values.

Alternatively, you can change the limits of integration for each substitution. If you're careful, both techniques will work.
 
  • #6
Mark44 said:
I would make two substitutions: u = y - 2, so du = dy, and then the trig substitution.

As far as your limits of integration, you can carry them along all the way through until you finally undo both substitutions, when your antiderivative will be in terms of y.

You can remind yourself that the limits of integration are y values by doing adding "y = " on the lower limit of integration, like so:
[tex] 18\int_{y = 0}^4 \sqrt{4- (2sin\theta-2)^2}2cos\theta d\theta [/tex]

That should help you remember that these are y values.

Alternatively, you can change the limits of integration for each substitution. If you're careful, both techniques will work.


Engineeringcalculus.jpg


Is this correct.
 
  • #7
Nyasha said:
Engineeringcalculus.jpg


Is this correct.

Sure, but you didn't finish evaluating it by substituting in the limits...

Incidentally, there is also a neat little trick to evaluating the integral [tex]\int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta[/tex]... If you sketch a graph of both sin^2 and cos^2 over that interval, you see they both have the same area underneath them

[tex]\implies \int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta=\frac{1}{2}\int_{-\pi/2}^{\pi/2} (\cos^2\theta+\sin^2\theta) d\theta=\frac{1}{2}\int_{-\pi/2}^{\pi/2} (1)d\theta=\frac{\pi}{2}[/tex]
 
  • #8
gabbagabbahey said:
Sure, but you didn't finish evaluating it by substituting in the limits...

Incidentally, there is also a neat little trick to evaluating the integral [tex]\int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta[/tex]... If you sketch a graph of both sin^2 and cos^2 over that interval, you see they both have the same area underneath them

[tex]\implies \int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta=\frac{1}{2}\int_{-\pi/2}^{\pi/2} (\cos^2\theta+\sin^2\theta) d\theta=\frac{1}{2}\int_{-\pi/2}^{\pi/2} (1)d\theta=\frac{\pi}{2}[/tex]


Thanks very much.
 

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