How do I Evaluate the Integral using Substitution and Simplification?

[Nyasha]

Homework Statement

$$18\int_0^4 \sqrt{4- (y-2)^2}dy$$

Homework Equations

According to the textbook l am supposed to use $$y=2sin\theta$$ for substitution

The Attempt at a Solution

$$y=2sin\theta$$

$$dy=2cos\theta d\theta$$

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


$$18\int_0^4 \sqrt{4- (4sin^2\theta-4sin\theta-4sin\theta+4)}2cos\theta d\theta$$

$$18\int_0^4 \sqrt{4- (4sin^2\theta-8sin\theta+4)}2cos\theta d\theta$$


( How do l simplify this integral )

 

[gabbagabbahey]

Homework Statement

$$18\int_0^4 \sqrt{4- (y-2)^2}dy$$

Homework Equations

According to the textbook l am supposed to use $$y=2sin\theta$$ for substitution

Ermmmm... how can $y$ ever equal 4 with this substitution?

 

[Nyasha]

Ermmmm... how can $y$ ever equal 4 with this substitution?

That is where l am stuck and also on how to simply this integral

 

[gabbagabbahey]

Try a different substitution!

...maybe $y=4\sin^2\theta$

 

[Mark44]

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:
$$18\int_{y = 0}^4 \sqrt{4- (2sin\theta-2)^2}2cos\theta d\theta$$

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.

 

[Nyasha]

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:
$$18\int_{y = 0}^4 \sqrt{4- (2sin\theta-2)^2}2cos\theta d\theta$$

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.

Is this correct.

 

[gabbagabbahey]

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 $$\int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta$$... 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

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

 

[Nyasha]

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 $$\int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta$$... 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

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

Thanks very much.