Integrate Function: Can't Find Answer Key - Help Needed

[okevino]

question in the attachment...

integrate this function from 1 to 2.

i can't find a way to do this quesiton
it's from previous year final exam so there's no answer key to it..

hope someone would teach me..
thanks..

 

[Mute]

Have you learned trigonometric substitution? Let $x = \sqrt(5)\sin\theta$, and use the pythagorean identity to write $5(1-\sin^2\theta) = 5\cos^2\theta$.

Under this change of variables, $dx = \sqrt(5)\cos\theta d\theta$. Your integral then becomes an integral over a constant times $1/\cos^2\theta = sec^2\theta$. If you know the integral of $sec^2\theta$, you're pretty much done then.

 

[okevino]

alright..
but where did the 3/2 go..

 

[okevino]

right now i have...

4/5 integral sec^2 theta..

if I'm right at this point
so...
integral of sec^2 is tan x
but i set x = root5 sin theta earlier..
how do i finish this question..

 

[Gib Z]

The Exponents canceled each other out.

$$\int \frac{4}{(5-x^2)^{3/2}} dx$$.

Now We do the substitution Mute said. $dx = \sqrt(5)\cos\theta d\theta$

So now the integral is $$\int {4\sqrt{5} \cos \theta}{5^{3/2}\cdot (\cos^2 \theta)^{3/2}} d\theta$$

We can take all constants out. When we have a power to a power, we multiply the powers. 3/2 times 2 is just 3.

$$\frac{4\sqrt{5}}{5^{3/2}} \int \frac{\cos \theta}{\cos^3 \theta} d\theta = \frac{4\sqrt{5}}{5^{3/2}} \int \frac{1}{\cos^2 \theta} d\theta$$. As Mute said, 1/cos^2 theta is sec^2 theta. And the integral of that is easy, you should know that.

 

[Gib Z]

O and the constant on the outside simplifies to 4/5.

 

[okevino]

that's exactly what i said ..
but if i take the integral from 1 to 2..
i dont' get the answer 6/5

cuz i think if you set x to some value.
the upper and lower bound needs to change..
but i dont' remember if there's such thing..

 

[Gib Z]

Well The integral is (4/5) tan theta.

x = sqrt{5} sin theta.
sin theta = x/(sqrt5)
theta = arcsin (x/sqrt5)
The integral is therefore 4/5 tan (arcsin x/sqrt5). Plug in x values and subtract.

 

[okevino]

my x values are 1 to 2..
if i sub into arcsinx it's undefined.

 

[Mute]

That would be a problem, except that you want to sub into arcsin(x/sqrt5), not arcsinx.

 

[okevino]

wow...
great...
got it...
thanks mute and GibZ..