MHB LCC 205 8.4.7 sine substitution (definite integral)

[karush]

$\tiny{LCC \ \ 205 \ \ \ 8.4.7 \ \ sine \ \ substitution }$

$\displaystyle
I=\int_0^{5/2} \frac{1}{\sqrt{25-x^2}}\ dx
=\frac{\pi}{6}
$

$$\displaystyle
x=5\sin\left({u}\right) \ \ \ dx=5\cos\left({u}\right) \ \ du
$$
$$\displaystyle
I=\int_0^{5/2}
\frac{1}{\sqrt{25-25 \sin^2 \left({u}\right)}}
\ \ 5\cos\left({u}\right)
\ du $$
Not sure if this setup is right

 

[MarkFL]

I edited the thread title to be a little easier on the eyes...

You are using a good substitution, but you need to change your limits of integration in accordance with your substitution, so that you have $u$-values there instead of $x$. :)

 

[karush]

$\tiny{LCC \ \ 205 \ \ \ 8.4.7 \ \ sine \ \ substitution }$
So if
$\displaystyle 0=5\sin\left({a}\right) \ \ \ a=\pi$

$\displaystyle \frac{5}{2}=5\sin\left({b}\right)\ \ \ b=\frac{7\pi}{6}$
Then

$$\displaystyle
I=\int_a^b
\frac{5\cos\left({u}\right)}{\sqrt{25-25 \sin^2 \left({u}\right)}}
\ \ du
\implies \int_a^b
\frac{5\cos\left({u}\right)}{5\sqrt{1- \sin^2 \left({u}\right)}}
\ \ du
\implies \int_a^b 1 \ \ du
$$

$I=\left[u\right]_a^b
= \frac{7\pi}{6}-\frac{6\pi}{6}
=\frac{\pi}{6}$
Hopefully

 

[MarkFL]

Note that:

$$\sin\left(\frac{7\pi}{6}\right)=-\frac{1}{2}$$

You have:

$$u(x)=\arcsin\left(\frac{x}{5}\right)$$

Now, observe that:

$$0\le x\le\frac{5}{2}$$

Hence:

$$0\le \frac{x}{5}\le\frac{1}{2}$$

Can you now get the correct limits of integration?

 

[karush]

How would that be in terms of u?

$\frac{\pi}{6}$

Is the correct answer

 

[MarkFL]

If we take the inverse sine of each value in the compound inequality, we get:

$$\arcsin(0)\le \arcsin\left(\frac{x}{5}\right)\le\arcsin\left(\frac{1}{2}\right)$$

Now simplify...:)

 

[karush]

If we take the inverse sine of each value in the compound inequality, we get:

$$\arcsin(0)\le \arcsin\left(\frac{x}{5}\right)\le\arcsin\left(\frac{1}{2}\right)$$

$$0\le\arcsin\left({\frac{x}{5}}\right)\le \frac{\pi}{6}$$
Why would this help

 

[MarkFL]

$$0\le\arcsin\left({\frac{x}{5}}\right)\le \frac{\pi}{6}$$
Why would this help

Because:

$$u(x)=\arcsin\left({\frac{x}{5}}\right)$$

And so we have:

$$0\le u\le\frac{\pi}{6}$$

And the integral then becomes:

$$I=\int_0^{\frac{\pi}{6}}\,du=\frac{\pi}{6}$$

 

[karush]

OK got it
Thanks for not giving up on me
The help here is awesome