MHB How Does Trig Substitution Simplify the Integral of 1/(25-t^2)^(3/2)?

[karush]

w,8.7.8 nmh{925}
$\displaystyle\int\frac{1}{\left(25-{t}^{2 }\right)^{3 /2 } }\ dt
=\frac{t}{25\sqrt{25-{t}^{2 }}}+C$

$\begin{align}\displaystyle
t& = {5\sin\left({u}\right)} &
dt&={5\cos\left({u}\right)} du&
\end{align}$

Then $ \displaystyle\int\dfrac{5\cos\left({u}\right)}
{\left(25-25\sin^2 \left({u}\right)\right)^{3 /2} }\ du$
?
Couldn't see the magic massage thing to avoid a trig substitution

 

[MarkFL]

I think a trig. substitution is a good strategy here...you should be able to now simplify to get:

$$I=\frac{1}{25}\int \sec^2(u)\,du$$

Can you proceed?

 

[karush]

Not sure but.

$$I=\frac{1}{25}\int \sec^2(u)\,du
=\frac{1}{25}\tan\left({u}\right)+C$$
Since
$\displaystyle
t=5\sin\left({u}\right)$
Then
$\displaystyle
u=\arcsin\left(\frac{t}{5}\right) $
So
$\displaystyle
I=\frac{t}{25\sqrt{25-{t}^{2}}}+C$

 

[karush]

https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy