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

In summary, the conversation discusses the use of trigonometric substitution to solve an integral problem. The integral in question is $\int\frac{1}{\left(25-{t}^{2}\right)^{3/2}}dt$, which can be simplified to $\frac{t}{25\sqrt{25-{t}^{2}}}+C$. The conversation also suggests the use of the substitution $t=5\sin(u)$ and the resulting integral can be solved by substituting back in for $u$.
  • #1
karush
Gold Member
MHB
3,269
5
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
 
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  • #2
I think a trig. substitution is a good strategy here...you should be able to now simplify to get:

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

Can you proceed?
 
  • #3
Not sure but.

\(\displaystyle I=\frac{1}{25}\int \sec^2(u)\,du
=\frac{1}{25}\tan\left({u}\right)+C\)
Since
$\displaystyle
t=5\sin\left({u}\right)$
Then
$\displaystyle
u=\arcsin\left(\frac{t}{5}\right) $
So
$\displaystyle
I=\frac{t}{25\sqrt{25-{t}^{2}}}+C$
 
  • #4
https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy
ssct.png
 

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

1. What is the meaning of "-w 8.7.8 inverse integral"?

The "-w 8.7.8 inverse integral" is a mathematical term that refers to the inverse integral of a function with respect to the variable w, where 8.7.8 is the specific value of w.

2. How is the "-w 8.7.8 inverse integral" calculated?

The "-w 8.7.8 inverse integral" can be calculated using various methods such as integration by parts, substitution, or the fundamental theorem of calculus. The specific method used depends on the function being integrated.

3. What is the significance of the "-w 8.7.8" value in the "-w 8.7.8 inverse integral"?

The "-w 8.7.8" value in the "-w 8.7.8 inverse integral" represents the specific variable (w) with respect to which the inverse integral is being calculated. This value is important as it determines the limits of integration and the specific integration technique to be used.

4. Can the "-w 8.7.8 inverse integral" be applied to any function?

Yes, the "-w 8.7.8 inverse integral" can be applied to any function that is integrable. However, the specific method used to calculate the inverse integral may vary depending on the complexity of the function.

5. How is the "-w 8.7.8 inverse integral" used in scientific research?

The "-w 8.7.8 inverse integral" is used in various fields of science, such as physics, engineering, and economics, to solve problems involving rates of change, optimization, and area under a curve. It is also used in modeling and predicting real-world phenomena.

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