Hello. Could you help me integrate this using trig sub? Thank you. I was stuck on a few steps. What does a = ? x = ? dx = ? `intx^3sqrt(x^2-25).`
Ok, this is gonna take a while.
The trig substitution we will make depends on how the stuff underneath the radical is arranged. Check out the table in the link I am posting below.
For the form `sqrt(x^2 - a^2)`
We use the substitution `x = asec(theta)`
In this case a = 5, so our substitution is `x = 5sec(theta)`
Then we derive to find `dx` and `d theta`
`dx = 5sec(theta)tan(theta) d theta`
Plug this into our integral at we get
`int (5sec(theta))^3 * sqrt((5sec(theta))^2 - 5^2) * 5 sec(theta)tan(theta) d theta`
We can factor out a 5^2 from each term under the radical to turn that into
`sqrt(5^2(sec(theta)^2 - 1))`
And using our trusty trig identities this turns into
`sqrt(5^2 * tan^2(theta))`
We will rewrite out equation without the radical now.
`int (5sec(theta))^3 * 5tan(theta) * 5sec(theta)tan(theta) d theta`
We combine like terms to get
`int 3125 sec^4(theta) tan^2(theta) d theta`
At this point we will want to solve our new integral using u substitution.
We will do this by turning sec^4 into sec^2*(tan^2 + 1)
`3125 int sec^2(theta) (tan^2(theta) + 1) tan^2(theta) d theta`
Now we let `u = tan(theta)`
so `du = sec^2 (theta) d theta`
Plug this all in, and we get
`3125 int (u^2 + 1)u^2 du`
integrate this to get
`3125(u^5/5 + u^3/3) + c` .
Now all we need to do is sub back in the two substitutions we made and we are finished!
`3125((tan^5(theta))/5 + (tan^3(theta))/3) + c`
Now sub in `x = 5sec(theta)`
But we do not have secants, we have tangents in our equation!
We will have to use triangle trig to figure out what tangent equals.
`x = 5sec(theta)`
`sec(theta) = x/5` so `cos(theta) = 5/x`
Adjacent = 5, hypotenuse = x, which implies that opposite = `sqrt(x^2 - 25)`
This means that `tan(theta) = sqrt(x^2 - 25) / 5`
Now we can finally plug that in for our tangents, and we get
`3125((sqrt(x^2-25)/5)^5/5 + (sqrt(x^2 - 25)/5)^3/3) + c`
And you can have fun simplifying that on your own.
Hope this helps!!!
`intx^3sqrt(x^2-25)dx` using a trigonometric substitution.
Let `x=5sectheta,dx=5secthetatantheta d theta`
`intx^3sqrt(x^2-25)dx=int(125sec^3theta)(5tantheta)(5secthetatantheta d theta)`
`=3125inttan^2theta(1+tan^2theta)sec^2theta d theta`
`=3125((tan^3 theta)/3+(tan^5 theta)/5)+C`