For the following integral find an appropriate trigonometric substitution of the form x=f(t) to simplify the integral.
INT((4x^2-3)^1.5) dx x=?
u = 4x^2 -3
(1/8)du = x dx
(1/8)INT u^(3/2) du
(1/8)(2/5)u^(5/2)
(1/20)(4x^2 -3)^(5/2)
I don't see why they demand a trigonometric solution. This seems like the logical one to use.
No it is required to use a trignometric substitution. x cannot be a variable in the final answer
you can do like this:
(4x^2-3)^1.5 = (4x^2-3).sq(4x^2-3)dx
Let 2x = Sq3.sec(det)
so, 4x^2 -3 = 3.sec^2(det)-3
= 3.tan^2(det)
new INT= INt(3^1.5tan^3(det)d(det)
Use Pythagore to solve the relation between x and the angle det.
To find an appropriate trigonometric substitution of the form x = f(t) for the given integral ∫(4x^2-3)^1.5 dx, we need to identify a trigonometric function that can eliminate the square root and simplify the integral.
Let's begin by analyzing the given integral: ∫(4x^2-3)^1.5 dx
We notice that the term (4x^2 - 3)^1.5 involves a squared expression inside the square root. This suggests that we can make a substitution using a trigonometric function that can eliminate the squared expression.
To proceed, we can let 4x^2 - 3 = u^2, where u is a trigonometric function of t.
Now, let's find x in terms of t by solving 4x^2 - 3 = u^2 for x:
4x^2 - 3 = u^2
4x^2 = u^2 + 3
x^2 = (u^2 + 3)/4
x = ±√[(u^2 + 3)/4]
Since the square root is always non-negative, we can ignore the negative sign and choose the positive sign. Therefore, x = √[(u^2 + 3)/4].
Now, we have expressed x in terms of t: x = √[(u^2 + 3)/4].
To find dx, we can take the derivative of x with respect to t:
dx/dt = d/dt [√[(u^2 + 3)/4]]
dx/dt = (1/2) * (u^2 + 3)^(-1/2) * 2u * du/dt
dx/dt = u / √(u^2 + 3) * du/dt
Now, we have dx in terms of u and du/dt.
Substituting x = √[(u^2 + 3)/4] and dx = u / √(u^2 + 3) * du/dt into the original integral, we get:
∫(4x^2 - 3)^1.5 dx = ∫[(4(√[(u^2 + 3)/4])^2 - 3)^1.5] * (u / √(u^2 + 3)) * du/dt
Simplifying further:
∫[(4(u^2 + 3)/4 - 3)^1.5] * (u / √(u^2 + 3)) * du/dt
∫[(u^2 + 3 - 3)^1.5] * (u / √(u^2 + 3)) * du/dt
∫[(u^2)^1.5] * (u / √(u^2 + 3)) * du/dt
∫[u^3] * (u / √(u^2 + 3)) * du/dt
Now, the integral is simplified with the appropriate trigonometric substitution.