For solving this question (y=t^2sin^-1t) can we use integration
no question was asked; all we were given was a function.
I apologize for the confusion. If we are only given the function y = t^2 sin^-1t, then we cannot use integration to "solve" the question because there is no question asked. However, we can still analyze the function and its properties.
First, note that the function is defined for -1 ≤ t ≤ 1, because the inverse sine of t is only defined for -1 ≤ t ≤ 1. The function is therefore restricted to this domain.
The function is also even, because the sine function is odd and the square function is even:
y(-t) = (-t)^2 sin^-1(-t) = t^2 sin^-1t = y(t)
Furthermore, we can find the derivative of the function using the chain rule:
dy/dt = 2t sin^-1t + t^2 / sqrt(1-t^2)
The derivative is defined for all t in the domain and is continuous in the domain, so the function is differentiable in the domain.
Finally, we can also find the limits of the function as t approaches the endpoints of the domain:
lim t→1- y(t) = lim t→1- t^2 sin^-1t = 0
lim t→-1+ y(t) = lim t→-1+ t^2 sin^-1t = π
Therefore, the function approaches 0 as t approaches 1 from the left, and approaches π as t approaches -1 from the right.
Yes, we can use integration to solve the equation y = t^2 * sin^(-1)(t).
To do so, we start by integrating both sides of the equation. However, before we can integrate, we need to rewrite sin^(-1)(t) in terms of a more familiar trigonometric function.
The inverse sine function, sin^(-1)(t) or arcsin(t), is the function that gives us the angle whose sine is t. In other words, if sin(x) = t, then arcsin(t) = x.
Now, let's solve the equation step by step:
1. Begin by substituting u for t^2: u = t^2.
2. Now, differentiate both sides of the equation with respect to t: du/dt = 2t.
3. To solve for dt, we rearrange the equation: dt = (1/2t) du.
4. Substitute these values back into the original equation: y = u * sin^(-1)(t).
5. Multiplying both sides of the equation by dt: y * dt = u * sin^(-1)(t) * dt.
6. Substitute for dt using step 3: y * (1/2t) du = u * sin^(-1)(t) * (1/2t) du.
7. Simplify the equation: (1/2) * (y/u) * du = sin^(-1)(t) * du.
8. Now, integrate both sides of the equation:
Integrating the left side: ∫(1/2) * (y/u) du = (1/2) * y * ∫(1/u) du = (1/2) * y * ln|u| + C.
Integrating the right side: ∫sin^(-1)(t) du = u * sin^(-1)(t) - ∫u * (1/(1-t^2)) dt.
9. Now, we have a u-term in the right integral that we can substitute back for t^2:
= u * sin^(-1)(t) - ∫u * (1/(1-u)) du.
10. Integrating the right side of the equation:
= u * sin^(-1)(t) - ∫(u/(1-u)) du.
To integrate ∫(u/(1-u)) du, we can use a partial fraction decomposition.
11. By expressing (u/(1-u)) as (1 + u)/(1-u), we can rewrite it as:
∫(u/(1-u)) du = ∫((1+u)/(1-u)) du.
12. Applying the partial fraction decomposition, we get:
(1+u)/(1-u) = A + (B/(1-u)).
To find A and B, we clear the denominators and compare the coefficients of like terms. We get:
1+u = (A(1-u)) + B.
Expanding and simplifying the equation:
1 + u = A - Au + B.
Matching coefficients, we find that A = 1 and B = 2, so:
∫((1+u)/(1-u)) du = ∫(1 + (2/(1-u))) du.
13. Now, we can integrate the right side of the equation:
∫((1+u)/(1-u)) du = ∫(1 + (2/(1-u))) du
= u + 2ln|1-u| + C'.
14. Substituting back into the original equation, we obtain:
(1/2) * y * ln|u| = u * sin^(-1)(t) - u - 2ln|1-u| + C'.
15. Finally, replace u with t^2:
(1/2) * y * ln|t^2| = t^2 * sin^(-1)(t) - t^2 - 2ln|1-t^2| + C'.
And that's the final solution obtained by integrating both sides of the equation y = t^2 * sin^(-1)(t).
Yes, we can use integration to solve this question. We can differentiate both sides with respect to t, using the chain rule:
dy/dt = 2tsin^-1t + t^2 d/dt(sin^-1t)
dy/dt = 2tsin^-1t + t^2 / sqrt(1-t^2)
Now, we can integrate both sides with respect to t, using substitution:
∫dy = ∫(2tsin^-1t + t^2 / sqrt(1-t^2)) dt
y = t^2 sin^-1t + ∫t^2 / sqrt(1-t^2) dt
The integral on the right-hand side can be solved using trigonometric substitution, by letting t = sin θ:
y = t^2 sin^-1t + ∫sin^2θ dθ
y = t^2 sin^-1t + (θ - sinθ cosθ) + C
Substituting back for t = sin θ, we get:
y = t^2 sin^-1t + (sin^-1t - t √(1-t^2)) + C
Therefore, the solution to the question is:
y = t^2 sin^-1t + (sin^-1t - t √(1-t^2)) + C