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
For solving this question (y=t^2sin^-1t) can we use integration
3 answers
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.
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.