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Question: How do you integrate \[{\sin ^{ - 1}}x\,dx\] from 0 to 1....

How do you integrate sin⁡−1x dx{\sin ^{ - 1}}x\,dx from 0 to 1.

Explanation

Solution

Here in this question we have to integrate the given function, where the function is of the form trigonometry. By using the standard integration formulas we integrate the given function. Since they have mentioned the limit points. It is a definite integral. Hence we determine the value.

Complete step by step solution:
In integration we have two different kinds. One is definite integral and another one is indefinite integral. In definite integral the limits points are mentioned. In indefinite integral the limit points are not mentioned.

Here this question belongs to the definite integral where the limits points are mentioned.
The function which we have to integrate is an inverse trigonometry function.
Now consider the given function
sin⁡−1x dx{\sin ^{ - 1}}x\,dx
On integrating the given function, here the lower limit is 0 and the upper limit is 1
∫01sin⁡−1x dx\int\limits_0^1 {{{\sin }^{ - 1}}x\,dx}
We integrate the above function by integration by parts.
Let u=sin⁡−1xu = {\sin ^{ - 1}}x and v=1v = 1
The formula of integration by parts is given by ∫udv=uv−∫vdu\int {udv = uv - \int {vdu} }
Therefore we have ⇒∫01sin⁡−1x dx=x sin⁡−1x]01−∫01x1−x2dx\Rightarrow \int\limits_0^1 {{{\sin }^{ - 1}}x\,dx} = \left. {x\,{{\sin }^{ - 1}}x} \right]_0^1 - \int\limits_0^1 {\dfrac{x}{{\sqrt {1 - {x^2}} }}dx}
On integrating the second term of the RHS we get
Let we integrate by substitution t=1−x2t = 1 - {x^2} and dt=−2x dxdt = - 2x\,dx, therefore we have
⇒∫01sin⁡−1x dx=π2+12∫101tdt\Rightarrow \int\limits_0^1 {{{\sin }^{ - 1}}x\,dx} = \dfrac{\pi }{2} + \dfrac{1}{2}\int\limits_1^0 {\dfrac{1}{{\sqrt t }}dt}
Interchange the limit points we get
⇒∫01sin⁡−1x dx=π2−12∫01t−12dt\Rightarrow \int\limits_0^1 {{{\sin }^{ - 1}}x\,dx} = \dfrac{\pi }{2} - \dfrac{1}{2}\int\limits_0^1 {{t^{\dfrac{{ - 1}}{2}}}dt}
On integrating
⇒∫01sin⁡−1x dx=π2−12(2t12)01\Rightarrow \int\limits_0^1 {{{\sin }^{ - 1}}x\,dx} = \dfrac{\pi }{2} - \dfrac{1}{2}\left( {2{t^{\dfrac{1}{2}}}} \right)_0^1
On substituting the limit points we get
⇒∫01sin⁡−1x dx=π2−1\Rightarrow \int\limits_0^1 {{{\sin }^{ - 1}}x\,dx} = \dfrac{\pi }{2} - 1
Hence we have integrated the given function and applied the limit points and obtained an answer.

There the answer for this given question ∫01sin⁡−1x dx=π2−1\int\limits_0^1 {{{\sin }^{ - 1}}x\,dx} = \dfrac{\pi }{2} - 1.

Note: In integration we have two kinds one is definite integral and other one is indefinite integral. This question comes under the definite integral. While integrating the function which is in the form of product or division form we use the integration by parts method. By applying the integration by parts we obtain the solution and then substitute the limit points to the function.