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Question: How do you integrate \[{\int {\left( {{{\sin }^{ - 1}}x} \right)} ^2}\] using integration by parts?...

How do you integrate ∫(sin⁡−1x)2{\int {\left( {{{\sin }^{ - 1}}x} \right)} ^2} using integration by parts?

Explanation

Solution

In the above given question, we are given the square of an inverse trigonometric function that is a sine function. We have to integrate the given expression using the method of integration by parts. In order to approach the solution, first we have to consider the given function as the combination of two functions as (sin⁡−1x)2⋅1{\left( {{{\sin }^{ - 1}}x} \right)^2} \cdot 1 , where (sin⁡−1x)2{\left( {{{\sin }^{ - 1}}x} \right)^2} is to be treated as the first function whereas 11 is to be treated as the second function.

Complete answer:
Given that, the square of an inverse trigonometric function as (sin⁡−1x)2{\left( {{{\sin }^{ - 1}}x} \right)^2} .
We have to integrate the above given expression using the method of integration by parts.
i.e. we have to integrate ∫(sin⁡−1x)2{\int {\left( {{{\sin }^{ - 1}}x} \right)} ^2} using the integration by parts method.
Since, the given function is ∫(sin⁡−1x)2{\int {\left( {{{\sin }^{ - 1}}x} \right)} ^2} .
Hence, we can also write the given function as ∫(sin⁡−1x)2⋅1\int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2} \cdot 1}
So, let
I=∫(sin⁡−1x)2⋅1dxI = \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2} \cdot 1} dx
Now let (sin⁡−1x)2{\left( {{{\sin }^{ - 1}}x} \right)^2} to be the first function and 11 to be the second function.
Now, integrating by the method of integration by parts we get,
I = {\left( {{{\sin }^{ - 1}}x} \right)^2}\int {1dx} - \int {\left\\{ {\dfrac{d}{{dx}}{{\left( {{{\sin }^{ - 1}}x} \right)}^2} \cdot \int 1 } \right\\}} dx
That gives us,
I=(sin⁡−1x)2⋅x−∫2sin⁡−1x1−x2⋅xdxI = {\left( {{{\sin }^{ - 1}}x} \right)^2} \cdot x - \int {\dfrac{{2{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}} \cdot xdx
We can also write it as,
I=x⋅(sin⁡−1x)2+∫sin⁡−1x⋅(−2x1−x2)dxI = x \cdot {\left( {{{\sin }^{ - 1}}x} \right)^2} + \int {{{\sin }^{ - 1}}x \cdot \left( {\dfrac{{ - 2x}}{{\sqrt {1 - {x^2}} }}} \right)} dx
Again integrating the second part using integration by parts, we get
I = x \cdot {\left( {{{\sin }^{ - 1}}x} \right)^2} + \left[ {{{\sin }^{ - 1}}x\int {\left( {\dfrac{{ - 2x}}{{\sqrt {1 - {x^2}} }}} \right)dx - \int {\left\\{ {\dfrac{d}{{dx}}{{\sin }^{ - 1}}x\int {\dfrac{{ - 2x}}{{\sqrt {1 - {x^2}} }}} dx} \right\\}dx} } } \right]
That gives us,
I=x⋅(sin⁡−1x)2+[sin⁡−1x⋅21−x2−∫11−x2⋅21−x2dx]I = x \cdot {\left( {{{\sin }^{ - 1}}x} \right)^2} + \left[ {{{\sin }^{ - 1}}x \cdot 2\sqrt {1 - {x^2}} - \int {\dfrac{1}{{\sqrt {1 - {x^2}} }} \cdot } 2\sqrt {1 - {x^2}} dx} \right]
On solving further, we get
I=x⋅(sin⁡−1x)2+21−x2sin⁡−1x−∫2dxI = x \cdot {\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - \int {2dx}
Hence, we get the integration as
I=x⋅(sin⁡−1x)2+21−x2sin⁡−1x−2x+CI = x \cdot {\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - 2x + C
Where CC is the integration constant.
That is the required integration.
Therefore, the value of the integral ∫(sin⁡−1x)2{\int {\left( {{{\sin }^{ - 1}}x} \right)} ^2} using integration by parts, is x⋅(sin⁡−1x)2+21−x2sin⁡−1x−2x+Cx \cdot {\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - 2x + C .

Note: Integration by parts is a technique for performing indefinite integration or definite integration by expanding the differential of a product of functions and expressing the original integral in terms of a known integral . Integration by parts is for functions that can be written as the product of another function and a third function's derivative.