Solveeit Logo
Login

Question

Question: Integrate the function \({{\left( {{\sin }^{-1}}x \right)}^{2}}\) ....

Integrate the function (sin⁡−1x)2{{\left( {{\sin }^{-1}}x \right)}^{2}} .

Explanation

Solution

We first rewrite the integrand as (sin⁡−1x)2.1{{\left( {{\sin }^{-1}}x \right)}^{2}}.1 . After that, we assume u as (sin⁡−1x)2{{\left( {{\sin }^{-1}}x \right)}^{2}} and v as 11 . Doing so, we now carry out the integration by parts according to ∫uvdx=u∫vdx−∫(dudx∫vdx)\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)} and gradually end up in solving it.

Complete step-by-step answer:
In this problem, we need to integrate the function (sin⁡−1x)2{{\left( {{\sin }^{-1}}x \right)}^{2}} . Integrating this function with the help of integration by parts will be the easiest method. According to the integration by parts rule, if there are two functions “u” and “v” of the variable “x”, then the integration of (uv)\left( uv \right) will be as follow,
∫uvdx=u∫vdx−∫(dudx∫vdx)\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}
Let I=∫(sin⁡−1x)2.1dxI=\int{{{\left( {{\sin }^{-1}}x \right)}^{2}}.1dx}
Now, we know the thumb rule of by parts, which is ILATE (Inverse Logarithmic Algebraic Trigonometric Exponential). This is the priority order of functions that are to be considered as the first function. If we compare the functions u and v with our problem, we can write u (the first function) as (sin⁡−1x)2{{\left( {{\sin }^{-1}}x \right)}^{2}} since it is Inverse and comes first in the list, and v (the second function) as 11 since being algebraic, it comes third in the list. Then, integrating their product by parts, we get,
\Rightarrow I={{\left( {{\sin }^{-1}}x \right)}^{2}}\int{1dx}-\int{\left\\{ \dfrac{d\left( {{\left( {{\sin }^{-1}}x \right)}^{2}} \right)}{dx}\left( \int{1dx} \right) \right\\}dx}....\left( i \right)
Now, the derivative of (sin⁡−1x)2{{\left( {{\sin }^{-1}}x \right)}^{2}} can be done as,
d((sin⁡−1x)2)dx=d((sin⁡−1x)2)d(sin⁡−1x)×d(sin⁡−1x)dx ⇒d((sin⁡−1x)2)dx=2(sin⁡−1x)×11−x2=2(sin⁡−1x)1−x2 \begin{aligned} & \dfrac{d\left( {{\left( {{\sin }^{-1}}x \right)}^{2}} \right)}{dx}=\dfrac{d\left( {{\left( {{\sin }^{-1}}x \right)}^{2}} \right)}{d\left( {{\sin }^{-1}}x \right)}\times \dfrac{d\left( {{\sin }^{-1}}x \right)}{dx} \\\ & \Rightarrow \dfrac{d\left( {{\left( {{\sin }^{-1}}x \right)}^{2}} \right)}{dx}=2\left( {{\sin }^{-1}}x \right)\times \dfrac{1}{\sqrt{1-{{x}^{2}}}}=\dfrac{2\left( {{\sin }^{-1}}x \right)}{\sqrt{1-{{x}^{2}}}} \\\ \end{aligned}
This value of d((sin⁡−1x)2)dx\dfrac{d\left( {{\left( {{\sin }^{-1}}x \right)}^{2}} \right)}{dx} can be inserted in the integral (i) and rewritten as,
\Rightarrow I={{\left( {{\sin }^{-1}}x \right)}^{2}}\int{1dx}-\int{\left\\{ \dfrac{2\left( {{\sin }^{-1}}x \right)}{\sqrt{1-{{x}^{2}}}}\left( \int{1dx} \right) \right\\}dx}
The integration of ∫1dx\int{1dx} is simply x. So, the above integral becomes,
\Rightarrow I={{\left( {{\sin }^{-1}}x \right)}^{2}}.x-\int{\left\\{ \dfrac{2\left( {{\sin }^{-1}}x \right)}{\sqrt{1-{{x}^{2}}}}.x \right\\}dx}
The above integral can be rearranged as,
\Rightarrow I=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\int{\left\\{ \left( {{\sin }^{-1}}x \right).\dfrac{-2x}{\sqrt{1-{{x}^{2}}}} \right\\}dx}
Again, integrating by parts the integrand \left\\{ \left( {{\sin }^{-1}}x \right).\dfrac{-2x}{\sqrt{1-{{x}^{2}}}} \right\\} , we get,

& \Rightarrow I=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ \left( {{\sin }^{-1}}x \right)\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}dx}-\int{\left\\{ \dfrac{d\left( {{\sin }^{-1}}x \right)}{dx}\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}dx} \right\\}dx} \right] \\\ & \Rightarrow I=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ 2\left( {{\sin }^{-1}}x \right)\int{\dfrac{-x}{\sqrt{1-{{x}^{2}}}}dx}-\int{\left\\{ \dfrac{d\left( {{\sin }^{-1}}x \right)}{dx}.2\int{\dfrac{-x}{\sqrt{1-{{x}^{2}}}}dx} \right\\}dx} \right] \\\ \end{aligned}$$ Now, the derivative $\dfrac{d\sqrt{1-{{x}^{2}}}}{dx}=\dfrac{1}{2\sqrt{1-{{x}^{2}}}}\times \left( -2x \right)=-\dfrac{x}{\sqrt{1-{{x}^{2}}}}$ . So, the above integrand can be written as, $$\Rightarrow I=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ 2\left( {{\sin }^{-1}}x \right)\int{d\left( \sqrt{1-{{x}^{2}}} \right)}-\int{\left\\{ \dfrac{d\left( {{\sin }^{-1}}x \right)}{dx}.2\int{d\left( \sqrt{1-{{x}^{2}}} \right)} \right\\}dx} \right]$$ Evaluating the above integral, we get, $$\Rightarrow I=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ 2\left( {{\sin }^{-1}}x \right).\sqrt{1-{{x}^{2}}}-\int{\left\\{ \dfrac{1}{\sqrt{1-{{x}^{2}}}}.2\sqrt{1-{{x}^{2}}} \right\\}dx} \right]$$ Simplifying the above integral, we get, $$\Rightarrow I=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ 2\left( {{\sin }^{-1}}x \right).\sqrt{1-{{x}^{2}}}-2\int{dx} \right]$$ The integration of $\int{1dx}$ is simply x. So, the above integral becomes, $$\Rightarrow I=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+2\left( {{\sin }^{-1}}x \right).\sqrt{1-{{x}^{2}}}-2x+c$$ Thus, we can conclude that the integration of ${{\left( {{\sin }^{-1}}x \right)}^{2}}$ is $$x{{\left( {{\sin }^{-1}}x \right)}^{2}}+2\left( {{\sin }^{-1}}x \right).\sqrt{1-{{x}^{2}}}-2x+c$$ . **Note:** Solving by the “by parts” method is the easiest. Other methods like the substitution method may give a result, but it will not be worth the time taken. This is a very long integration, so we must be very careful in the calculations and should remember to put the integration constant in the end.