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Question: How do you integrate \[\int {{{\sin }^{ - 1}}x} \] by integration by parts method?...

How do you integrate ∫sin⁡−1x\int {{{\sin }^{ - 1}}x} by integration by parts method?

Explanation

Solution

Here we will first form the integral equation in such a way that we can apply the integration by parts. Then we will use the formula of the integration by parts and simplify the equation to get the integration of the required equation.

Complete step by step solution:
Given function is ∫sin⁡−1x\int {{{\sin }^{ - 1}}x} .
We can write the above equation as
I=∫(sin⁡−1x⋅1)dxI = \int {\left( {{{\sin }^{ - 1}}x \cdot 1} \right)dx}
Now we will use the basic formula of the integration by parts i.e. ∫(u×v)dx=u∫vdx−∫(u′∫vdx)dx\int {\left( {u \times v} \right)dx} = u\int {vdx} - \int {\left( {u'\int {vdx} } \right)dx} . Therefore by applying this we get
⇒I=∫(sin⁡−1x⋅1)dx=sin⁡−1x∫1dx−∫((ddxsin⁡−1x)∫1dx)dx\Rightarrow I = \int {\left( {{{\sin }^{ - 1}}x \cdot 1} \right)dx} = {\sin ^{ - 1}}x\int {1dx} - \int {\left( {\left( {\dfrac{d}{{dx}}{{\sin }^{ - 1}}x} \right)\int {1dx} } \right)dx}
We know that the differentiation of the sin⁡−1x{\sin ^{ - 1}}x is equal to 11−x2\dfrac{1}{{\sqrt {1 - {x^2}} }} i.e. ddxsin⁡−1x=11−x2\dfrac{d}{{dx}}{\sin ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}. Therefore by putting this value in the equation, we get
⇒I=sin⁡−1x∫1dx−∫((11−x2)∫1dx)dx\Rightarrow I = {\sin ^{ - 1}}x\int {1dx} - \int {\left( {\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right)\int {1dx} } \right)dx}
Now by using the simple integration we will get the simplified equation. Therefore, we get
⇒I=sin⁡−1x×x−∫((11−x2)×x)dx\Rightarrow I = {\sin ^{ - 1}}x \times x - \int {\left( {\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right) \times x} \right)dx}
⇒I=xsin⁡−1x−∫x1−x2dx\Rightarrow I = x{\sin ^{ - 1}}x - \int {\dfrac{x}{{\sqrt {1 - {x^2}} }}dx}
Now we will substitute t=1−x2t = 1 - {x^2} and by differentiating it we get dt=−2xdxdt = - 2xdx. From this we will get the value of xdxxdx i.e. xdx=−dt2xdx = - \dfrac{{dt}}{2}. Therefore, the equation becomes
⇒I=xsin⁡−1x−∫1t(−dt2)\Rightarrow I = x{\sin ^{ - 1}}x - \int {\dfrac{1}{{\sqrt t }}} \left( { - \dfrac{{dt}}{2}} \right)
⇒I=xsin⁡−1x+12∫1tdt\Rightarrow I = x{\sin ^{ - 1}}x + \dfrac{1}{2}\int {\dfrac{1}{{\sqrt t }}dt}
Now by using the simple integration we will get the simplified form of the above equation. Therefore, we get
⇒I=xsin⁡−1x+12(t−12+1−12+1)+C\Rightarrow I = x{\sin ^{ - 1}}x + \dfrac{1}{2}\left( {\dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}}} \right) + C
Adding and subtracting the terms, we get
⇒I=xsin⁡−1x+12(t1212)+C\Rightarrow I = x{\sin ^{ - 1}}x + \dfrac{1}{2}\left( {\dfrac{{{t^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right) + C
⇒I=xsin⁡−1x+12(2t)+C\Rightarrow I = x{\sin ^{ - 1}}x + \dfrac{1}{2}\left( {2\sqrt t } \right) + C
Now we will open the brackets in the equation. Therefore, we get
⇒I=xsin⁡−1x+t+C\Rightarrow I = x{\sin ^{ - 1}}x + \sqrt t + C
Substituting the value of tt in the equation, we get
⇒I=xsin⁡−1x+1−x2+C\Rightarrow I = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C

Hence the value of the given integration ∫sin⁡−1x\int {{{\sin }^{ - 1}}x} is equal to xsin⁡−1x+1−x2+Cx{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C.

Note:
Integration is defined as the summation of all the discrete data. Differentiation is the opposite of integration i.e. differentiation of the integration is equal to the value of the function or vice versa. Here we should note that we need to convert the given integrand such that we can apply a suitable integration formula. We should know the basic formula of the integration by parts of an equation to solve this question. We must not forget to put the constant term CC after the integration of an equation.