Question

If \[{I_1} = \int {{{\sin }^{ - 1}}} xdx\] and \[{I_2} = \int {{{\sin }^{ - 1}}} \sqrt {\left( {1 - {x^2}} \right)} dx\], then choose the correct option.
A. \[{I_1} = {I_2}\]
B. \[{I_2} = \dfrac{\pi }{{2{I_1}}}\]
C. \[{I_1} + {I_2} = \left( {\dfrac{\pi }{2}} \right)x + C\]
D. \[{I_1} + {I_2} = \dfrac{\pi }{2}\]
E. None of these

Answer 1

HintThis is a simple formula-based question, we need to substitution procedure and simplify the given expression \[{\sin ^{ - 1}}\sqrt {\left( {1 - {x^2}} \right)} \] and use integration addition and we will get the required solution.

Formula used:
We substitute \[{\sin ^{ - 1}}\sqrt {\left( {1 - {x^2}} \right)} = \theta \]
Integration addition formula \[\int {f(x)dx} + \int {g(x)dx} = \int {\left( {f(x) + g(x)} \right)dx} \]
Trigonometric summation formula \[{\sin ^2}x + {\cos ^2}x = 1\]
Inverse formula of trigonometry \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\]

Complete step by step Solution:
We are given the equations
\[{I_1} = \int {{{\sin }^{ - 1}}} xdx\]……………..(1)
And
\[{I_2} = \int {{{\sin }^{ - 1}}} \sqrt {\left( {1 - {x^2}} \right)} dx\] ………………(2)
Now we substitute \[{\sin ^{ - 1}}\sqrt {\left( {1 - {x^2}} \right)} = \theta \] and simplify
\[{\sin ^{ - 1}}\sqrt {\left( {1 - {x^2}} \right)} = \theta \]
Taking \[\sin \] both side of the above equation and we get
\[ \Rightarrow \sin \left( {{{\sin }^{ - 1}}\sqrt {\left( {1 - {x^2}} \right)} } \right) = \sin \theta \]
\[ \Rightarrow \sqrt {\left( {1 - {x^2}} \right)} = \sin \theta \]
\[ \Rightarrow \left( {1 - {x^2}} \right) = {\sin ^2}\theta \]
\[ \Rightarrow {x^2} = 1 - {\sin ^2}\theta \]………….(3)
We know that \[{\sin ^2}x + {\cos ^2}x = 1\]
Substitute \[1 - {\sin ^2}x = {\cos ^2}x\] in above equation (3) and we get
\[ \Rightarrow {x^2} = {\cos ^2}\theta \]
Taking square root in both sides of the above equation and we get
\[ \Rightarrow x = \cos \theta \]
Taking \[{\cos ^{ - 1}}\] in both sides of the above equation and we get
\[ \Rightarrow {\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {\cos \theta } \right)\]
\[ \Rightarrow {\cos ^{ - 1}}x = \theta \]
Now equation (2), we get
\[{I_2} = \int {{{\sin }^{ - 1}}} \sqrt {\left( {1 - {x^2}} \right)} dx\]
\[{I_2} = \int \theta dx\]
Substitute \[{\cos ^{ - 1}}x = \theta \] and we get
\[ \Rightarrow {I_2} = \int {{{\cos }^{ - 1}}} xdx\]
Further, as we can see the \[{{\rm{I}}_1}\] and \[{{\rm{I}}_2}\] are not equal, so option A is not correct.
Next, we can confirm option B is incorrect just by comparing the derived value with the provided one.
Next, we will add \[{{\rm{I}}_1}\]and \[{{\rm{I}}_2}\], we get
\[{I_1} + {I_2} = \int {{{\sin }^{ - 1}}} xdx + \int {{{\cos }^{ - 1}}} xdx\]
Use the formula of integration addition and we get
\[ = \int {\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right)dx} \]……………..(4)
From the inverse trigonometric formula, we get \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\]
Substitute this value in equation (4) and we get
\[ = \int {\dfrac{\pi }{2}dx} \]
\[ = \left( {\dfrac{\pi }{2}} \right)\int d x\]
Integrating and we get
\[\left( {\dfrac{\pi }{2}} \right)x + C\], where \[C\] is the constant of integration.

Option C is the correct option.

Note: Many students make mistake with the substitution. After taking \[{\sin ^{ - 1}}\sqrt {\left( {1 - {x^2}} \right)} = \theta \], if we try to find \[d\theta \] for doing the integration directly then this solution becomes much lengthier. We need to simplify the substitution part first and then integrating the given part.