Question

Solve the following integral \[\int {{{\sin }^{ - 1}}\left( {2x} \right)dx} \].

Answer 1

Hint: n this question we have to solve the given integral. In order to solve this question, first of all we will assume \[{\sin ^{ - 1}}2x = t\] . After that we will differentiate it and transfer the given integral in terms of \[t\] . After that we will use the concept of integration by parts i.e., \[\int {\left( {u \cdot v} \right)dx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } } {\text{ }}dx\] where \[u\] is the first function and \[v\] is the second function, to solve the given integral. And finally, we will substitute the value of \[t\]. Hence, we get the required result.

Complete step by step answer:
We have to solve the integral: \[\int {{{\sin }^{ - 1}}\left( {2x} \right)dx} \]
Let us consider the given integral as,
\[I = \int {{{\sin }^{ - 1}}\left( {2x} \right)dx} {\text{ }} - - - \left( i \right)\]
Now let us assume
\[{\sin ^{ - 1}}2x = t{\text{ }} - - - \left( {ii} \right)\]
Now we know that
If \[{\sin ^{ - 1}}x = a\] then \[x = \sin a\]
Therefore, from equation \[\left( {ii} \right)\] we have
\[2x = \sin t{\text{ }} - - - \left( {iii} \right)\]
\[ \Rightarrow x = \dfrac{{\sin t}}{2}{\text{ }} - - - \left( {iv} \right)\]
Now on differentiating equation \[\left( {iv} \right)\] with respect to \[x\] we get
\[dx = \dfrac{{\cos t}}{2}dt\]
Now on substituting the values in the equation \[\left( i \right)\] we get
\[I = \int {t \cdot \dfrac{{\cos t}}{2}dt} \]
\[ \Rightarrow I = \dfrac{1}{2}\int {t \cdot \cos tdt} {\text{ }} - - - \left( v \right)\]

Now using integration by parts,
We know that
\[\int {\left( {u \cdot v} \right)dx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } } {\text{ }}dx\]
From equation \[\left( v \right)\] we have
\[u = t\] and \[v = \cos t\]
Therefore, we have
\[I = \dfrac{1}{2}\int {\left( {t \cdot \cos t} \right)dt{\text{ }} = {\text{ }}\dfrac{1}{2}\left[ {t\int {\cos tdt - \int {\left( {\dfrac{{dt}}{{dt}}\int {\cos tdt} } \right)} } dt} \right]} + c\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {t\int {\cos tdt - \int {\left( {\dfrac{{dt}}{{dt}}\int {\cos tdt} } \right)} } dt} \right] + c{\text{ }} - - - \left( {vi} \right)\]

Now we know that
\[\int {\cos t} {\text{ }}dt = \sin t\]
\[\dfrac{{dt}}{{dt}} = 1\]
Therefore, from the equation \[\left( {vi} \right)\] we get
\[ \Rightarrow I = \dfrac{1}{2}\left[ {t \cdot \sin t - \int {1 \cdot \sin t{\text{ }}dt} } \right] + c\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {t \cdot \sin t - \int {\sin t{\text{ }}dt} } \right] + c\]
We know that
\[\int {\sin t} {\text{ }}dt = - \cos t\]
Therefore, we get
\[ \Rightarrow I = \dfrac{1}{2}\left[ {t \cdot \sin t - \left( { - \cos t} \right)} \right] + c\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {t \cdot \sin t + \cos t} \right] + c\]

Now back substituting the value from equation \[\left( {ii} \right)\] and equation \[\left( {iii} \right)\] we get
\[ \Rightarrow I = \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\left( {2x} \right) \cdot 2x + \cos t} \right] + c{\text{ }} - - - \left( {vii} \right)\]
As we know that
\[\cos x = \sqrt {1 - {{\sin }^2}x} \]
Therefore, \[\cos t = \sqrt {1 - {{\sin }^2}t} \]
\[ \Rightarrow \cos t = \sqrt {1 - 4{x^2}} \]
Therefore, from equation \[\left( {vii} \right)\] we get
\[ \therefore I = \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\left( {2x} \right) \cdot 2x + \sqrt {1 - 4{x^2}} } \right] + c{\text{ }}\]

Hence, the value of the integral \[\int {{{\sin }^{ - 1}}\left( {2x} \right)dx} \] is \[\dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\left( {2x} \right) \cdot 2x + \sqrt {1 - 4{x^2}} } \right] + c{\text{ }}\].

Note: The given integral is an indefinite integral. So, always remember while calculating the indefinite integral never forget to add constant \[c\] in the final result. Also remember you cannot choose the first and second functions in the formula of integration by parts as you like. Always choose them by using “ILATE” where
-$I$ stands for Inverse trigonometric functions
-$L$ stands for logarithmic functions
-$A$ stands for Algebraic functions
-$T$ stands for Trigonometric functions
-$E$ stands for exponent function