Question

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

Answer 1

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