Question

Evaluate the integral\(\int {{{\sin }^{ - 1}}\sqrt x .dx} \)

Short answer

\(\int {{{{\mathop{\rm Sin}\nolimits} }^{ - 1}}\sqrt x .dx = } \) -\(\frac{{{{\sin }^{ - 1}}\sqrt x (1 - 2x)}}{2}\)+\(\frac{{\sqrt x \sqrt {1 - x} }}{2} + c\) is answer of integral

Step-by-step solution

Step  1- To find the integral \(\int {{{\sin }^{ - 1}}\sqrt x .dx} \)

Let \(\sqrt x = {\mathop{\rm Sin}\nolimits} t\)

\(x = {{\mathop{\rm Sin}\nolimits} ^2}t\)

\(dx = 2{\mathop{\rm Sin}\nolimits} t*{\mathop{\rm Cos}\nolimits} t*dt\),\(t = {{\mathop{\rm Sin}\nolimits} ^{ - 1}}\sqrt x \)

Using\({{\mathop{\rm Sin}\nolimits} ^{ - 1}}(\sin t) = t\)and \(\)\(2{\mathop{\rm Sin}\nolimits} t*{\mathop{\rm Cos}\nolimits} t = {\mathop{\rm Sin}\nolimits} 2t\)

=\(\int {t*{\mathop{\rm Sin}\nolimits} 2t.dt} \)

Where t is first function and \({\mathop{\rm Sin}\nolimits} 2t\) is second function

Step  2-   Using By Parts Method that is ILATE

=\(\begin{aligned}{l}t\int {{\mathop{\rm Sin}\nolimits} 2t.dt} - \int {(\frac{{dt}}{{dt}}\int {{\mathop{\rm Sin}\nolimits} 2t} } ).dt\\\end{aligned}\)

=(-)\(\begin{aligned}{l}\frac{{{\mathop{\rm Cos}\nolimits} 2t}}{2} - \int 1 *(\frac{{{\mathop{\rm Cos}\nolimits} 2t}}{2}).dt\\\end{aligned}\)

=\(( - )t*\frac{{{\mathop{\rm Cos}\nolimits} 2t}}{2} + \frac{{{\mathop{\rm Sin}\nolimits} 2t}}{4} + c\)

Using,

\(\begin{aligned}{l}{\mathop{\rm Cos}\nolimits} 2t = 1 - 2{{\mathop{\rm Sin}\nolimits} ^2}t,\\{\mathop{\rm Sin}\nolimits} 2t = 2{\mathop{\rm Sin}\nolimits} t*{\mathop{\rm Cos}\nolimits} t\\\end{aligned}\)

=\(\frac{{{{\sin }^{ - 1}}\sqrt x (1 - 2x)}}{2} + \)\(\frac{{2\sqrt x \sqrt {1 - x} }}{4} + c\)
Hence, \(\int {{{{\mathop{\rm Sin}\nolimits} }^{ - 1}}\sqrt x .dx = } \) -\(\frac{{{{\sin }^{ - 1}}\sqrt x (1 - 2x)}}{2}\)+\(\frac{{\sqrt x \sqrt {1 - x} }}{2} + c\)