Question

integrate (x sin^-1 x)/(√{1-x²}) dx

simply,

$\displaystyle\sf\int\dfrac{x\:sin^{-1}\:x}{\sqrt{1-x^2}}\:dx$​

Answer 1

kindly check the attached pictures for stepwise solution and answer.

Additional Information :

$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}</p><p></p><p></p><p>$

$\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\;\dfrac{d}{dx}(sin\;x)=cosx \\\\ \circ \;\dfrac{d}{dx}(cos\;x) = -sinx \\\\ \circ \; \dfrac{d}{dx}(tan\;x) = sec^{2}x \\\\ \circ\; \dfrac{d}{dx}(cot\;x) = -csc^{2}x \\\\ \circ \; \dfrac{d}{dx}(sec\;x) = secx \cdot tanx \\\\ \circ \; \dfrac{d}{dx}(csc\;x) = -cscx \cdot cotx \\\\ \circ\; \dfrac{d}{dx}(sinh\;x)=coshx \\\\ \circ\; \dfrac{d}{dx}(cosh\;x)= sinhx \\\\ \circ\;\dfrac{d}{dx}(tanh\;x)=sech^{2}h \\\\ \circ\;\dfrac{d}{dx}(coth\;x)=-csch^{2}x \\\\ \circ\;\dfrac{d}{dx}(sech\;x) =-sechx \cdot tanhx \\\\ \circ\;\dfrac{d}{dx}(csch\;x) = -cschx \cdot cothx\end{minipage}}}</p><p></p><p></p><p>$

Answer 2

${ \underline {\sf{ \large{Solution}}}}$

We Have,

$\displaystyle\sf\int\dfrac{x\:sin^{-1}\:x}{\sqrt{1-x^2}}\:dx$

Let,

${ \displaystyle \rm{t = \sin^{ - 1}x }}$

${ \implies\displaystyle\rm{ \frac{dt}{dx} = \frac{1}{ \sqrt{1 - x^{2} } } }}$

${ \implies\displaystyle\rm{ dt= \frac{x}{ \sqrt{1 - x^{2} } } }}$

While,

${ \implies \displaystyle \rm{x = \sin t }}$

Then,

${ \implies \displaystyle \rm{I = \int t \sin t \: dt}}$

${ \implies \displaystyle \rm{t \int t \sin t \: dt - \int \: \bigg( \frac{d}{dt} (t \bigg) \int \sin t \: dt)dt}}$

Using Integration of Parts,

${ \implies \displaystyle \rm{t( - \cos \: t) - \int \: 1.( \cos t )\: dt}}$

${ \implies \displaystyle \rm{ - t( \cos \: t) + \int \: \cos t \: dt}}$

${ \implies \displaystyle \rm{ - t \cos t + \sin t + c}}$

Now,

${ \boxed{ \rm{ \cos t = \sqrt{1 - \sin ^{2}t } = \sqrt{1 - x^{2} } }}}$

So,

${ \implies \displaystyle \rm{I=-sin^{-1}x\sqrt{1-x^2+C}}}$

Thus,

${ \implies { \boxed{ \rm{x - \sin \: ^{ - 1} x \sqrt{1 - x ^{2} + c } }}}}$

________________________

Additional Information :

$\bigstar \: \underline{ \boxed{ \rm\dfrac{d}{dx}( {x}^{n} ) =n {x}^{n - 1} }}$

$\bigstar \:\underline{ \boxed{ \rm\dfrac{d}{dx}( C ) =0}}$

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}(C \times f) = C\dfrac{d}{dx}f}}$

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}(f \pm g) = \dfrac{d}{dx}f \pm \dfrac{d}{dx}g}}$

$\bigstar \boxed{\displaystyle \rm \int {x}^{n} \: dx = \dfrac{ {x}^{n + 1} }{n + 1} + C}$

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}(fg) =f \dfrac{d}{dx}g + g \dfrac{d}{dx}f }}$