$∫\frac{xsin^{-1}x}{\sqrt{1-x^2}}dx=$ |
$\sqrt{1-x^2}cos \, x+ x +C$ where C is arbitrary constant $\sqrt{1-x^2}sin \, x+ x +C$ where C is arbitrary constant $\sqrt{1-x^2}sin^{-1} \, x+ x +C$ where C is arbitrary constant $-\sqrt{1-x^2}sin^{-1} \, x+ x +C$ where C is arbitrary constant |
$-\sqrt{1-x^2}sin^{-1} \, x+ x +C$ where C is arbitrary constant |
The correct answer is Option (4) → $-\sqrt{1-x^2}sin^{-1} \, x+ x +C$ where C is arbitrary constant $∫\frac{x\sin^{-1}x}{\sqrt{1-x^2}}dx$ let $y=\sin^{-1}x⇒\sin y =x$ $dy=\frac{dx}{\sqrt{1-x^2}}$ $\int y\sin y dy$ Using $\int uvdx=u\int vdx=\int u'\int v dxdx$ $=y\int \sin ydy-\int\frac{d(y)}{dy}\int\sin ydydy$ $=-y\cos y+\int\cos y dy$ $=-y\cos y+\sin y+C$ $≡-\sqrt{1-x^2}\sin^{-1}+x+C$ |