BeMath-2x-cos-1-x-dx-

Question Number 108095 by bemath last updated on 14/Aug/20

\frac{\infty B e M a t h \infty}{♠}

\int 2 x \cos^{- 1} (x) d x

Answered by bemath last updated on 14/Aug/20

Answered by Dwaipayan Shikari last updated on 14/Aug/20

x^{2} {c o s}^{- 1} x + \int \frac{x^{2}}{\sqrt{1 - x^{2}}} d x

x^{2} {c o s}^{- 1} x - \int \frac{1 - x^{2}}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - x^{2}}}

x^{2} {c o s}^{- 1} x - \int \sqrt{1 - x^{2}} + {s i n}^{- 1} x

x^{2} {c o s}^{- 1} x + {s i n}^{- 1} x - \frac{x}{2} \sqrt{1 - x^{2}} - \frac{1}{2} {s i n}^{- 1} x + C

x^{2} {c o s}^{- 1} x + \frac{1}{2} {s i n}^{- 1} x - \frac{x}{2} \sqrt{1 - x^{2}} + C

Answered by mathmax by abdo last updated on 14/Aug/20

I = 2 \int x arcosx dx by parts

I = 2 {\frac{x^{2}}{2} arcosx + \int \frac{x^{2}}{2} \frac{dx}{\sqrt{1 - x^{2}}}} = x^{2} arcosx + \int \frac{x^{2}}{\sqrt{1 - x^{2}}} dx} we have

\int \frac{x^{2}}{\sqrt{1 - x^{2}}} dx =_{x = sint} \int \frac{\sin^{2} t}{cost} cost dt = \int \sin^{2} t dt

= \int \frac{1 - \cos (2 t)}{2} dt = \frac{t}{2} - \frac{1}{4} \sin (2 t) + C

= \frac{t}{2} - \frac{1}{2} sint cost + C = \frac{arcsinx}{2} - \frac{x}{2} \sqrt{1 - x^{2}} + C \Rightarrow

I = x^{2} arcosx + \frac{arcsinx}{2} - \frac{x}{2} \sqrt{1 - x^{2}} + C