Question

Integrate the function.
$\int xco{s}^{-1}xdx.$

Answer 1

Let $I=\int xco{s}^{-1}xdx$
Put $co{s}^{-1}x=t\Rightarrow x=cost\Rightarrow dx=-sintdt$
$\therefore I=\int xco{s}^{-1}xdx=-\int tcost.sintdt=-\frac{1}{2}\int t.2sintcostdt=-\frac{1}{2}\int t.sin2tdt(\because 2sinxcosx=sin2x)$
On taking t as first function and sin 2t as second function and integrating by parts, we get
$I=-\frac{1}{2}[t\int sin2tdt-\int \left\{\frac{d}{dt}\right(t).\int sin2tdt\}dt]=-\frac{1}{2}[t\frac{(-cos2t)}{2}+\int 1.\frac{cos2t}{2}dt]=\frac{1}{4}tcos2t-\frac{1}{4}\int cos2tdt=\frac{1}{4}tcos2t-\frac{1}{4}\frac{sin2t}{2}+C=\frac{1}{4}tcos2t-\frac{1}{8}sin2t+C=\frac{1}{4}t(2co{s}^{2}t-1)-\frac{1}{8}.2sintcost+C[\because si{n}^{2}x+co{s}^{2}x=1\Rightarrow sinx=\sqrt{1-co{s}^{2}x}]\therefore \int xco{s}^{-1}xdx=\frac{1}{4}co{s}^{-1}x(2x^2-1)-\frac{1}{4}(1-x^2{)}^{\frac{1}{2}}x+C[putco{s}^{-1}x=tandcost=x]=\frac{1}{4}(2x^2-1)co{s}^{-1}x-\frac{1}{4}x\sqrt{1-x^2}+C$