Evaluate:$\int\limits_0^1 x\sqrt{\large\frac{1-x^2}{1+x^2}}$$dx 1 Answer Toolbox: • \int \limits _a^b f(x)dx=F(b)-F(a) Step 1: Given \int\limits_0^1 x\sqrt{\large\frac{1-x^2}{1+x^2}}$$dx$
Let $x^2=t$ on differentiating w.r.t x we get,
$2xdx=dt \qquad => xdx=\large\frac{dt}{2}$
on substituting t and dt,
Therefore $I=\int\limits_0^1 \sqrt{\large\frac{1-t}{1+t}}$$dt/2 =\large\frac{1}{2} \int\limits_0^1 \sqrt{\frac{1-t}{1+t}} Multiply and divide by the conjugate 1-t I=\large\frac{1}{2} \int\limits_0^1 \sqrt {\frac{(1-t)(1-t)}{(1+t)(1-t)}}$$dt=\large\frac{1}{2}$$\int\limits_0^1 \sqrt {(1-t)^2}dt =\large\frac{1}{2} \int\limits_0^1 \frac{(1-t)}{\sqrt {1-t^2}}$$dt$
Step 2:
On seperating the term we get
$\large\frac{1}{2} \bigg\{\bigg[\int\limits_0^1 \frac{1}{\sqrt {1-t^2}}$$dt\bigg]-\bigg[\int\limits_0^1 \large\frac{t}{\sqrt {1-t^2}}$$dt \bigg] \bigg\}$
Consider $\large\frac{t}{\sqrt {1-t^2}}$. Here let $1-t^2=u$
on differentiating we get $-2tdt=du$
$=>tdt=\large\frac{-du}{2}$
On substituting for u and du
we know we substitute u, the limit of t also changes
when t=0 and when t=1
when u=1 and when u=0
Therefore $\int\limits_0^1 \large\frac{t}{\sqrt {1-t^2}}$$dt=-\int\limits_0^1\large\frac{du/2}{\sqrt u} The negative symbol can be removed by changing the limits hence \int\limits_0^1 \large\frac{t}{\sqrt {1-t^2}}$$dt =\large\frac{1}{2}\int\limits_0^1 \frac{du}{\sqrt u}$
Therefore $I=\large\frac{1}{2}\bigg[\int\limits_0^1 \frac{t}{\sqrt {1-t^2}} $$dt -\large\frac{1}{2}\int\limits_0^1 \frac{du}{\sqrt u}\bigg] Step 3: =\large\frac{dx}{\sqrt {a^2-x^2}}=$$\sin ^{-1}(x/a)$
Therefore $I=\large\frac{1}{2} $$[\sin^{-1}(t)]_0^1-\large\frac{1}{4}\bigg[\frac{u^{-1/2+1}}{-1/2+1}\bigg]_0^1 =\large\frac{1}{2}$$(sin ^{-1}(t))_0^1-\large\frac{1}{4}$$(-2 \sqrt u)_0^1 on applying the limits I=\large\frac{1}{2}$$[\sin^{-1}(1)-\sin^{-1}(0)]+\large\frac{2}{4}[(\sqrt 1)-0]$
But $\sin^{-1}(1)=\large\frac{\pi}{2}$
Therefore $I=\large\frac{1}{2}$$(\pi/2)+\large\frac{2}{4}$
$=\large\frac{\pi}{4}+\frac{1}{2}$