Oberlaudacu

2021-12-28

Evaluate the following integrals.
${\int }_{0}^{\frac{1}{2}}\frac{{\mathrm{sin}}^{-1}x}{\sqrt{1-{x}^{2}}}dx$

### Answer & Explanation

soanooooo40

Step 1
Given:
$I={\int }_{0}^{\frac{1}{2}}\frac{{\mathrm{sin}}^{-1}x}{\sqrt{1-{x}^{2}}}dx$...(1)
for evaluating given integral we substitute
${\mathrm{sin}}^{-1}x=t$...(2)
if, x=0 then t=0
and
if, $x=\frac{1}{2}$ then $t=\frac{\pi }{6}$
now differentiating equation (2) with respect to x
so,

$\frac{1}{\sqrt{1-{x}^{2}}}=\frac{dt}{dx}$
$\frac{dx}{\sqrt{1-{x}^{2}}}=dt$
Step 2
now replacing ${\mathrm{sin}}^{-1}x$ with t, $\frac{dx}{\sqrt{1-{x}^{2}}}$ with dt and changing limits also in equation (1)
so,

$={\left[\frac{{t}^{2}}{2}\right]}_{0}^{\frac{\pi }{6}}$
$=\frac{1}{2}\left[{\left(\frac{\pi }{6}\right)}^{2}-{\left(0\right)}^{2}\right]$
$=\frac{1}{2}\left(\frac{{\pi }^{2}}{36}\right)$
$=\frac{{\pi }^{2}}{72}$
hence, given integral is equal to $\frac{{\pi }^{2}}{72}$.

Dabanka4v

Make the substitution, and find the corresponding limits in terms of u.

$a={\mathrm{sin}}^{-1}0=0$
$b={\mathrm{sin}}^{-1}\frac{1}{2}=\frac{\pi }{6}$
${\int }_{0}^{\frac{1}{2}}\frac{{\mathrm{sin}}^{-1}x}{\sqrt{1-{x}^{2}}}dx={\int }_{0}^{\frac{\pi }{6}}udu$
$={\left[\frac{1}{2}{u}^{2}\right]}_{0}^{\frac{\pi }{6}}$
$=\frac{1}{2}\left[\frac{{\pi }^{2}}{36}-0\right]$
$=\frac{{\pi }^{2}}{72}$
Result:
$\frac{{\pi }^{2}}{72}$

Vasquez

We need to evaluate the definite integral
${\int }_{0}^{\frac{1}{2}}\frac{{\mathrm{sin}}^{-1}x}{\sqrt{1-{x}^{2}}}dx$
We will use substitution.
Substitute $u={\mathrm{sin}}^{-1}x⇒du=\frac{1}{\sqrt{1-{x}^{2}}}dx;$
$⇒{\int }_{0}^{\frac{1}{2}}udu$
Integrate by applying $\int {x}^{n}=\frac{{x}^{n+1}}{n+1};$
$⇒\frac{{u}^{2}}{2}$
Revert the substitution and evaluate;
$⇒\frac{\left({\mathrm{sin}}^{-1}x{\right)}^{2}}{2}{|}_{0}^{\frac{1}{2}}=\frac{\left({\mathrm{sin}}^{-1}\frac{1}{2}{\right)}^{2}}{2}-\frac{\left({\mathrm{sin}}^{-1}0{\right)}^{2}}{2}$
$=\frac{\left(\frac{\pi }{6}{\right)}^{2}}{2}-\frac{{0}^{2}}{2}$
$=\frac{\frac{{\pi }^{2}}{36}}{2}$
$=\frac{{\pi }^{2}}{72}$
Result:
${\int }_{0}^{\frac{1}{2}}\frac{{\mathrm{sin}}^{-1}x}{\sqrt{1-{x}^{2}}}dx=\frac{{\pi }^{2}}{72}$

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