Question

Solve given definite integral.

${\int }_{1/4}^{1/2} \frac{1}{x^2\sqrt{1-x^2}}dx$

Short answer

$-\sqrt{3}+\sqrt{15}$

Step-by-step solution

Step1. Given Information

The integral is as follows.

${\int }_{\frac{1}{4}}^{\frac{1}{2}} \frac{1}{x^2\sqrt{1-x^2}}dx$

The objective is to solve the integral.

Step2. Assumption

To solve the integral, let $x=\sin \left(u\right)$

Now, derivation of above equation is solved below.

$x=\sin \left(u\right)$

$dx=\cos \left(u\right)du$

Step3. Solution

Now use the identity, $-{\sin }^2\left(u\right)+1={\cos }^2\left(u\right)$

So,

$\frac{1}{x^2\sqrt{1-x^2}}=\frac{1}{\sqrt{-{\sin }^2\left(u\right)+1}{\sin }^2\left(u\right)}$

$\begin{array}{r}=\frac{1}{\sqrt{{\cos }^2\left(u\right)}{\sin }^2\left(u\right)}\\ =\frac{1}{\cos \left(u\right){\sin }^2\left(u\right)}\end{array}$

Step4. Solution

The integral is solved below

$\begin{array}{r}\int \frac{1}{x^2\sqrt{1-x^2}}dx\\ =\int \left(\frac{1}{{\sin }^2\left(u\right)}\right)du\\ =\int {\csc }^2\left(u\right)du\\ =-\cot \left(u\right)\\ =-\cot \left({\sin }^{-1}\left(x\right)\right)\\ =-\frac{1}{x}\sqrt{-x^2+1}\end{array}$

Step5. Solution

The definite integral is solved below.

${\int }_{\frac{1}{4}}^{\frac{1}{2}} \frac{1}{x^2\sqrt{1-x^2}}dx$

$={\left[-\frac{1}{x}\sqrt{-x^2+1}\right]}_{\frac{1}{4}}^{\frac{1}{2}}$

$=\left[-\frac{1}{\left(\frac{1}{2}\right)}\sqrt{-{\left(\frac{1}{2}\right)}^2+1}\right]-\left[-\frac{1}{\left(\frac{1}{4}\right)}\sqrt{-{\left(\frac{1}{4}\right)}^2+1}\right]$

$=-\sqrt{3}+\sqrt{15}$

Therefore, the value is$=-\sqrt{3}+\sqrt{15}$