Question

Solve the following definite integral :-

$\sf {\displaystyle \sf \int \limits^{1/2}_ {0}\ \dfrac{dx}{(1\ +\ x^2) \sqrt{1\ -\ x^2}}}$

Answer 1

Step-by-step explanation:

Topic :-

Definite Integrals

Given :-

We are given that,

$\sf : \implies \Large {\displaystyle \sf \int \limits^{1/2}_ {0}\ \footnotesize \dfrac{dx}{(1\ +\ x^2) \sqrt{1\ -\ x^2}}}$

To find :-

We have to find the value of the given definite integral.

Solution :-

In order to solve this definite integral, we have to substitute the value of x = 1/t and then further, we will apply the integration formulae given as below :-

$\sf : \implies \Large {\displaystyle \sf \int \limits^{x}_ {0}\ \footnotesize \dfrac{dx}{x^2\ +\ a^2}\ =\ \dfrac{1}{a}\ tan^{-1}\ \dfrac{x}{a}}$

Calculations :-

Given definite integral in the question,

$\sf : \implies I\ =\ \Large {\displaystyle \sf \int \limits^{1/2}_ {0}\ \footnotesize \dfrac{dx}{(1\ +\ x^2) \sqrt{1\ -\ x^2}}}$

By putting the value of x = 1/t and differentiating it, dx = -dt/t², we will get,

$\sf : \implies \Large {\displaystyle \sf \int \limits^{1/2}_ {0}\ \footnotesize \dfrac{\dfrac{-dt}{t^2}}{\bigg(1\ +\ \dfrac{1}{t^2} \bigg)\ \sqrt{1\ -\ \dfrac{1}{t^2}}}}$

Now, by taking LCM in denominator and solving it, we will get,

$\sf : \implies \Large {\displaystyle \sf \int \limits^{1/2}_ {0}\ \footnotesize \dfrac{\dfrac{-dt}{t^2}}{\bigg(\dfrac{t^2\ +\ 1}{t^2} \bigg)\ \sqrt{\dfrac{t^2\ -\ 1}{t^2}}}}$

On further solving it, it will be deduced to,

$\sf : \implies \Large {\displaystyle \sf \int \limits^{1/2}_ {0}\ \footnotesize \dfrac{-tdt}{(1\ +\ t^2) \sqrt{t^2\ -\ 1}}}$

In this case, we need to have further substitution by substituting t² - 1 = u²,

$\sf : \implies {t^2\ =\ u^2\ +\ 1}$

Differentiating it we will find, tdt = udu. Now, substituting all these values, the integral will become,

$\sf : \implies \Large {\displaystyle \sf \int \limits^{1/2}_ {0}\ \footnotesize \dfrac{-udu}{(1\ +\ u^2\ +\ 1) \sqrt{u^2}}}$

As, we will solve the integral in terms of u, we will determine,

$\sf : \implies \Large {\displaystyle \sf \int \limits^{1/2}_ {0}\ \footnotesize \dfrac{-du}{u^2\ +\ 2}}$

Now, we will use the the integration formulae which can be stated as,

$\sf : \implies \Large {\displaystyle \sf \int \limits^{x}_ {0}\ \footnotesize \dfrac{dx}{x^2\ +\ a^2}\ =\ \dfrac{1}{a}\ tan^{-1}\ \dfrac{x}{a}}$

So, our integral will be in terms of u will become,

$\sf : \implies {I\ =\ \dfrac{1}{\sqrt{2}}\ tan^{-1}\ \dfrac{(-u)}{\sqrt{2}} \bigg|^{1/2}_ {0}}$

$\sf : \implies {\dfrac{-1}{\sqrt{2}}\ tan^{-1}\ \dfrac{\sqrt{t^2\ -\ 1}}{\sqrt{2}} \bigg|^{1/2}_ {0}}$ [On putting the value of u],

Putting t² = 1/x² and we will get,

$\sf : \implies {\dfrac{-1}{\sqrt{2}}\ tan^{-1}\ \dfrac{\sqrt{1\ -\ x^2}}{x \sqrt{2}} \bigg|^{1/2}_ {0}}$

Applying both upper and lower limits and due to (-) term, we will get,

$\sf : \implies {I\ =\ \dfrac{1}{\sqrt{2}}\ tan^{-1} \sqrt{\dfrac{2}{3}}}$

Answer :-

Hence, the value of the definite integral is,

$\sf : \implies {\dfrac{1}{\sqrt{2}}\ tan^{-1} \sqrt{\dfrac{2}{3}}}$

Note :-

▪︎To solve the definite integrals of the form $\sf {\dfrac{1}{(x^2\ +\ a^2)\ (\sqrt{x^2\ -\ a^2})}}$ , we have to suppose the square root term be another variable and then we will further solve it.

▪︎It can be solved by another type of substitution i.e, x = tana

$\: \: \: \: \sf {\dfrac{1}{(x^2\ +\ a^2)\ (\sqrt{x^2\ -\ a^2})}}$