Question

Find the value of k if the value of integral \[\int\limits_0^{\dfrac{1}{2}} {\dfrac{{{x^2}}}{{{{\left( {1 - {x^2}} \right)}^{\dfrac{3}{2}}}}}} dx\] is \[\dfrac{k}{6}\].
A. \[2\sqrt 3 + \pi \]
B. \[3\sqrt 2 + \pi \]
C. \[3\sqrt 2 - \pi \]
D. \[2\sqrt 3 - \pi \]

Answer 1

Hint: In the given integral, substitute \[x\] with \[\sin \theta \] and then solve the definite integral further to get the answer of integral. After that, we will compare with the given value then we will get the required answer.

Formula used:
Integration formula \[\int\limits_a^b {xdx} = \left[ {\dfrac{{{x^2}}}{2}} \right]_a^b\]
Integration of \[{\tan ^2}x\] is \[\tan x - x\] . i.e., \[\int {{{\tan }^2}x} = \tan x - x + c\] , c is integrating constant.

Complete step by step solution:
The given integral is:
\[\int\limits_0^{\dfrac{1}{2}} {\dfrac{{{x^2}}}{{{{\left( {1 - {x^2}} \right)}^{\dfrac{3}{2}}}}}} dx\]
Put \[x = \sin \theta \], then \[dx = \cos \theta d\theta \]
Also, the lower limit becomes \[0\] and the upper limit becomes \[\dfrac{\pi }{6}\]
Therefore, the expression becomes:
\[ = \int\limits_0^{\dfrac{1}{2}} {\dfrac{{{{\left( {\sin \theta } \right)}^2}}}{{{{\left( {1 - {{\left( {\sin \theta } \right)}^2}} \right)}^{\dfrac{3}{2}}}}}} \cos \theta d\theta \]
\[ = \int\limits_0^{\dfrac{1}{2}} {\dfrac{{{{\left( {\sin \theta } \right)}^2}}}{{{{\left( {{{\cos }^2}\theta } \right)}^{\dfrac{3}{2}}}}}} \cos \theta d\theta \]
\[ = \int\limits_0^{\dfrac{\pi }{6}} {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^3}\theta }}} \cos \theta d\theta \]
Solving further, we get:
 \[ = \int\limits_0^{\dfrac{\pi }{6}} {{{\tan }^2}} \theta d\theta \]
Now, putting limits, we get:
= \[\left[ {\tan \theta - \theta } \right]_0^{\dfrac{\pi }{6}}\]
\[ \Rightarrow \left( {\dfrac{1}{{\sqrt 3 }}} \right) - \left( {\dfrac{\pi }{6}} \right)\]
According to the question, The answer of integral is \[\dfrac{k}{6}\]
Now, equating \[ \Rightarrow \left( {\dfrac{1}{{\sqrt 3 }}} \right) - \left( {\dfrac{\pi }{6}} \right)\] with \[\dfrac{k}{6}\], we get:
\[ \Rightarrow \left( {\dfrac{1}{{\sqrt 3 }}} \right) - \left( {\dfrac{\pi }{6}} \right)\] = \[\dfrac{k}{6}\]
\[k = 2\sqrt 3 - \pi \]

The correct answer is option D.

Note: By altering the independent variable \[x\] to \[z\], an integer provided as \[f\left( x \right)dx\] can be transformed into another form in the integration by substitution. To do this, simply substitute\[x = k\left( z \right)\]. Think about \[I = f\left( x \right)dx\]. In order for \[\dfrac{{dx}}{{dz}} = k'\left( z \right)\]or\[dx = k'\left( z \right)dz\], replace \[x\] with \[k\left( z \right)\]. I thus equals \[\int {f\left( x \right)} dx = f\left[ {k\left( z \right)k'\left( z \right)dz} \right]\]