Question

If we have an integral function as \[f(x) = \int {\dfrac{{dx}}{{\left[ {{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}} \right]}}} \].and $f\left( 0 \right) = 0$, then the value of $f\left( 1 \right)$is

Answer 1

Hint: To solve this problem you should know basic trigonometrical and integral formulae. In this problem we are going to use a substitution method to solve it. In the substitution method we substitute the given variable by any other variable to make it simple.
Trigonometric Identities:
1. $\dfrac{1}{{\sec \theta }} = \cos \theta $
2. $1 + {\tan ^2}\theta = {\sec ^2}\theta $

Complete step-by-step solution:
Given function: \[f(x) = \int {\dfrac{{dx}}{{\left[ {{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}} \right]}}} \] ……………………. (1)
To solve this problem:
As in the given question, the denominator is ${\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}}$ .So in order to simplify the given function,
Let, $x = \tan y$
Differentiating both sides, we get
\[dx = {\sec ^2}ydy\]
Putting $x = \tan y$ and \[dx = {\sec ^2}ydy\] in equation (1),
$f(\tan y) = \int {\dfrac{{{{\sec }^2}ydy}}{{{{\left( {1 + {{\tan }^2}y} \right)}^{\dfrac{3}{2}}}}}} $
Using the trigonometric identity $1 + {\tan ^2}\theta = {\sec ^2}\theta $ ,we get
 $f(\tan y) = \int {\dfrac{{{{\sec }^2}ydy}}{{{{\left( {{{\sec }^2}y} \right)}^{\dfrac{3}{2}}}}}} $
Solving powers in the denominator, we get
$f(\tan y) = \int {\dfrac{{{{\sec }^2}ydy}}{{{{\sec }^3}y}}} $
Dividing numerator and denominator by ${\sec ^2}y$ ,we get
 $f(\tan y) = \int {\dfrac{{dy}}{{\sec y}}} $
Using another trigonometric identity $\dfrac{1}{{\sec \theta }} = \cos \theta $ ,we get
$f(\tan y) = \int {\cos ydy} $
Integrating $\cos y$
$f(\tan y) = \sin y + c$
Now, putting the value of y in above function, we get
$f(x) = \sin \left( {{{\tan }^{ - 1}}x} \right) + c$ …………………… (2)
But it is given in the question that $f\left( 0 \right) = 0$. So
$f(0) = \sin \left( {{{\tan }^{ - 1}}0} \right) + c$
 $0 = \sin 0 + c$
c=0
Putting c=0 in equation (2), we get
$f(x) = \sin \left( {{{\tan }^{ - 1}}x} \right)$
Now, according to the given question we need to find $f\left( 1 \right)$.So putting $f = 1$ in above equation, we get
$f(1) = \sin \left( {{{\tan }^{ - 1}}1} \right)$
Putting value of ${\tan ^{ - 1}}1$, we get
 $f(1)=\sin {45}^{\circ}$
Putting the value of $\sin {45^ \circ }$ ,we get
$f\left( 1 \right) = \dfrac{1}{{\sqrt 2 }}$
Hence from the above calculation, we get
$f\left( 1 \right) = \dfrac{1}{{\sqrt 2 }}$

Note: As we can see from the above solution, we had put $x = \tan y$ in the given integral equation in order to simplify it before integrating.
Alternatively, we can also put $x = \cot y$ in order to simplify it. When we shall put $x = \cot y$ in the given integral equation we will get our integrand in the form of $\cos ecx$.