Question

The value of $\int\limits_{0}^{1}{\dfrac{dx}{x+\sqrt{1-{{x}^{2}}}}}$ is
A. $\dfrac{\pi }{3}$
B. $\dfrac{\pi }{2}$
C. $\dfrac{1}{2}$
D. $\dfrac{\pi }{4}$

Answer 1

Hint: In this question, we are to find the given integral. For this, the variable substitution method is applied in the given integral. So, that the required integral will be obtained.



Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ (lower limit) and $b$ (upper limit).
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a3) $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
  & \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
 & \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$
6) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\dfrac{\pi }{4}}}$
7) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\tan x)}{f(\tan x)+f(\cot x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cot x)}{f(\tan x)+f(\cot x)}dx=\dfrac{\pi }{4}}}$



Complete step by step solution:Given integral is
$I=\int\limits_{0}^{1}{\dfrac{dx}{x+\sqrt{1-{{x}^{2}}}}}$
Substituting $x=\sin \theta $
Then, $\theta ={{\sin }^{-1}}x$ and $dx=\cos \theta d\theta $
If $x=0$, then $\theta ={{\sin }^{-1}}x={{\sin }^{-1}}(0)=0$
If $x=1$, then $\theta ={{\sin }^{-1}}x={{\sin }^{-1}}(1)=\dfrac{\pi }{2}$
Then, the given integral become
$\begin{align}
  & I=\int\limits_{0}^{1}{\dfrac{dx}{x+\sqrt{1-{{x}^{2}}}}} \\
 & \text{ }=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos \theta d\theta }{\sin \theta +\sqrt{1-{{\sin }^{2}}\theta }}} \\
 & \text{ }=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos \theta }{\sin \theta +\cos \theta }}d\theta \\
\end{align}$
Since the obtained integral is in the form of \[\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx}\], we can write
$I=\int\limits_{0}^{1}{\dfrac{dx}{x+\sqrt{1-{{x}^{2}}}}}==\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos \theta }{\sin \theta +\cos \theta }}d\theta =\dfrac{\pi }{4}$



Option ‘D’ is correct



Note: Here, we applied the substitution method to evaluate the integral. Here we may forget to change the limits. It is just that the limits should be altered as per the substituting variable. This method of solving an integral is an easy way of evaluating a definite integral. This integral can also be solved by using the property $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$. Since we know that $\sin (\dfrac{\pi }{2}-x)=\cos x$, we get $2I=\int\limits_{0}^{\dfrac{\pi }{2}}{d\theta }=\dfrac{\pi }{2}\Rightarrow I=\dfrac{\pi }{4}$. In this way, we can calculate the given integral. In any case, we need to remember that the interval of the integral must be in the form of $[0, a]$.