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Question: What is the integral of, \(\left( \dfrac{1}{\sqrt{4-{{x}^{2}}}} \right)\) ?...

What is the integral of, (14x2)\left( \dfrac{1}{\sqrt{4-{{x}^{2}}}} \right) ?

Explanation

Solution

We can see in our problem that, if we use a suitable substitute for our variable ‘x’ in the denominator, the problem could be easily solved. Thus, we will use the substitution method of solving an integral to get the answer to our question. This will make our problem easier and less calculative.

Complete step-by-step answer:
Let us denote the expression given to us with ‘y’, such that we need to find the integral of ‘y’ over very small change in ‘x’. Mathematically, this could be written as:
I=y.dx or,I=14x2.dx \begin{aligned} & \Rightarrow I=\int{y.dx} \\\ & or,I=\int{\dfrac{1}{\sqrt{4-{{x}^{2}}}}.dx} \\\ \end{aligned}
Where, “I” is the result of this integral.

Let us now substitute the variable ‘x’ with the term ‘2sinθ2\sin \theta ’. Thus, the differential of ‘x’ shall also change. This can be calculated as follows:
dx=d(2sinθ) dx=2cosθdθ \begin{aligned} & \Rightarrow dx=d\left( 2\sin \theta \right) \\\ & \therefore dx=2\cos \theta d\theta \\\ \end{aligned}
Thus, putting the new substituted values of ‘x’ and ‘θ\theta ’, our new equation becomes:
I=14(2sinθ)2.2cosθdθ I=2cosθ44sin2θdθ I=2cosθ21sin2θdθ \begin{aligned} & \Rightarrow I=\int{\dfrac{1}{\sqrt{4-{{\left( 2\sin \theta \right)}^{2}}}}.2\cos \theta d}\theta \\\ & \Rightarrow I=\int{\dfrac{2\cos \theta }{\sqrt{4-4{{\sin }^{2}}\theta }}}d\theta \\\ & \Rightarrow I=\int{\dfrac{2\cos \theta }{2\sqrt{1-{{\sin }^{2}}\theta }}}d\theta \\\ \end{aligned}

Using the trigonometric identity:
1sin2θ=cosθ\Rightarrow \sqrt{1-{{\sin }^{2}}\theta }=\cos \theta
We can simplify our equation to:
I=2cosθ2cosθdθ I=dθ I=θ+c \begin{aligned} & \Rightarrow I=\int{\dfrac{2\cos \theta }{2\cos \theta }}d\theta \\\ & \Rightarrow I=\int{d\theta } \\\ & \therefore I=\theta +c \\\ \end{aligned}
Where, ‘c’ is the constant of integration.

Now, we need to convert it into the variable which was originally presented to us, that is, ‘x’. Thus, on re-substituting, we get the equation as:
x=2cosθ x2=cosθ θ=cos1(x2) \begin{aligned} & \Rightarrow x=2\cos \theta \\\ & \Rightarrow \dfrac{x}{2}=\cos \theta \\\ & \therefore \theta ={{\cos }^{-1}}\left( \dfrac{x}{2} \right) \\\ \end{aligned}
Thus, using this in our above result, we get the final integral as: [cos1(x2)+c]\left[ {{\cos }^{-1}}\left( \dfrac{x}{2} \right)+c \right].
Hence, the integral of, (14x2)\left( \dfrac{1}{\sqrt{4-{{x}^{2}}}} \right) comes out to be [cos1(x2)+c]\left[ {{\cos }^{-1}}\left( \dfrac{x}{2} \right)+c \right].

Note: Whenever using the substitution method to find out the integral of an expression, we should always be careful as to what to substitute in the place of our original variable. The substitute should be such that it reduces the complexity in our equation. This makes our new equation easy to understand and solve. The new equation sometimes might even become a standard integral just like in this case.