Question

How do you use substitution to integrate $\sqrt{4-{{x}^{2}}}dx?$

Answer 1

Hint: We are required to find the integration of the function $\sqrt{4-{{x}^{2}}}dx$ . For this we use the method of trigonometric substitution. We substitute $x=2\sin \theta $ and simplify the integral to a trigonometric term. We then integrate this by using the basic trigonometric formulae and integration formulae. For the solution now obtained, we have the answer in terms of $\theta $ but we need to represent it in terms of x. In order to do this, we use the inverse sine function $\theta =\arcsin \left( \dfrac{x}{2} \right).$ We then will have our answers in terms of x.

Complete step by step solution:
To solve the question $\sqrt{4-{{x}^{2}}}dx$ by integration using substitution, we first represent the integral of the question as,
$= \int{\sqrt{4-{{x}^{2}}}dx}$
Let us assume $x=2\sin \theta ,$ differentiating both sides, $dx=2\cos \theta d\theta .$
We now make the substitution for x in the above integral.
$= \int{\sqrt{4-{{\left( 2\sin \theta \right)}^{2}}}dx}$
Squaring the term inside,
$= \int{\sqrt{4-4{{\sin }^{2}}\theta }dx}$
Taking the 4 common out, we get
$= \int{\sqrt{4\left( 1-{{\sin }^{2}}\theta \right)}dx}$
We know the trigonometric identity that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.$ Rearranging the terms,
$= {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $
Substituting this in the above equation, and substituting for dx,
$= \int{\sqrt{4{{\cos }^{2}}\theta }.2\cos \theta d\theta }$
Taking the terms outside the root,
$= \int{2\cos \theta 2\cos \theta d\theta }$
Multiplying the terms and taking the constant outside,
$= 4.\int{{{\cos }^{2}}\theta d\theta }$
We use another trigonometric identity given as,
$= {{\cos }^{2}}\theta =\dfrac{1+\cos \left( 2\theta \right)}{2}$
Substituting this in the above equation,
$= 4.\int{\dfrac{1+\cos \left( 2\theta \right)}{2}}d\theta $
Dividing 4 by 2 and splitting the integral as two separate integrals,
$= 2\int{d\theta }+2.\int{\cos \left( 2\theta \right)}d\theta $
Integral of $d\theta $ is $\theta .$ The integral of $\cos \left( 2\theta \right)$ is given as $\dfrac{1}{2}\sin \left( 2\theta \right).$ Substituting these,
$= 2\theta +2.\dfrac{1}{2}\sin \left( 2\theta \right)+c$
Cancelling the 2 and $\dfrac{1}{2},$
$= 2\theta +\sin \left( 2\theta \right)+c$
Here, c is the constant of integration and since the solution is in terms of $\theta ,$ we need to convert it in terms of x.
We do this by rearranging this equation $x=2\sin \theta $ as,
$= \theta =\arcsin \dfrac{x}{2}$
We also know that $\sin \left( 2\theta \right)=2\sin \theta \cos \theta $ , and substituting in the solution,
$= 2\theta +2\sin \theta \cos \theta +c$
From the above equation $x=2\sin \theta ,$ we know that $\sin \theta =\dfrac{x}{2}$ and from ${{\cos }^{2}}\theta =\sqrt{4-4{{\sin }^{2}}\theta }=\sqrt{4-{{x}^{2}}},$ we substitute all these to get
 $= 2\arcsin \left( \dfrac{x}{2} \right)+\dfrac{x\sqrt{4-{{x}^{2}}}}{2}+c$
Hence, the solution to the above question is $2\arcsin \left( \dfrac{x}{2} \right)+\dfrac{x\sqrt{4-{{x}^{2}}}}{2}+c.$

Note: It is important to know the basic integration formulae and their properties. Knowing the relation between the different trigonometric functions is important too. We need to note that arcsin is nothing but the inverse sine function. Also, since this is an indefinite integral, we do not have limits. Integrals with limits are called definite integrals.