Solveeit Logo
Login

Question

Question: How do you use substitution to integrate \(\sqrt{4-{{x}^{2}}}dx?\)...

How do you use substitution to integrate 4−x2dx?\sqrt{4-{{x}^{2}}}dx?

Explanation

Solution

We are required to find the integration of the function 4−x2dx\sqrt{4-{{x}^{2}}}dx . For this we use the method of trigonometric substitution. We substitute x=2sin⁡θx=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 θ\theta but we need to represent it in terms of x. In order to do this, we use the inverse sine function θ=arcsin⁡(x2).\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 4−x2dx\sqrt{4-{{x}^{2}}}dx by integration using substitution, we first represent the integral of the question as,
=∫4−x2dx= \int{\sqrt{4-{{x}^{2}}}dx}
Let us assume x=2sin⁡θ,x=2\sin \theta , differentiating both sides, dx=2cos⁡θdθ.dx=2\cos \theta d\theta .
We now make the substitution for x in the above integral.
=∫4−(2sin⁡θ)2dx= \int{\sqrt{4-{{\left( 2\sin \theta \right)}^{2}}}dx}
Squaring the term inside,
=∫4−4sin⁡2θdx= \int{\sqrt{4-4{{\sin }^{2}}\theta }dx}
Taking the 4 common out, we get
=∫4(1−sin⁡2θ)dx= \int{\sqrt{4\left( 1-{{\sin }^{2}}\theta \right)}dx}
We know the trigonometric identity that sin⁡2θ+cos⁡2θ=1.{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1. Rearranging the terms,
=cos⁡2θ=1−sin⁡2θ= {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta
Substituting this in the above equation, and substituting for dx,
=∫4cos⁡2θ.2cos⁡θdθ= \int{\sqrt{4{{\cos }^{2}}\theta }.2\cos \theta d\theta }
Taking the terms outside the root,
=∫2cos⁡θ2cos⁡θdθ= \int{2\cos \theta 2\cos \theta d\theta }
Multiplying the terms and taking the constant outside,
=4.∫cos⁡2θdθ= 4.\int{{{\cos }^{2}}\theta d\theta }
We use another trigonometric identity given as,
=cos⁡2θ=1+cos⁡(2θ)2= {{\cos }^{2}}\theta =\dfrac{1+\cos \left( 2\theta \right)}{2}
Substituting this in the above equation,
=4.∫1+cos⁡(2θ)2dθ= 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∫dθ+2.∫cos⁡(2θ)dθ= 2\int{d\theta }+2.\int{\cos \left( 2\theta \right)}d\theta
Integral of dθd\theta is θ.\theta . The integral of cos⁡(2θ)\cos \left( 2\theta \right) is given as 12sin⁡(2θ).\dfrac{1}{2}\sin \left( 2\theta \right). Substituting these,
=2θ+2.12sin⁡(2θ)+c= 2\theta +2.\dfrac{1}{2}\sin \left( 2\theta \right)+c
Cancelling the 2 and 12,\dfrac{1}{2},
=2θ+sin⁡(2θ)+c= 2\theta +\sin \left( 2\theta \right)+c
Here, c is the constant of integration and since the solution is in terms of θ,\theta , we need to convert it in terms of x.
We do this by rearranging this equation x=2sin⁡θx=2\sin \theta as,
=θ=arcsin⁡x2= \theta =\arcsin \dfrac{x}{2}
We also know that sin⁡(2θ)=2sin⁡θcos⁡θ\sin \left( 2\theta \right)=2\sin \theta \cos \theta , and substituting in the solution,
=2θ+2sin⁡θcos⁡θ+c= 2\theta +2\sin \theta \cos \theta +c
From the above equation x=2sin⁡θ,x=2\sin \theta , we know that sin⁡θ=x2\sin \theta =\dfrac{x}{2} and from cos⁡2θ=4−4sin⁡2θ=4−x2,{{\cos }^{2}}\theta =\sqrt{4-4{{\sin }^{2}}\theta }=\sqrt{4-{{x}^{2}}}, we substitute all these to get
=2arcsin⁡(x2)+x4−x22+c= 2\arcsin \left( \dfrac{x}{2} \right)+\dfrac{x\sqrt{4-{{x}^{2}}}}{2}+c
Hence, the solution to the above question is 2arcsin⁡(x2)+x4−x22+c.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.