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1

Solved example of integrals with radicals

$\int\sqrt{4-x^2}dx$
2

We can solve the integral $\int\sqrt{4-x^2}dx$ by applying integration method of trigonometric substitution using the substitution

$x=2\sin\left(\theta \right)$

Differentiate both sides of the equation $x=2\sin\left(\theta \right)$

$dx=\frac{d}{d\theta}\left(2\sin\left(\theta \right)\right)$

Find the derivative

$\frac{d}{d\theta}\left(2\sin\left(\theta \right)\right)$

The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function

$2\frac{d}{d\theta}\left(\sin\left(\theta \right)\right)$

The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if ${f(x) = \sin(x)}$, then ${f'(x) = \cos(x)\cdot D_x(x)}$

$2\cos\left(\theta \right)$
3

Now, in order to rewrite $d\theta$ in terms of $dx$, we need to find the derivative of $x$. We need to calculate $dx$, we can do that by deriving the equation above

$dx=2\cos\left(\theta \right)d\theta$
4

Substituting in the original integral, we get

$\int2\sqrt{4-4\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
5

Factor by the greatest common divisor $4$

$\int2\sqrt{4\left(1-\sin\left(\theta \right)^2\right)}\cos\left(\theta \right)d\theta$

$\int2\sqrt{4-2^2\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$

Calculate the power $2^2$

$\int2\sqrt{4-1\cdot 4\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$

Multiply $-1$ times $4$

$\int2\sqrt{4-4\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$

$\int2\sqrt{4}\sqrt{1-\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$

Calculate the power $\sqrt{4}$

$\int2\cdot 2\sqrt{1-\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$

Multiply $2$ times $2$

$\int4\sqrt{1-\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
6

The power of a product is equal to the product of it's factors raised to the same power

$\int4\sqrt{1-\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
7

Applying the trigonometric identity: $1-\sin\left(\theta\right)^2=\cos\left(\theta\right)^2$

$\int4\cos\left(\theta \right)^2d\theta$
8

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$4\int\cos\left(\theta \right)^2d\theta$
9

Apply the trigonometric identity: $\cos\left(x\right)^2$$=\frac{1+\cos\left(2x\right)}{2}$, where $x=\theta $

$4\int\frac{1+\cos\left(2\theta \right)}{2}d\theta$

$4\left(\frac{1}{2}\right)\int\left(1+\cos\left(2\theta \right)\right)d\theta$

Divide $1$ by $2$

$4\cdot \frac{1}{2}\int\left(1+\cos\left(2\theta \right)\right)d\theta$

Multiply $4$ times $\frac{1}{2}$

$2\int\left(1+\cos\left(2\theta \right)\right)d\theta$
10

Take the constant $\frac{1}{2}$ out of the integral

$2\int\left(1+\cos\left(2\theta \right)\right)d\theta$
11

Expand the integral $\int\left(1+\cos\left(2\theta \right)\right)d\theta$

$2\left(\int1d\theta+\int\cos\left(2\theta \right)d\theta\right)$

$2\left(1\theta +\int\cos\left(2\theta \right)d\theta\right)$

Any expression multiplied by $1$ is equal to itself

$2\left(\theta +\int\cos\left(2\theta \right)d\theta\right)$
12

The integral of a constant is equal to the constant times the integral's variable

$2\left(\theta +\int\cos\left(2\theta \right)d\theta\right)$

$2\left(\theta +\frac{1}{2}\sin\left(2\theta \right)\right)$

Divide $1$ by $2$

$2\left(\theta +\frac{1}{2}\sin\left(2\theta \right)\right)$

We can solve the integral $\int\cos\left(2\theta \right)d\theta$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2\theta $ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=2\theta $

Now, in order to rewrite $d\theta$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=2d\theta$

Isolate $d\theta$ in the previous equation

$\frac{du}{2}=d\theta$

Substituting $u$ and $d\theta$ in the integral and simplify

$\int\frac{\cos\left(u\right)}{2}du$

Take the constant $\frac{1}{2}$ out of the integral

$\frac{1}{2}\int\cos\left(u\right)du$

Apply the integral of the cosine function: $\int\cos(x)dx=\sin(x)$

$\frac{1}{2}\sin\left(u\right)$

Replace $u$ with the value that we assigned to it in the beginning: $2\theta $

$\frac{1}{2}\sin\left(2\theta \right)$
13

Apply the formula: $\int\cos\left(ax\right)dx$$=\frac{1}{a}\sin\left(ax\right)$, where $a=2$ and $x=\theta $

$2\left(\theta +\frac{1}{2}\sin\left(2\theta \right)\right)$
14

Express the variable $\theta$ in terms of the original variable $x$

$2\left(\arcsin\left(\frac{x}{2}\right)+\frac{1}{2}\sin\left(2\theta \right)\right)$

$2\arcsin\left(\frac{x}{2}\right)+2\cdot \frac{1}{2}\sin\left(2\theta \right)$

Multiply $2$ times $\frac{1}{2}$

$2\arcsin\left(\frac{x}{2}\right)+1\sin\left(2\theta \right)$

Any expression multiplied by $1$ is equal to itself

$2\arcsin\left(\frac{x}{2}\right)+\sin\left(2\theta \right)$
15

Solve the product $2\left(\arcsin\left(\frac{x}{2}\right)+\frac{1}{2}\sin\left(2\theta \right)\right)$

$2\arcsin\left(\frac{x}{2}\right)+\sin\left(2\theta \right)$
16

Using the sine double-angle identity: $\sin\left(2\theta\right)=2\sin\left(\theta\right)\cos\left(\theta\right)$

$2\arcsin\left(\frac{x}{2}\right)+2\sin\left(\theta \right)\cos\left(\theta \right)$
17

Express the variable $\theta$ in terms of the original variable $x$

$2\arcsin\left(\frac{x}{2}\right)+2\left(\frac{x}{2}\right)\left(\frac{\sqrt{4-x^2}}{2}\right)$
18

Multiplying the fraction by $2$

$2\arcsin\left(\frac{x}{2}\right)+\frac{2x}{2}\frac{\sqrt{4-x^2}}{2}$
19

Take $\frac{2}{2}$ out of the fraction

$2\arcsin\left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{2}$
20

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$2\arcsin\left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{2}+C_0$

Final Answer

$2\arcsin\left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{2}+C_0$

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