Integrals with radicals Calculator

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Example

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Solved example of integrals with radicals

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

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

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

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 ${f'(x) = \cos(x)\cdot D_x(x)}$

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

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

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

Substituting in the original integral, we get

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

Factor by the greatest common divisor $4$

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

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

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

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

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

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

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

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

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

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

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

The integral of the sum of two or more functions is equal to the sum of their integrals

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

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

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

Expand the integral

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

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

$2\theta $

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

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

Any expression multiplied by $1$ is equal to itself

$2\arcsin\left(\frac{x}{2}\right)$

The integral $2\int1d\theta$ results in: $2\arcsin\left(\frac{x}{2}\right)$

$2\arcsin\left(\frac{x}{2}\right)$

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

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

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

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

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

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

Multiplying the fraction by $2$

$\frac{2x}{2}\frac{\sqrt{4-x^2}}{2}$

Simplify the fraction by $2$

$\frac{x\sqrt{4-x^2}}{2}$

The integral $2\int\cos\left(2\theta \right)d\theta$ results in: $\frac{x\sqrt{4-x^2}}{2}$

$\frac{x\sqrt{4-x^2}}{2}$

Gather the results of all integrals

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

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