Integrals with radicals Calculator

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Example

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Solved example of integrals with radicals

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

Solve the integral $\int\sqrt{4-x^2}dx$ by trigonometric substitution using the substitution

$\begin{matrix}x=2\sin\left(\theta \right) \\ dx=2\cos\left(\theta \right)d\theta\end{matrix}$

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: $1-\sin\left(\theta\right)^2=\cos\left(\theta\right)^2$

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

Cancel exponents $2$ and $0.5$

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

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 $

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

Split the fraction $\frac{1+\cos\left(2\theta \right)}{2}$ inside the integral, in two terms with common denominator $2$

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

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

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

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

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

Take the constant out of the integral

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

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

$4\left(\frac{1}{2}\theta +\frac{1}{4}\sin\left(2\theta \right)\right)$

Apply the formula: $\int\cos\left(x\right)^2dx$$=\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)$, where $x=\theta $

$4\left(\frac{1}{2}\theta +\frac{1}{4}\sin\left(2\theta \right)\right)$

Using the sine double-angle identity

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

Expressing the result of the integral in terms of the original variable

$4\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{\frac{1}{2}x\sqrt{4-x^2}}{4}\right)$

Take $\frac{\frac{1}{2}}{4}$ out of the fraction

$4\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{1}{8}x\sqrt{4-x^2}\right)$

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

$4\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{1}{8}x\sqrt{4-x^2}\right)+C_0$

Answer