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Calculatrice Intégrales avec radicaux

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1

Here, we show you a step-by-step solved example of integrals with radicals. This solution was automatically generated by our smart calculator:

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

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

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

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

Multiply $-1$ times $4$

$\int2\sqrt{4-4\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
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 the polynomial $4-4\sin\left(\theta \right)^2$ by it's greatest common factor (GCF): $4$

$\int2\sqrt{4\left(1-\sin\left(\theta \right)^2\right)}\cos\left(\theta \right)d\theta$
6

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

$\int2\cdot 2\sqrt{1-\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
7

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

$\int2\cdot 2\sqrt{\cos\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
8

The integral of a function times a constant ($2$) is equal to the constant times the integral of the function

$2\int\sqrt{\cos\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
9

Simplify $\sqrt{\cos\left(\theta \right)^2}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $\frac{1}{2}$

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

When multiplying two powers that have the same base ($\cos\left(\theta \right)$), you can add the exponents

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

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

$2\left(\frac{1}{2}\theta +\frac{1}{4}\sin\left(2\theta \right)\right)$
12

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

$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{1}{4}\sin\left(2\theta \right)\right)$
13

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

$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+2\left(\frac{1}{4}\right)\sin\left(\theta \right)\cos\left(\theta \right)\right)$
14

Multiply the fraction and term in $2\left(\frac{1}{4}\right)\sin\left(\theta \right)\cos\left(\theta \right)$

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

Multiplying fractions $\frac{1}{2} \times \frac{x}{2}$

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

Multiplying fractions $\frac{x}{4} \times \frac{\sqrt{4-x^2}}{2}$

$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{8}\right)$
15

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

$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{8}\right)$
16

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

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

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

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

Multiplying the fraction by $2$

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

Take $\frac{2}{8}$ out of the fraction

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

Multiply the fraction and term in $2\cdot \left(\frac{1}{2}\right)\arcsin\left(\frac{x}{2}\right)$

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

Any expression multiplied by $1$ is equal to itself

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

Divide $2$ by $2$

$\arcsin\left(\frac{x}{2}\right)+\frac{1}{4}x\sqrt{4-x^2}+C_0$
17

Expand and simplify

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

Réponse finale au problème

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

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