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Here, we show you a step-by-step solved example of integrals with radicals. This solution was automatically generated by our smart calculator:
$\int\sqrt{4-x^2}dx$
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We can solve the integral $\int\sqrt{4-x^2}dx$ by applying integration method of trigonometric substitution using the substitution
$x=2\sin\left(\theta \right)$
Intermediate steps
Differentiate both sides of the equation $x=2\sin\left(\theta \right)$
$dx=\frac{d}{d\theta}\left(2\sin\left(\theta \right)\right)$
$\frac{d}{d\theta}\left(2\sin\left(\theta \right)\right)$
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 ${f'(x) = \cos(x)\cdot D_x(x)}$
$2\cos\left(\theta \right)$
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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
$dx=2\cos\left(\theta \right)d\theta$
Explain this step further
Intermediate steps
The power of a product is equal to the product of it's factors raised to the same power
$\int2\sqrt{4- 4\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
$\int2\sqrt{4-4\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
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Substituting in the original integral, we get
$\int2\sqrt{4-4\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
Explain this step further
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Factor the polynomial $4-4\sin\left(\theta \right)^2$ by it's greatest common factor (GCF): $4$
$\int2\sqrt{4\left(1-\sin\left(\theta \right)^2\right)}\cos\left(\theta \right)d\theta$
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The power of a product is equal to the product of it's factors raised to the same power
$\int2\cdot 2\sqrt{1-\sin\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
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Applying the trigonometric identity: $1-\sin\left(\theta \right)^2 = \cos\left(\theta \right)^2$
$\int2\cdot 2\sqrt{\cos\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
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The integral of a function times a constant ($2$) is equal to the constant times the integral of the function
$2\int\sqrt{\cos\left(\theta \right)^2}\cos\left(\theta \right)d\theta$
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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 $\frac{1}{2}$
$2\int\cos\left(\theta \right)^{2\cdot \left(\frac{1}{2}\right)}\cos\left(\theta \right)d\theta$
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Multiply the fraction and term in $2\cdot \left(\frac{1}{2}\right)$
$2\int\cos\left(\theta \right)^{\frac{2}{2}}\cos\left(\theta \right)d\theta$
$2\int\cos\left(\theta \right)\cos\left(\theta \right)d\theta$
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When multiplying two powers that have the same base ($\cos\left(\theta \right)$), you can add the exponents
$2\int\cos\left(\theta \right)^2d\theta$
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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 $
$2\left(\frac{1}{2}\theta +\frac{1}{4}\sin\left(2\theta \right)\right)$
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Express the variable $\theta$ in terms of the original variable $x$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{1}{4}\sin\left(2\theta \right)\right)$
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Using the sine double-angle identity: $\sin\left(2\theta\right)=2\sin\left(\theta\right)\cos\left(\theta\right)$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\left(\frac{1}{4}\right)\cdot 2\sin\left(\theta \right)\cos\left(\theta \right)\right)$
Intermediate steps
Multiply the fraction and term in $\left(\frac{1}{4}\right)\cdot 2\sin\left(\theta \right)\cos\left(\theta \right)$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{2\cdot 1}{4}\sin\left(\theta \right)\cos\left(\theta \right)\right)$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{2}{4}\sin\left(\theta \right)\cos\left(\theta \right)\right)$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{1}{2}\sin\left(\theta \right)\cos\left(\theta \right)\right)$
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Multiply the fraction and term in $\left(\frac{1}{4}\right)\cdot 2\sin\left(\theta \right)\cos\left(\theta \right)$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{1}{2}\sin\left(\theta \right)\cos\left(\theta \right)\right)$
Explain this step further
Intermediate steps
Express the variable $\theta$ in terms of the original variable $x$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{1}{2}\frac{x}{2}\frac{\sqrt{4-x^2}}{2}\right)$
Multiplying fractions $\frac{1}{2} \times \frac{x}{2}$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{\sqrt{4-x^2}}{2}\frac{1x}{2\cdot 2}\right)$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{\sqrt{4-x^2}}{2}\frac{1x}{4}\right)$
Any expression multiplied by $1$ is equal to itself
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{\sqrt{4-x^2}}{2}\frac{x}{4}\right)$
Multiplying fractions $\frac{x}{4} \times \frac{\sqrt{4-x^2}}{2}$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{4\cdot 2}\right)$
Multiplying fractions $\frac{1}{2} \times \frac{x}{2}$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{\sqrt{4-x^2}}{2}\frac{1x}{2\cdot 2}\right)$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{\sqrt{4-x^2}}{2}\frac{1x}{4}\right)$
Any expression multiplied by $1$ is equal to itself
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{\sqrt{4-x^2}}{2}\frac{x}{4}\right)$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{8}\right)$
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Express the variable $\theta$ in terms of the original variable $x$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{8}\right)$
Explain this step further
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As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$2\left(\frac{1}{2}\arcsin\left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{8}\right)+C_0$
Intermediate steps
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
$\frac{2}{2}\arcsin\left(\frac{x}{2}\right)+\frac{1}{4}x\sqrt{4-x^2}+C_0$
$\arcsin\left(\frac{x}{2}\right)+\frac{1}{4}x\sqrt{4-x^2}+C_0$
$\arcsin\left(\frac{x}{2}\right)+\frac{1}{4}x\sqrt{4-x^2}+C_0$
Explain this step further
Final answer to the problem
$\arcsin\left(\frac{x}{2}\right)+\frac{1}{4}x\sqrt{4-x^2}+C_0$